\(\int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [275]

Optimal result

Integrand size = 28, antiderivative size = 475 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 i (e+f x)^3}{3 a d}-\frac {i f (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d^2}+\frac {f^3 \text {arctanh}(\sin (c+d x))}{a d^4}+\frac {2 f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{a d^2}+\frac {f^3 \log (\cos (c+d x))}{a d^4}+\frac {i f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^3}-\frac {i f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {2 i f^2 (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{a d^3}-\frac {f^3 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^4}+\frac {f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}+\frac {f^3 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{a d^4}-\frac {f^2 (e+f x) \sec (c+d x)}{a d^3}-\frac {f (e+f x)^2 \sec ^2(c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^3(c+d x)}{3 a d}+\frac {f^2 (e+f x) \tan (c+d x)}{a d^3}+\frac {2 (e+f x)^3 \tan (c+d x)}{3 a d}+\frac {f (e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec ^2(c+d x) \tan (c+d x)}{3 a d} \] Output:

-I*f*(f*x+e)^2*arctan(exp(I*(d*x+c)))/a/d^2+I*f^2*(f*x+e)*polylog(2,-I*exp 
(I*(d*x+c)))/a/d^3+f^3*arctanh(sin(d*x+c))/a/d^4+2*f*(f*x+e)^2*ln(1+exp(2* 
I*(d*x+c)))/a/d^2+f^3*ln(cos(d*x+c))/a/d^4-2*I*f^2*(f*x+e)*polylog(2,-exp( 
2*I*(d*x+c)))/a/d^3-I*f^2*(f*x+e)*polylog(2,I*exp(I*(d*x+c)))/a/d^3-2/3*I* 
(f*x+e)^3/a/d-f^3*polylog(3,-I*exp(I*(d*x+c)))/a/d^4+f^3*polylog(3,I*exp(I 
*(d*x+c)))/a/d^4+f^3*polylog(3,-exp(2*I*(d*x+c)))/a/d^4-f^2*(f*x+e)*sec(d* 
x+c)/a/d^3-1/2*f*(f*x+e)^2*sec(d*x+c)^2/a/d^2-1/3*(f*x+e)^3*sec(d*x+c)^3/a 
/d+f^2*(f*x+e)*tan(d*x+c)/a/d^3+2/3*(f*x+e)^3*tan(d*x+c)/a/d+1/2*f*(f*x+e) 
^2*sec(d*x+c)*tan(d*x+c)/a/d^2+1/3*(f*x+e)^3*sec(d*x+c)^2*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. \(1173\) vs. \(2(475)=950\).

Time = 10.16 (sec) , antiderivative size = 1173, normalized size of antiderivative = 2.47 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((e + f*x)^3*Sec[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

((d^3*(e + f*x)^3)/(-I + E^(I*c)) + 3*d^2*f*(e + f*x)^2*Log[1 - I/E^(I*(c 
+ d*x))] + 6*f^2*(I*d*(e + f*x)*PolyLog[2, I/E^(I*(c + d*x))] + f*PolyLog[ 
3, I/E^(I*(c + d*x))]))/(2*a*d^4) - (f*(Cos[c] + I*Sin[c])*(5*d^2*e^2*x*Co 
s[c] + 4*f^2*x*Cos[c] + 5*d^2*e*f*x^2*Cos[c] + (5*d^2*f^2*x^3*(Cos[c] - I* 
Sin[c]))/3 - (5*I)*d^2*e^2*x*Sin[c] - (4*I)*f^2*x*Sin[c] - (5*I)*d^2*e*f*x 
^2*Sin[c] + 10*e*f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - 
I*(1 + Sin[c])) + 10*f^2*x*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(C 
os[c] - I*(1 + Sin[c])) - 10*d*e*f*x*Log[1 + I*Cos[c + d*x] + Sin[c + d*x] 
]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - 5*d*f^2*x^2*Log[1 + I*Co 
s[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - 
 ((5*d^2*e^2 + 4*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 - (10*f^2*PolyLog[3, (-I)*Cos[c + d* 
x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d + (5*d 
^2*e^2 + 4*f^2)*x*(I*Cos[c] + Sin[c])*(Cos[c] + I*(1 + Sin[c]))))/(2*a*d^3 
*(Cos[c] + I*(1 + Sin[c]))) + (e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d*x)/2] + 
 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/(2*a*d*(Cos[c/2] - Sin[c 
/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (e^3*Sin[(d*x)/2] + 3*e^ 
2*f*x*Sin[(d*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2])/(3*a 
*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (- 
(d*e^3*Cos[c/2]) - 3*e^2*f*Cos[c/2] - 3*d*e^2*f*x*Cos[c/2] - 6*e*f^2*x*...
 

Rubi [A] (verified)

Time = 2.74 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5042, 3042, 4674, 3042, 4672, 25, 3042, 3956, 4202, 2620, 3011, 2720, 4909, 3042, 4674, 3042, 4257, 4669, 3011, 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.

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

\(\Big \downarrow \) 5042

\(\displaystyle \frac {\int (e+f x)^3 \sec ^4(c+d x)dx}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x)^3 \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}-\frac {\int (e+f x)^3 \sec ^3(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 4909

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[((e + f*x)^3*Sec[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
 

Output:

(-1/2*(f*(e + f*x)^2*Sec[c + d*x]^2)/d^2 + ((e + f*x)^3*Sec[c + d*x]^2*Tan 
[c + d*x])/(3*d) + (f^2*((f*Log[Cos[c + d*x]])/d^2 + ((e + f*x)*Tan[c + d* 
x])/d))/d^2 + (2*((-3*f*(((I/3)*(e + f*x)^3)/f - (2*I)*(((-1/2*I)*(e + f*x 
)^2*Log[1 + E^((2*I)*(c + d*x))])/d + (I*f*(((I/2)*(e + f*x)*PolyLog[2, -E 
^((2*I)*(c + d*x))])/d - (f*PolyLog[3, -E^((2*I)*(c + d*x))])/(4*d^2)))/d) 
))/d + ((e + f*x)^3*Tan[c + d*x])/d))/3)/a - (((e + f*x)^3*Sec[c + d*x]^3) 
/(3*d) - (f*((f^2*ArcTanh[Sin[c + d*x]])/d^3 + (((-2*I)*(e + f*x)^2*ArcTan 
[E^(I*(c + d*x))])/d + (2*f*((I*(e + f*x)*PolyLog[2, (-I)*E^(I*(c + d*x))] 
)/d - (f*PolyLog[3, (-I)*E^(I*(c + d*x))])/d^2))/d - (2*f*((I*(e + f*x)*Po 
lyLog[2, I*E^(I*(c + d*x))])/d - (f*PolyLog[3, I*E^(I*(c + d*x))])/d^2))/d 
)/2 - (f*(e + f*x)*Sec[c + d*x])/d^2 + ((e + f*x)^2*Sec[c + d*x]*Tan[c + d 
*x])/(2*d)))/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 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 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 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, 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 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (438 ) = 876\).

Time = 2.73 (sec) , antiderivative size = 1135, normalized size of antiderivative = 2.39

method	result	size
risch	\(\text {Expression too large to display}\)	\(1135\)

Input:

int((f*x+e)^3*sec(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

3/a/d^2*e*f^2*ln(1+I*exp(I*(d*x+c)))*x+3/a/d^3*e*f^2*ln(1+I*exp(I*(d*x+c)) 
)*c+4*I/a/d^3*c^2*f^3*x-4*I/a/d*e*f^2*x^2-I/a/d^2*e^2*f*arctan(exp(I*(d*x+ 
c)))-4*I/a/d^3*c^2*e*f^2-I/a/d^4*f^3*c^2*arctan(exp(I*(d*x+c)))-5*I/a/d^3* 
e*f^2*polylog(2,I*exp(I*(d*x+c)))-3*I/a/d^3*e*f^2*polylog(2,-I*exp(I*(d*x+ 
c)))-5*I/a/d^3*f^3*polylog(2,I*exp(I*(d*x+c)))*x-3*I/a/d^3*f^3*polylog(2,- 
I*exp(I*(d*x+c)))*x-3/2/a/d^4*c^2*f^3*ln(1+I*exp(I*(d*x+c)))-8*I/a/d^2*e*f 
^2*c*x+2*I/a/d^3*e*f^2*c*arctan(exp(I*(d*x+c)))-4/3*I/a/d*x^3*f^3-5/2/d^4/ 
a*c^2*f^3*ln(1-I*exp(I*(d*x+c)))-4/d^2/a*e^2*f*ln(exp(I*(d*x+c)))-4/d^4/a* 
c^2*f^3*ln(exp(I*(d*x+c)))+2/d^4/a*c^2*f^3*ln(exp(2*I*(d*x+c))+1)+5/2/d^2/ 
a*f^3*ln(1-I*exp(I*(d*x+c)))*x^2+5*f^3*polylog(3,I*exp(I*(d*x+c)))/a/d^4+8 
/3*I/a/d^4*c^3*f^3-2*I/a/d^4*f^3*arctan(exp(I*(d*x+c)))+5/d^2/a*e*f^2*ln(1 
-I*exp(I*(d*x+c)))*x+3*f^3*polylog(3,-I*exp(I*(d*x+c)))/a/d^4-2/a/d^4*f^3* 
ln(exp(I*(d*x+c)))+1/a/d^4*f^3*ln(exp(2*I*(d*x+c))+1)+5/d^3/a*e*f^2*ln(1-I 
*exp(I*(d*x+c)))*c-4/d^3/a*c*e*f^2*ln(exp(2*I*(d*x+c))+1)+8/d^3/a*c*f^2*e* 
ln(exp(I*(d*x+c)))+2/d^2/a*e^2*f*ln(exp(2*I*(d*x+c))+1)-1/3*(8*d^2*f^3*x^3 
*exp(I*(d*x+c))+6*I*f^3*x*exp(2*I*(d*x+c))+12*I*d^2*e^2*f*x+3*I*d*f^3*x^2* 
exp(3*I*(d*x+c))+6*f^3*x*exp(3*I*(d*x+c))+24*d^2*e*f^2*x^2*exp(I*(d*x+c))+ 
24*d^2*e^2*f*x*exp(I*(d*x+c))+3*I*d*e^2*f*exp(I*(d*x+c))+12*I*d^2*e*f^2*x^ 
2+6*I*d*e*f^2*x*exp(3*I*(d*x+c))+6*I*d*e*f^2*x*exp(I*(d*x+c))+6*I*f^3*x+8* 
d^2*e^3*exp(I*(d*x+c))+3*I*d*f^3*x^2*exp(I*(d*x+c))+4*I*d^2*e^3+3*I*d*e...
 

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. 1531 vs. \(2 (426) = 852\).

Time = 0.16 (sec) , antiderivative size = 1531, normalized size of antiderivative = 3.22 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")
 

Output:

1/12*(4*d^3*f^3*x^3 + 12*d^3*e*f^2*x^2 + 12*d^3*e^2*f*x + 4*d^3*e^3 - 4*(2 
*d^3*f^3*x^3 + 6*d^3*e*f^2*x^2 + 2*d^3*e^3 + 3*d*e*f^2 + 3*(2*d^3*e^2*f + 
d*f^3)*x)*cos(d*x + c)^2 - 6*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*cos 
(d*x + c) - 18*((-I*d*f^3*x - I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (-I*d 
*f^3*x - I*d*e*f^2)*cos(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) - 3 
0*((I*d*f^3*x + I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (I*d*f^3*x + I*d*e* 
f^2)*cos(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) - 18*((I*d*f^3*x + 
 I*d*e*f^2)*cos(d*x + c)*sin(d*x + c) + (I*d*f^3*x + I*d*e*f^2)*cos(d*x + 
c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) - 30*((-I*d*f^3*x - I*d*e*f^2)*c 
os(d*x + c)*sin(d*x + c) + (-I*d*f^3*x - I*d*e*f^2)*cos(d*x + c))*dilog(-I 
*cos(d*x + c) - sin(d*x + c)) + 3*((5*d^2*e^2*f - 10*c*d*e*f^2 + (5*c^2 + 
4)*f^3)*cos(d*x + c)*sin(d*x + c) + (5*d^2*e^2*f - 10*c*d*e*f^2 + (5*c^2 + 
 4)*f^3)*cos(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + 9*((d^2*e^ 
2*f - 2*c*d*e*f^2 + c^2*f^3)*cos(d*x + c)*sin(d*x + c) + (d^2*e^2*f - 2*c* 
d*e*f^2 + c^2*f^3)*cos(d*x + c))*log(cos(d*x + c) - I*sin(d*x + c) + I) + 
15*((d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c)*sin 
(d*x + c) + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x 
+ c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) + 9*((d^2*f^3*x^2 + 2*d^2*e*f 
^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c)*sin(d*x + c) + (d^2*f^3*x^2 + 2 
*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c))*log(I*cos(d*x + c) ...
 

Sympy [F]

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

integrate((f*x+e)**3*sec(d*x+c)**2/(a+a*sin(d*x+c)),x)
 

Output:

(Integral(e**3*sec(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f**3*x**3 
*sec(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(3*e*f**2*x**2*sec(c + d 
*x)**2/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*sec(c + d*x)**2/(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. 5130 vs. \(2 (426) = 852\).

Time = 0.99 (sec) , antiderivative size = 5130, normalized size of antiderivative = 10.80 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/12*(24*c^2*e*f^2*(sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(co 
s(d*x + c) + 1)^2 + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1)/(a*d^2 + 2* 
a*d^2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*d^2*sin(d*x + c)^3/(cos(d*x + 
c) + 1)^3 - a*d^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + 6*(4*(8*(d*x + c) 
*cos(d*x + c) - sin(3*d*x + 3*c) - sin(d*x + c))*cos(4*d*x + 4*c) + 16*(2* 
d*x + 4*(d*x + c)*sin(d*x + c) + 2*c + cos(d*x + c))*cos(3*d*x + 3*c) + 8* 
cos(3*d*x + 3*c)^2 + 8*cos(d*x + c)^2 + 5*(2*(2*sin(3*d*x + 3*c) + 2*sin(d 
*x + c) + 1)*cos(4*d*x + 4*c) - cos(4*d*x + 4*c)^2 - 4*cos(3*d*x + 3*c)^2 
- 8*cos(3*d*x + 3*c)*cos(d*x + c) - 4*cos(d*x + c)^2 - 4*(cos(3*d*x + 3*c) 
 + cos(d*x + c))*sin(4*d*x + 4*c) - sin(4*d*x + 4*c)^2 - 4*(2*sin(d*x + c) 
 + 1)*sin(3*d*x + 3*c) - 4*sin(3*d*x + 3*c)^2 - 4*sin(d*x + c)^2 - 4*sin(d 
*x + c) - 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3 
*(2*(2*sin(3*d*x + 3*c) + 2*sin(d*x + c) + 1)*cos(4*d*x + 4*c) - cos(4*d*x 
 + 4*c)^2 - 4*cos(3*d*x + 3*c)^2 - 8*cos(3*d*x + 3*c)*cos(d*x + c) - 4*cos 
(d*x + c)^2 - 4*(cos(3*d*x + 3*c) + cos(d*x + c))*sin(4*d*x + 4*c) - sin(4 
*d*x + 4*c)^2 - 4*(2*sin(d*x + c) + 1)*sin(3*d*x + 3*c) - 4*sin(3*d*x + 3* 
c)^2 - 4*sin(d*x + c)^2 - 4*sin(d*x + c) - 1)*log(cos(d*x + c)^2 + sin(d*x 
 + c)^2 - 2*sin(d*x + c) + 1) + 4*(4*d*x + 8*(d*x + c)*sin(d*x + c) + 4*c 
+ cos(3*d*x + 3*c) + cos(d*x + c))*sin(4*d*x + 4*c) - 4*(16*(d*x + c)*cos( 
d*x + c) - 4*sin(d*x + c) - 1)*sin(3*d*x + 3*c) + 8*sin(3*d*x + 3*c)^2 ...
 

Giac [F]

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

integrate((f*x+e)^3*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)^3*sec(d*x + c)^2/(a*sin(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \] Input:

int((e + f*x)^3/(cos(c + d*x)^2*(a + a*sin(c + d*x))),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {too large to display} \] Input:

int((f*x+e)^3*sec(d*x+c)^2/(a+a*sin(d*x+c)),x)
 

Output:

(576*cos(c + d*x)*int(x**2/(tan((c + d*x)/2)**6 + 2*tan((c + d*x)/2)**5 - 
tan((c + d*x)/2)**4 - 4*tan((c + d*x)/2)**3 - tan((c + d*x)/2)**2 + 2*tan( 
(c + d*x)/2) + 1),x)*sin(c + d*x)*d**3*f**3 + 576*cos(c + d*x)*int(x**2/(t 
an((c + d*x)/2)**6 + 2*tan((c + d*x)/2)**5 - tan((c + d*x)/2)**4 - 4*tan(( 
c + d*x)/2)**3 - tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 1),x)*d**3*f** 
3 - 1296*cos(c + d*x)*int((tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**6 + 2 
*tan((c + d*x)/2)**5 - tan((c + d*x)/2)**4 - 4*tan((c + d*x)/2)**3 - tan(( 
c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 1),x)*sin(c + d*x)*d**3*f**3 - 1296* 
cos(c + d*x)*int((tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**6 + 2*tan((c + 
 d*x)/2)**5 - tan((c + d*x)/2)**4 - 4*tan((c + d*x)/2)**3 - tan((c + d*x)/ 
2)**2 + 2*tan((c + d*x)/2) + 1),x)*d**3*f**3 - 2592*cos(c + d*x)*int((tan( 
(c + d*x)/2)*x)/(tan((c + d*x)/2)**6 + 2*tan((c + d*x)/2)**5 - tan((c + d* 
x)/2)**4 - 4*tan((c + d*x)/2)**3 - tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2 
) + 1),x)*sin(c + d*x)*d**3*e*f**2 + 8208*cos(c + d*x)*int((tan((c + d*x)/ 
2)*x)/(tan((c + d*x)/2)**6 + 2*tan((c + d*x)/2)**5 - tan((c + d*x)/2)**4 - 
 4*tan((c + d*x)/2)**3 - tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 1),x)* 
sin(c + d*x)*d**2*f**3 - 2592*cos(c + d*x)*int((tan((c + d*x)/2)*x)/(tan(( 
c + d*x)/2)**6 + 2*tan((c + d*x)/2)**5 - tan((c + d*x)/2)**4 - 4*tan((c + 
d*x)/2)**3 - tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 1),x)*d**3*e*f**2 
+ 8208*cos(c + d*x)*int((tan((c + d*x)/2)*x)/(tan((c + d*x)/2)**6 + 2*t...