\(\int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx\) [269]

Optimal result

Integrand size = 26, antiderivative size = 502 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \arctan \left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac {3 f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}-\frac {3 i f^3 \operatorname {PolyLog}\left (4,-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,i e^{i (c+d x)}\right )}{a d^4}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:

3*I*f^3*polylog(2,-I*exp(I*(d*x+c)))/a/d^4-3/2*I*f^3*polylog(2,-exp(2*I*(d 
*x+c)))/a/d^4-3/2*I*f*(f*x+e)^2/a/d^2+3*f^2*(f*x+e)*ln(1+exp(2*I*(d*x+c))) 
/a/d^3-I*(f*x+e)^3*arctan(exp(I*(d*x+c)))/a/d-3/2*I*f*(f*x+e)^2*polylog(2, 
I*exp(I*(d*x+c)))/a/d^2+3*I*f^3*polylog(4,I*exp(I*(d*x+c)))/a/d^4-3*I*f^3* 
polylog(4,-I*exp(I*(d*x+c)))/a/d^4-6*I*f^2*(f*x+e)*arctan(exp(I*(d*x+c)))/ 
a/d^3-3*f^2*(f*x+e)*polylog(3,-I*exp(I*(d*x+c)))/a/d^3+3*f^2*(f*x+e)*polyl 
og(3,I*exp(I*(d*x+c)))/a/d^3-3*I*f^3*polylog(2,I*exp(I*(d*x+c)))/a/d^4+3/2 
*I*f*(f*x+e)^2*polylog(2,-I*exp(I*(d*x+c)))/a/d^2-3/2*f*(f*x+e)^2*sec(d*x+ 
c)/a/d^2-1/2*(f*x+e)^3*sec(d*x+c)^2/a/d+3/2*f*(f*x+e)^2*tan(d*x+c)/a/d^2+1 
/2*(f*x+e)^3*sec(d*x+c)*tan(d*x+c)/a/d
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1025\) vs. \(2(502)=1004\).

Time = 10.31 (sec) , antiderivative size = 1025, normalized size of antiderivative = 2.04 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((e + f*x)^3*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]
 

Output:

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))/(8*a*(Cos[c/2] - Sin[c/2]) 
*(Cos[c/2] + Sin[c/2])) - ((Cos[c] + I*Sin[c])*(((e + f*x)^3*Log[1 - I*Cos 
[c + d*x] - Sin[c + d*x]]*(1 - I*Cos[c] - Sin[c]))/d + ((e + f*x)^4*(Cos[c 
] - I*Sin[c]))/(4*f) + (3*f*(d^2*(e + f*x)^2*PolyLog[2, I*Cos[c + d*x] + S 
in[c + d*x]] - (2*I)*d*f*(e + f*x)*PolyLog[3, I*Cos[c + d*x] + Sin[c + d*x 
]] - 2*f^2*PolyLog[4, I*Cos[c + d*x] + Sin[c + d*x]])*(Cos[c] + I*(-1 + Si 
n[c]))*(I*Cos[c] + Sin[c]))/d^4))/(2*a*(Cos[c] + I*(-1 + Sin[c]))) - ((Cos 
[c] + I*Sin[c])*(((12*f^2 + d^2*(e + f*x)^2)^2*(Cos[c] - I*Sin[c]))/(4*d^2 
*f) + (3*f*(d^2*e^2 + 4*f^2)*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]* 
(Cos[c] - I*Sin[c])*(1 - I*Cos[c] + Sin[c]))/d^2 + 6*e*f^2*x*PolyLog[2, (- 
I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(1 - I*Cos[c] + Sin[c] 
) + 3*f^3*x^2*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 
+ Sin[c])) - (6*f^3*PolyLog[4, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - 
 I*(1 + Sin[c])))/d^2 - (3*f*(d^2*e^2 + 4*f^2)*x*Log[1 + I*Cos[c + d*x] + 
Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d - 3*d*e*f^2 
*x^2*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + 
I*(1 + Sin[c])) - d*f^3*x^3*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] 
 - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - (e*(d^2*e^2 + 12*f^2)*Log[Cos[c + 
 d*x] + I*(1 + Sin[c + d*x])]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c]) 
))/d - (6*e*f^2*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - ...
 

Rubi [A] (verified)

Time = 2.93 (sec) , antiderivative size = 496, normalized size of antiderivative = 0.99, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {5042, 3042, 4674, 3042, 4669, 2715, 2838, 3011, 4909, 3042, 4672, 25, 3042, 4202, 2620, 2715, 2838, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(e+f x)^3 \sec (c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 5042

\(\displaystyle \frac {\int (e+f x)^3 \sec ^3(c+d x)dx}{a}-\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x)^3 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a}-\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\frac {3 f^2 \int (e+f x) \sec (c+d x)dx}{d^2}+\frac {1}{2} \int (e+f x)^3 \sec (c+d x)dx-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 f^2 \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )dx}{d^2}+\frac {1}{2} \int (e+f x)^3 \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4669

\(\displaystyle -\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3 f^2 \left (-\frac {f \int \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {f \int \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}\right )}{d^2}+\frac {1}{2} \left (-\frac {3 f \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {3 f \int (e+f x)^2 \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3 f^2 \left (\frac {i f \int e^{-i (c+d x)} \log \left (1-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {i f \int e^{-i (c+d x)} \log \left (1+i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}\right )}{d^2}+\frac {1}{2} \left (-\frac {3 f \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {3 f \int (e+f x)^2 \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {1}{2} \left (-\frac {3 f \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {3 f \int (e+f x)^2 \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 4909

\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \sec ^2(c+d x)dx}{2 d}}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{2 d}}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 4672

\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {2 f \int -((e+f x) \tan (c+d x))dx}{d}+\frac {(e+f x)^2 \tan (c+d x)}{d}\right )}{2 d}}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \int (e+f x) \tan (c+d x)dx}{d}\right )}{2 d}}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \int (e+f x) \tan (c+d x)dx}{d}\right )}{2 d}}{a}+\frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \int \frac {e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}}dx\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (\frac {i f \int \log \left (1+e^{2 i (c+d x)}\right )dx}{2 d}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (\frac {f \int e^{-2 i (c+d x)} \log \left (1+e^{2 i (c+d x)}\right )de^{2 i (c+d x)}}{4 d^2}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (-\frac {f \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (-\frac {f \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (-\frac {f \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\frac {3 f^2 \left (-\frac {2 i (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {i f \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^2}-\frac {i f \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^2}\right )}{d^2}+\frac {1}{2} \left (-\frac {2 i (e+f x)^3 \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,-i e^{i (c+d x)}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,i e^{i (c+d x)}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d}\right )}{d}\right )}{d}\right )-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 d^2}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 d}}{a}-\frac {\frac {(e+f x)^3 \sec ^2(c+d x)}{2 d}-\frac {3 f \left (\frac {(e+f x)^2 \tan (c+d x)}{d}-\frac {2 f \left (\frac {i (e+f x)^2}{2 f}-2 i \left (-\frac {f \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{4 d^2}-\frac {i (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{2 d}\right )\right )}{d}\right )}{2 d}}{a}\)

Input:

Int[((e + f*x)^3*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]
 

Output:

((3*f^2*(((-2*I)*(e + f*x)*ArcTan[E^(I*(c + d*x))])/d + (I*f*PolyLog[2, (- 
I)*E^(I*(c + d*x))])/d^2 - (I*f*PolyLog[2, I*E^(I*(c + d*x))])/d^2))/d^2 + 
 (((-2*I)*(e + f*x)^3*ArcTan[E^(I*(c + d*x))])/d + (3*f*((I*(e + f*x)^2*Po 
lyLog[2, (-I)*E^(I*(c + d*x))])/d - ((2*I)*f*(((-I)*(e + f*x)*PolyLog[3, ( 
-I)*E^(I*(c + d*x))])/d + (f*PolyLog[4, (-I)*E^(I*(c + d*x))])/d^2))/d))/d 
 - (3*f*((I*(e + f*x)^2*PolyLog[2, I*E^(I*(c + d*x))])/d - ((2*I)*f*(((-I) 
*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/d + (f*PolyLog[4, I*E^(I*(c + d* 
x))])/d^2))/d))/d)/2 - (3*f*(e + f*x)^2*Sec[c + d*x])/(2*d^2) + ((e + f*x) 
^3*Sec[c + d*x]*Tan[c + d*x])/(2*d))/a - (((e + f*x)^3*Sec[c + d*x]^2)/(2* 
d) - (3*f*((-2*f*(((I/2)*(e + f*x)^2)/f - (2*I)*(((-1/2*I)*(e + f*x)*Log[1 
 + E^((2*I)*(c + d*x))])/d - (f*PolyLog[2, -E^((2*I)*(c + d*x))])/(4*d^2)) 
))/d + ((e + f*x)^2*Tan[c + d*x])/d))/(2*d))/a
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

rule 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

rule 5042

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. 
)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(e + f*x)^m*Sec[c + 
 d*x]^(n + 2), x], x] - Simp[1/b   Int[(e + f*x)^m*Sec[c + d*x]^(n + 1)*Tan 
[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && EqQ[a 
^2 - b^2, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1195 vs. \(2 (444 ) = 888\).

Time = 2.51 (sec) , antiderivative size = 1196, normalized size of antiderivative = 2.38

method	result	size
risch	\(\text {Expression too large to display}\)	\(1196\)

Input:

int((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

3*I*f^3*polylog(4,I*exp(I*(d*x+c)))/a/d^4-3*I*f^3*polylog(4,-I*exp(I*(d*x+ 
c)))/a/d^4-3*I/a/d^2*e*f^2*polylog(2,I*exp(I*(d*x+c)))*x+3*I/a/d^2*e*f^2*p 
olylog(2,-I*exp(I*(d*x+c)))*x+3*I/a/d^2*e^2*f*c*arctan(exp(I*(d*x+c)))-3*I 
/a/d^3*e*f^2*c^2*arctan(exp(I*(d*x+c)))-I*(d*exp(I*(d*x+c))*f^3*x^3+3*d*ex 
p(I*(d*x+c))*e*f^2*x^2+3*d*exp(I*(d*x+c))*e^2*f*x+d*exp(I*(d*x+c))*e^3+3*f 
^3*x^2-3*I*f^3*x^2*exp(I*(d*x+c))+6*e*f^2*x-6*I*e*f^2*x*exp(I*(d*x+c))+3*e 
^2*f-3*I*e^2*f*exp(I*(d*x+c)))/d^2/(exp(I*(d*x+c))+I)^2/a+3/2*I/a/d^2*f^3* 
polylog(2,-I*exp(I*(d*x+c)))*x^2-3/2*I/a/d^2*e^2*f*polylog(2,I*exp(I*(d*x+ 
c)))+3/2*I/a/d^2*e^2*f*polylog(2,-I*exp(I*(d*x+c)))-3/2*I/a/d^2*f^3*polylo 
g(2,I*exp(I*(d*x+c)))*x^2+6*I/a/d^4*f^3*c*arctan(exp(I*(d*x+c)))-6*I/a/d^3 
*e*f^2*arctan(exp(I*(d*x+c)))+I/a/d^4*f^3*c^3*arctan(exp(I*(d*x+c)))+3/2/a 
/d^3*e*f^2*ln(1+I*exp(I*(d*x+c)))*c^2+3/2/a/d*e^2*f*ln(1-I*exp(I*(d*x+c))) 
*x+3/2/a/d^2*e^2*f*ln(1-I*exp(I*(d*x+c)))*c-3/2/a/d*e^2*f*ln(1+I*exp(I*(d* 
x+c)))*x-3/2/a/d^2*e^2*f*ln(1+I*exp(I*(d*x+c)))*c+3/2/a/d*e*f^2*ln(1-I*exp 
(I*(d*x+c)))*x^2-3/2/a/d*e*f^2*ln(1+I*exp(I*(d*x+c)))*x^2-3/2/a/d^3*e*f^2* 
ln(1-I*exp(I*(d*x+c)))*c^2-6*I/a/d^3*f^3*c*x-6/a/d^3*e*f^2*ln(exp(I*(d*x+c 
)))+3/a/d^3*e*f^2*ln(exp(2*I*(d*x+c))+1)+3/a/d^3*e*f^2*polylog(3,I*exp(I*( 
d*x+c)))-3/a/d^3*e*f^2*polylog(3,-I*exp(I*(d*x+c)))+1/2/a/d^4*c^3*f^3*ln(1 
-I*exp(I*(d*x+c)))-1/2/a/d^4*c^3*f^3*ln(1+I*exp(I*(d*x+c)))+6/a/d^3*f^3*ln 
(1-I*exp(I*(d*x+c)))*x+6/a/d^4*f^3*ln(1-I*exp(I*(d*x+c)))*c+6/a/d^4*f^3...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1884 vs. \(2 (421) = 842\).

Time = 0.25 (sec) , antiderivative size = 1884, normalized size of antiderivative = 3.75 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
 

Output:

-1/4*(2*d^3*f^3*x^3 + 6*d^3*e*f^2*x^2 + 6*d^3*e^2*f*x + 2*d^3*e^3 + 6*(d^2 
*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*cos(d*x + c) + 3*(I*d^2*f^3*x^2 + 2* 
I*d^2*e*f^2*x + I*d^2*e^2*f + (I*d^2*f^3*x^2 + 2*I*d^2*e*f^2*x + I*d^2*e^2 
*f)*sin(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) + 3*(I*d^2*f^3*x^2 
+ 2*I*d^2*e*f^2*x + I*d^2*e^2*f + 4*I*f^3 + (I*d^2*f^3*x^2 + 2*I*d^2*e*f^2 
*x + I*d^2*e^2*f + 4*I*f^3)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + 
 c)) + 3*(-I*d^2*f^3*x^2 - 2*I*d^2*e*f^2*x - I*d^2*e^2*f + (-I*d^2*f^3*x^2 
 - 2*I*d^2*e*f^2*x - I*d^2*e^2*f)*sin(d*x + c))*dilog(-I*cos(d*x + c) + si 
n(d*x + c)) + 3*(-I*d^2*f^3*x^2 - 2*I*d^2*e*f^2*x - I*d^2*e^2*f - 4*I*f^3 
+ (-I*d^2*f^3*x^2 - 2*I*d^2*e*f^2*x - I*d^2*e^2*f - 4*I*f^3)*sin(d*x + c)) 
*dilog(-I*cos(d*x + c) - sin(d*x + c)) - (d^3*e^3 - 3*c*d^2*e^2*f + 3*(c^2 
 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3 + (d^3*e^3 - 3*c*d^2*e^2*f + 3*(c^2 + 4)* 
d*e*f^2 - (c^3 + 12*c)*f^3)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c 
) + I) + (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3 + (d^3*e^3 - 3 
*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*sin(d*x + c))*log(cos(d*x + c) - I 
*sin(d*x + c) + I) - (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*c*d^2*e^2*f - 3*c^ 
2*d*e*f^2 + (c^3 + 12*c)*f^3 + 3*(d^3*e^2*f + 4*d*f^3)*x + (d^3*f^3*x^3 + 
3*d^3*e*f^2*x^2 + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + (c^3 + 12*c)*f^3 + 3*(d^ 
3*e^2*f + 4*d*f^3)*x)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) 
 + (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*d^3*e^2*f*x + 3*c*d^2*e^2*f - 3*c...
 

Sympy [F]

\[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{3} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate((f*x+e)**3*sec(d*x+c)/(a+a*sin(d*x+c)),x)
 

Output:

(Integral(e**3*sec(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*se 
c(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e*f**2*x**2*sec(c + d*x)/(s 
in(c + d*x) + 1), x) + Integral(3*e**2*f*x*sec(c + d*x)/(sin(c + d*x) + 1) 
, x))/a
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3854 vs. \(2 (421) = 842\).

Time = 0.86 (sec) , antiderivative size = 3854, normalized size of antiderivative = 7.68 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
 

Output:

1/4*(3*c*e^2*f*(2/(a*d*sin(d*x + c) + a*d) - log(sin(d*x + c) + 1)/(a*d) + 
 log(sin(d*x + c) - 1)/(a*d)) + e^3*(log(sin(d*x + c) + 1)/a - log(sin(d*x 
 + c) - 1)/a - 2/(a*sin(d*x + c) + a)) - 4*(12*d^2*e^2*f - 24*c*d*e*f^2 + 
12*c^2*f^3 + 2*(3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3 - (3*(c^2 + 4)*d*e* 
f^2 - (c^3 + 12*c)*f^3)*cos(2*d*x + 2*c) + 2*(3*(-I*c^2 - 4*I)*d*e*f^2 + ( 
I*c^3 + 12*I*c)*f^3)*cos(d*x + c) + (3*(-I*c^2 - 4*I)*d*e*f^2 + (I*c^3 + 1 
2*I*c)*f^3)*sin(2*d*x + 2*c) + 2*(3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3)* 
sin(d*x + c))*arctan2(sin(d*x + c) + 1, cos(d*x + c)) - 2*(3*c^2*d*e*f^2 - 
 c^3*f^3 - (3*c^2*d*e*f^2 - c^3*f^3)*cos(2*d*x + 2*c) - 2*(3*I*c^2*d*e*f^2 
 - I*c^3*f^3)*cos(d*x + c) - (3*I*c^2*d*e*f^2 - I*c^3*f^3)*sin(2*d*x + 2*c 
) + 2*(3*c^2*d*e*f^2 - c^3*f^3)*sin(d*x + c))*arctan2(sin(d*x + c) - 1, co 
s(d*x + c)) - 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^ 
2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c) - ((d*x + c)^3*f^3 + 3*(d 
*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)* 
(d*x + c))*cos(2*d*x + 2*c) - 2*(I*(d*x + c)^3*f^3 + 3*(I*d*e*f^2 - I*c*f^ 
3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + (I*c^2 + 4*I)*f^3)*(d*x 
+ c))*cos(d*x + c) - (I*(d*x + c)^3*f^3 + 3*(I*d*e*f^2 - I*c*f^3)*(d*x + c 
)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + (I*c^2 + 4*I)*f^3)*(d*x + c))*sin(2 
*d*x + 2*c) + 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^ 
2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c))*sin(d*x + c))*arctan2...
 

Giac [F]

\[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sec \left (d x + c\right )}{a \sin \left (d x + c\right ) + a} \,d x } \] Input:

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)^3*sec(d*x + c)/(a*sin(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \] Input:

int((e + f*x)^3/(cos(c + d*x)*(a + a*sin(c + d*x))),x)
                                                                                    
                                                                                    
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {too large to display} \] Input:

int((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x)
 

Output:

( - 12*cos(c + d*x)*d**3*e**2*f*x - 12*cos(c + d*x)*d**3*e*f**2*x**2 - 4*c 
os(c + d*x)*d**3*f**3*x**3 - 12*cos(c + d*x)*d**2*e**2*f + 48*cos(c + d*x) 
*d**2*e*f**2*x + 24*cos(c + d*x)*d**2*f**3*x**2 + 24*cos(c + d*x)*d*e*f**2 
 - 120*cos(c + d*x)*d*f**3*x - 48*cos(c + d*x)*f**3 - 48*int(x**2/(tan((c 
+ d*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)*sin(c + 
d*x)*d**3*f**3 - 48*int(x**2/(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**3 
- 2*tan((c + d*x)/2) - 1),x)*d**3*f**3 - 16*int((tan((c + d*x)/2)*x**3)/(t 
an((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)*si 
n(c + d*x)*d**4*f**3 - 16*int((tan((c + d*x)/2)*x**3)/(tan((c + d*x)/2)**4 
 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)*d**4*f**3 - 48*int(( 
tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*ta 
n((c + d*x)/2) - 1),x)*sin(c + d*x)*d**4*e*f**2 + 48*int((tan((c + d*x)/2) 
*x**2)/(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 
 1),x)*sin(c + d*x)*d**3*f**3 - 48*int((tan((c + d*x)/2)*x**2)/(tan((c + d 
*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)*d**4*e*f**2 
 + 48*int((tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2 
)**3 - 2*tan((c + d*x)/2) - 1),x)*d**3*f**3 - 48*int((tan((c + d*x)/2)*x)/ 
(tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)* 
sin(c + d*x)*d**4*e**2*f + 96*int((tan((c + d*x)/2)*x)/(tan((c + d*x)/2)** 
4 + 2*tan((c + d*x)/2)**3 - 2*tan((c + d*x)/2) - 1),x)*sin(c + d*x)*d**...