\(\int \frac {(e+f x)^2 \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [282]

Optimal result

Integrand size = 28, antiderivative size = 431 \[ \int \frac {(e+f x)^2 \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{4 a d}+\frac {5 f^2 \text {arctanh}(\sin (c+d x))}{6 a d^3}+\frac {f^2 \log (\cos (c+d x))}{3 a d^3}+\frac {3 i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{4 a d^2}-\frac {3 i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{4 a d^2}-\frac {3 f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{4 a d^3}+\frac {3 f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{4 a d^3}-\frac {3 f (e+f x) \sec (c+d x)}{4 a d^2}-\frac {f^2 \sec ^2(c+d x)}{12 a d^3}-\frac {f (e+f x) \sec ^3(c+d x)}{6 a d^2}-\frac {(e+f x)^2 \sec ^4(c+d x)}{4 a d}+\frac {f (e+f x) \tan (c+d x)}{3 a d^2}+\frac {f^2 \sec (c+d x) \tan (c+d x)}{12 a d^3}+\frac {3 (e+f x)^2 \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {f (e+f x) \sec ^2(c+d x) \tan (c+d x)}{6 a d^2}+\frac {(e+f x)^2 \sec ^3(c+d x) \tan (c+d x)}{4 a d} \] Output:

-3/4*I*(f*x+e)^2*arctan(exp(I*(d*x+c)))/a/d+5/6*f^2*arctanh(sin(d*x+c))/a/ 
d^3+1/3*f^2*ln(cos(d*x+c))/a/d^3-3/4*I*f*(f*x+e)*polylog(2,I*exp(I*(d*x+c) 
))/a/d^2+3/4*I*f*(f*x+e)*polylog(2,-I*exp(I*(d*x+c)))/a/d^2-3/4*f^2*polylo 
g(3,-I*exp(I*(d*x+c)))/a/d^3+3/4*f^2*polylog(3,I*exp(I*(d*x+c)))/a/d^3-3/4 
*f*(f*x+e)*sec(d*x+c)/a/d^2-1/12*f^2*sec(d*x+c)^2/a/d^3-1/6*f*(f*x+e)*sec( 
d*x+c)^3/a/d^2-1/4*(f*x+e)^2*sec(d*x+c)^4/a/d+1/3*f*(f*x+e)*tan(d*x+c)/a/d 
^2+1/12*f^2*sec(d*x+c)*tan(d*x+c)/a/d^3+3/8*(f*x+e)^2*sec(d*x+c)*tan(d*x+c 
)/a/d+1/6*f*(f*x+e)*sec(d*x+c)^2*tan(d*x+c)/a/d^2+1/4*(f*x+e)^2*sec(d*x+c) 
^3*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. \(1579\) vs. \(2(431)=862\).

Time = 10.01 (sec) , antiderivative size = 1579, normalized size of antiderivative = 3.66 \[ \int \frac {(e+f x)^2 \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/8*((Cos[c] + I*Sin[c])*(3*d^2*e^2*x*Cos[c] + 4*f^2*x*Cos[c] + 3*d^2*e*f 
*x^2*Cos[c] + 6*e*f*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I* 
(-1 + Sin[c])) + 6*f^2*x*PolyLog[2, I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] 
 - I*(-1 + Sin[c])) + d^2*f^2*x^3*(Cos[c] - I*Sin[c]) + ((3*d^2*e^2 + 4*f^ 
2)*Log[-Cos[c + d*x] - I*(-1 + Sin[c + d*x])]*(Cos[c] + I*(-1 + Sin[c]))*( 
Cos[c] - I*Sin[c]))/d + 6*d*e*f*x*Log[1 - I*Cos[c + d*x] - Sin[c + d*x]]*( 
Cos[c] + I*(-1 + Sin[c]))*(Cos[c] - I*Sin[c]) + 3*d*f^2*x^2*Log[1 - I*Cos[ 
c + d*x] - Sin[c + d*x]]*(Cos[c] + I*(-1 + Sin[c]))*(Cos[c] - I*Sin[c]) + 
(6*f^2*PolyLog[3, I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] + I*(-1 + Sin[c]) 
)*(Cos[c] - I*Sin[c]))/d - (3*I)*d^2*e^2*x*Sin[c] - (4*I)*f^2*x*Sin[c] - ( 
3*I)*d^2*e*f*x^2*Sin[c] + (3*d^2*e^2 + 4*f^2)*x*(Cos[c] - I*Sin[c])*(-1 - 
I*Cos[c] + Sin[c])))/(a*d^2*(Cos[c] + I*(-1 + Sin[c]))) - ((Cos[c] + I*Sin 
[c])*(9*d^2*e^2*x*Cos[c] + 28*f^2*x*Cos[c] + 9*d^2*e*f*x^2*Cos[c] + 3*d^2* 
f^2*x^3*Cos[c] - (9*I)*d^2*e^2*x*Sin[c] - (28*I)*f^2*x*Sin[c] - (9*I)*d^2* 
e*f*x^2*Sin[c] - (3*I)*d^2*f^2*x^3*Sin[c] + 18*e*f*PolyLog[2, (-I)*Cos[c + 
 d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) + 18*f^2*x*PolyLog[2, (-I) 
*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) - 18*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])) - 9*d*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])) - ((9*d^2*e^2 + 28*f^2)*Log[Cos[c + d*x]...
 

Rubi [A] (verified)

Time = 2.81 (sec) , antiderivative size = 421, normalized size of antiderivative = 0.98, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5042, 3042, 4674, 3042, 4255, 3042, 4257, 4674, 3042, 4257, 4669, 3011, 2720, 4909, 3042, 4673, 3042, 4672, 25, 3042, 3956, 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.

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

\(\Big \downarrow \) 5042

\(\displaystyle \frac {\int (e+f x)^2 \sec ^5(c+d x)dx}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\frac {f^2 \int \sec ^3(c+d x)dx}{6 d^2}+\frac {3}{4} \int (e+f x)^2 \sec ^3(c+d x)dx-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{6 d^2}+\frac {3}{4} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {f^2 \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}+\frac {3}{4} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f^2 \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}+\frac {3}{4} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {3}{4} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\frac {3}{4} \left (\frac {f^2 \int \sec (c+d x)dx}{d^2}+\frac {1}{2} \int (e+f x)^2 \sec (c+d x)dx-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} \left (\frac {f^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{d^2}+\frac {1}{2} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int (e+f x)^2 \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}-\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}\)

\(\Big \downarrow \) 4669

\(\displaystyle -\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 f \int (e+f x) \log \left (1-i e^{i (c+d x)}\right )dx}{d}+\frac {2 f \int (e+f x) \log \left (1+i e^{i (c+d x)}\right )dx}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )dx}{d}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {\int (e+f x)^2 \sec ^4(c+d x) \tan (c+d x)dx}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 4909

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \int (e+f x) \sec ^4(c+d x)dx}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 4673

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \int (e+f x) \sec ^2(c+d x)dx-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \int (e+f x) \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 4672

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \left (\frac {f \int -\tan (c+d x)dx}{d}+\frac {(e+f x) \tan (c+d x)}{d}\right )-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \left (\frac {(e+f x) \tan (c+d x)}{d}-\frac {f \int \tan (c+d x)dx}{d}\right )-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \left (\frac {(e+f x) \tan (c+d x)}{d}-\frac {f \int \tan (c+d x)dx}{d}\right )-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )de^{i (c+d x)}}{d^2}\right )}{d}-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {\frac {(e+f x)^2 \sec ^4(c+d x)}{4 d}-\frac {f \left (\frac {2}{3} \left (\frac {f \log (\cos (c+d x))}{d^2}+\frac {(e+f x) \tan (c+d x)}{d}\right )-\frac {f \sec ^2(c+d x)}{6 d^2}+\frac {(e+f x) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{2 d}}{a}+\frac {\frac {3}{4} \left (\frac {1}{2} \left (-\frac {2 i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d}+\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^2}\right )}{d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^2}\right )}{d}\right )+\frac {f^2 \text {arctanh}(\sin (c+d x))}{d^3}-\frac {f (e+f x) \sec (c+d x)}{d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {f^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{6 d^2}-\frac {f (e+f x) \sec ^3(c+d x)}{6 d^2}+\frac {(e+f x)^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}}{a}\)

Input:

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

Output:

(-1/6*(f*(e + f*x)*Sec[c + d*x]^3)/d^2 + ((e + f*x)^2*Sec[c + d*x]^3*Tan[c 
 + d*x])/(4*d) + (f^2*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + 
 d*x])/(2*d)))/(6*d^2) + (3*((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)*PolyLog[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)))/4)/a - (((e + f*x)^2*Sec[c + d*x]^4)/(4*d) - 
(f*(-1/6*(f*Sec[c + d*x]^2)/d^2 + ((e + f*x)*Sec[c + d*x]^2*Tan[c + d*x])/ 
(3*d) + (2*((f*Log[Cos[c + d*x]])/d^2 + ((e + f*x)*Tan[c + d*x])/d))/3))/( 
2*d))/a
 

Defintions of rubi rules used

rule 25

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

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 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

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 4673

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

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. 1034 vs. \(2 (382 ) = 764\).

Time = 2.02 (sec) , antiderivative size = 1035, normalized size of antiderivative = 2.40

method	result	size
risch	\(\text {Expression too large to display}\)	\(1035\)

Input:

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

Output:

-1/12*I*(9*d^2*f^2*x^2*exp(5*I*(d*x+c))+9*exp(I*(d*x+c))*d^2*f^2*x^2+6*exp 
(3*I*(d*x+c))*d^2*f^2*x^2+36*exp(4*I*(d*x+c))*d*f^2*x+44*exp(2*I*(d*x+c))* 
d*f^2*x+36*exp(4*I*(d*x+c))*d*e*f+44*exp(2*I*(d*x+c))*d*e*f+9*d^2*e^2*exp( 
5*I*(d*x+c))+4*f^2*exp(3*I*(d*x+c))+2*f^2*exp(I*(d*x+c))-18*I*d*f^2*x*exp( 
5*I*(d*x+c))-18*I*d*e*f*exp(5*I*(d*x+c))-18*I*d^2*f^2*x^2*exp(2*I*(d*x+c)) 
-36*I*d^2*e*f*x*exp(2*I*(d*x+c))+36*I*d^2*e*f*x*exp(4*I*(d*x+c))+8*d*f^2*x 
+2*I*d*f^2*x*exp(I*(d*x+c))+2*I*d*e*f*exp(I*(d*x+c))-16*I*d*f^2*x*exp(3*I* 
(d*x+c))-16*I*d*e*f*exp(3*I*(d*x+c))+2*f^2*exp(5*I*(d*x+c))+9*exp(I*(d*x+c 
))*d^2*e^2+6*exp(3*I*(d*x+c))*d^2*e^2+18*exp(I*(d*x+c))*d^2*e*f*x+12*exp(3 
*I*(d*x+c))*d^2*e*f*x-18*I*d^2*e^2*exp(2*I*(d*x+c))+18*I*d^2*e^2*exp(4*I*( 
d*x+c))+8*d*e*f+18*I*d^2*f^2*x^2*exp(4*I*(d*x+c))+18*d^2*e*f*x*exp(5*I*(d* 
x+c)))/(exp(I*(d*x+c))+I)^4/d^3/(exp(I*(d*x+c))-I)^2/a+1/3/a/d^3*f^2*ln(ex 
p(2*I*(d*x+c))+1)+3/8/d/a*f^2*ln(1-I*exp(I*(d*x+c)))*x^2-5/3*I/a/d^3*f^2*a 
rctan(exp(I*(d*x+c)))-3/4/a/d*ln(1+I*exp(I*(d*x+c)))*e*f*x-3/4*I/a/d^2*f^2 
*polylog(2,I*exp(I*(d*x+c)))*x+3/4/d/a*e*f*ln(1-I*exp(I*(d*x+c)))*x+3/4*I/ 
a/d^2*f^2*polylog(2,-I*exp(I*(d*x+c)))*x-3/4*I/a/d^2*e*f*polylog(2,I*exp(I 
*(d*x+c)))+3/4*I/a/d^2*e*f*polylog(2,-I*exp(I*(d*x+c)))-3/4*I/a/d*e^2*arct 
an(exp(I*(d*x+c)))-3/4*I/a/d^3*f^2*c^2*arctan(exp(I*(d*x+c)))+3/2*I/a/d^2* 
e*f*c*arctan(exp(I*(d*x+c)))+3/4/d^2/a*e*f*ln(1-I*exp(I*(d*x+c)))*c-2/3/a/ 
d^3*f^2*ln(exp(I*(d*x+c)))-3/4*f^2*polylog(3,-I*exp(I*(d*x+c)))/a/d^3-3...
 

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. 1517 vs. \(2 (373) = 746\).

Time = 0.19 (sec) , antiderivative size = 1517, normalized size of antiderivative = 3.52 \[ \int \frac {(e+f x)^2 \sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(6*d^2*f^2*x^2 + 12*d^2*e*f*x + 6*d^2*e^2 - 16*(d*f^2*x + d*e*f)*cos( 
d*x + c)^3 - 2*(9*d^2*f^2*x^2 + 18*d^2*e*f*x + 9*d^2*e^2 + 2*f^2)*cos(d*x 
+ c)^2 - 28*(d*f^2*x + d*e*f)*cos(d*x + c) - 18*((I*d*f^2*x + I*d*e*f)*cos 
(d*x + c)^2*sin(d*x + c) + (I*d*f^2*x + I*d*e*f)*cos(d*x + c)^2)*dilog(I*c 
os(d*x + c) + sin(d*x + c)) - 18*((I*d*f^2*x + I*d*e*f)*cos(d*x + c)^2*sin 
(d*x + c) + (I*d*f^2*x + I*d*e*f)*cos(d*x + c)^2)*dilog(I*cos(d*x + c) - s 
in(d*x + c)) - 18*((-I*d*f^2*x - I*d*e*f)*cos(d*x + c)^2*sin(d*x + c) + (- 
I*d*f^2*x - I*d*e*f)*cos(d*x + c)^2)*dilog(-I*cos(d*x + c) + sin(d*x + c)) 
 - 18*((-I*d*f^2*x - I*d*e*f)*cos(d*x + c)^2*sin(d*x + c) + (-I*d*f^2*x - 
I*d*e*f)*cos(d*x + c)^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + ((9*d^2*e 
^2 - 18*c*d*e*f + (9*c^2 + 28)*f^2)*cos(d*x + c)^2*sin(d*x + c) + (9*d^2*e 
^2 - 18*c*d*e*f + (9*c^2 + 28)*f^2)*cos(d*x + c)^2)*log(cos(d*x + c) + I*s 
in(d*x + c) + I) - 3*((3*d^2*e^2 - 6*c*d*e*f + (3*c^2 + 4)*f^2)*cos(d*x + 
c)^2*sin(d*x + c) + (3*d^2*e^2 - 6*c*d*e*f + (3*c^2 + 4)*f^2)*cos(d*x + c) 
^2)*log(cos(d*x + c) - I*sin(d*x + c) + I) + 9*((d^2*f^2*x^2 + 2*d^2*e*f*x 
 + 2*c*d*e*f - c^2*f^2)*cos(d*x + c)^2*sin(d*x + c) + (d^2*f^2*x^2 + 2*d^2 
*e*f*x + 2*c*d*e*f - c^2*f^2)*cos(d*x + c)^2)*log(I*cos(d*x + c) + sin(d*x 
 + c) + 1) - 9*((d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*cos(d*x 
+ c)^2*sin(d*x + c) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*co 
s(d*x + c)^2)*log(I*cos(d*x + c) - sin(d*x + c) + 1) + 9*((d^2*f^2*x^2 ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

( - 4472*cos(c + d*x)*sin(c + d*x)**2*d*e*f - 4472*cos(c + d*x)*sin(c + d* 
x)**2*d*f**2*x - 3816*cos(c + d*x)*sin(c + d*x)**2*f**2 + 1872*cos(c + d*x 
)*sin(c + d*x)*d**2*e*f*x + 936*cos(c + d*x)*sin(c + d*x)*d**2*f**2*x**2 + 
 1144*cos(c + d*x)*sin(c + d*x)*d*e*f + 2872*cos(c + d*x)*sin(c + d*x)*d*f 
**2*x - 504*cos(c + d*x)*sin(c + d*x)*f**2 + 3744*cos(c + d*x)*d**2*e*f*x 
+ 1872*cos(c + d*x)*d**2*f**2*x**2 + 6448*cos(c + d*x)*d*e*f + 9904*cos(c 
+ d*x)*d*f**2*x + 3456*cos(c + d*x)*f**2 - 14976*int((tan((c + d*x)/2)*x** 
2)/(tan((c + d*x)/2)**8 + 2*tan((c + d*x)/2)**7 - 2*tan((c + d*x)/2)**6 - 
6*tan((c + d*x)/2)**5 + 6*tan((c + d*x)/2)**3 + 2*tan((c + d*x)/2)**2 - 2* 
tan((c + d*x)/2) - 1),x)*sin(c + d*x)**3*d**3*f**2 - 14976*int((tan((c + d 
*x)/2)*x**2)/(tan((c + d*x)/2)**8 + 2*tan((c + d*x)/2)**7 - 2*tan((c + d*x 
)/2)**6 - 6*tan((c + d*x)/2)**5 + 6*tan((c + d*x)/2)**3 + 2*tan((c + d*x)/ 
2)**2 - 2*tan((c + d*x)/2) - 1),x)*sin(c + d*x)**2*d**3*f**2 + 14976*int(( 
tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**8 + 2*tan((c + d*x)/2)**7 - 2*ta 
n((c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5 + 6*tan((c + d*x)/2)**3 + 2*tan( 
(c + d*x)/2)**2 - 2*tan((c + d*x)/2) - 1),x)*sin(c + d*x)*d**3*f**2 + 1497 
6*int((tan((c + d*x)/2)*x**2)/(tan((c + d*x)/2)**8 + 2*tan((c + d*x)/2)**7 
 - 2*tan((c + d*x)/2)**6 - 6*tan((c + d*x)/2)**5 + 6*tan((c + d*x)/2)**3 + 
 2*tan((c + d*x)/2)**2 - 2*tan((c + d*x)/2) - 1),x)*d**3*f**2 - 29952*int( 
(tan((c + d*x)/2)*x)/(tan((c + d*x)/2)**8 + 2*tan((c + d*x)/2)**7 - 2*t...