\(\int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [310]

Optimal result

Integrand size = 28, antiderivative size = 923 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 b^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 b^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i a f^2 (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 b f^3 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 b f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 i b^2 f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i b^2 f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}+\frac {3 a f^3 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^4}-\frac {6 b^2 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 10.31 (sec) , antiderivative size = 1438, normalized size of antiderivative = 1.56 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(f*(((2*I)*a*(e + f*x)^3)/f + (3*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log 
[1 - I/E^(I*(c + d*x))])/d + (3*(a + b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[ 
1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x)* 
PolyLog[2, (-I)/E^(I*(c + d*x))] + f*PolyLog[3, (-I)/E^(I*(c + d*x))]))/d^ 
3 + (6*(a - b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x)*PolyLog[2, I/E^(I*(c + d 
*x))] + f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3))/((a^2 - b^2)*d*(1 + E^((2* 
I)*c))) + (b^2*(2*Sqrt[-a^2 + b^2]*d^3*e^3*ArcTan[(I*a + b*E^(I*(c + d*x)) 
)/Sqrt[a^2 - b^2]] + 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 - (b*E^(I*(c + d* 
x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 
- (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + Sqrt[a^2 - b^2]*d^3*f 
^3*x^3*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 3*Sqrt[a 
^2 - b^2]*d^3*e^2*f*x*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]) 
] - 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqr 
t[-a^2 + b^2])] - Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1 + (b*E^(I*(c + d*x)))/ 
(I*a + Sqrt[-a^2 + b^2])] - (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLo 
g[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + (3*I)*Sqrt[a^2 - b 
^2]*d^2*f*(e + f*x)^2*PolyLog[2, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + 
b^2]))] + 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a 
 + Sqrt[-a^2 + b^2])] + 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, (b*E^(I*(c + 
d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog...
 

Rubi [A] (verified)

Time = 3.44 (sec) , antiderivative size = 774, normalized size of antiderivative = 0.84, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5044, 3042, 3804, 2694, 27, 2620, 3011, 7163, 2720, 7143, 7293, 2009}

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.

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

\(\Big \downarrow \) 5044

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*b^2*(((-1/2*I)*b*(((e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt 
[a^2 - b^2])])/(b*d) - (3*f*((I*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x) 
))/(a - Sqrt[a^2 - b^2])])/d - ((2*I)*f*(((-I)*(e + f*x)*PolyLog[3, (I*b*E 
^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d + (f*PolyLog[4, (I*b*E^(I*(c + d 
*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/d))/(b*d)))/Sqrt[a^2 - b^2] + ((I/2)*b 
*(((e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) 
 - (3*f*((I*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b 
^2])])/d - ((2*I)*f*(((-I)*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + 
 Sqrt[a^2 - b^2])])/d + (f*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 
- b^2])])/d^2))/d))/(b*d)))/Sqrt[a^2 - b^2]))/(a^2 - b^2) + (((-I)*a*(e + 
f*x)^3)/d - ((6*I)*b*f*(e + f*x)^2*ArcTan[E^(I*(c + d*x))])/d^2 + (3*a*f*( 
e + f*x)^2*Log[1 + E^((2*I)*(c + d*x))])/d^2 + ((6*I)*b*f^2*(e + f*x)*Poly 
Log[2, (-I)*E^(I*(c + d*x))])/d^3 - ((6*I)*b*f^2*(e + f*x)*PolyLog[2, I*E^ 
(I*(c + d*x))])/d^3 - ((3*I)*a*f^2*(e + f*x)*PolyLog[2, -E^((2*I)*(c + d*x 
))])/d^3 - (6*b*f^3*PolyLog[3, (-I)*E^(I*(c + d*x))])/d^4 + (6*b*f^3*PolyL 
og[3, I*E^(I*(c + d*x))])/d^4 + (3*a*f^3*PolyLog[3, -E^((2*I)*(c + d*x))]) 
/(2*d^4) - (b*(e + f*x)^3*Sec[c + d*x])/d + (a*(e + f*x)^3*Tan[c + d*x])/d 
)/(a^2 - b^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && 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 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5044

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

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \sec \left (d x +c \right )^{2}}{a +b \sin \left (d x +c \right )}d x\]

Input:

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

Output:

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

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. 4116 vs. \(2 (810) = 1620\).

Time = 0.45 (sec) , antiderivative size = 4116, normalized size of antiderivative = 4.46 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

( - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*a**3*b**2*d**3*e**3 + 24*cos(c + d*x)*int(x**2/(tan((c + d*x)/ 
2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c + d*x)/2)**4*a - 4*tan((c + d*x 
)/2)**3*b - tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*a**6*b*d* 
*3*f**3 - 48*cos(c + d*x)*int(x**2/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x 
)/2)**5*b - tan((c + d*x)/2)**4*a - 4*tan((c + d*x)/2)**3*b - tan((c + d*x 
)/2)**2*a + 2*tan((c + d*x)/2)*b + a),x)*a**4*b**3*d**3*f**3 + 24*cos(c + 
d*x)*int(x**2/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c + 
d*x)/2)**4*a - 4*tan((c + d*x)/2)**3*b - tan((c + d*x)/2)**2*a + 2*tan((c 
+ d*x)/2)*b + a),x)*a**2*b**5*d**3*f**3 - 32*cos(c + d*x)*int((tan((c + d* 
x)/2)**3*x**3)/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c + 
 d*x)/2)**4*a - 4*tan((c + d*x)/2)**3*b - tan((c + d*x)/2)**2*a + 2*tan((c 
 + d*x)/2)*b + a),x)*a**4*b**3*d**4*f**3 + 64*cos(c + d*x)*int((tan((c + d 
*x)/2)**3*x**3)/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c 
+ d*x)/2)**4*a - 4*tan((c + d*x)/2)**3*b - tan((c + d*x)/2)**2*a + 2*tan(( 
c + d*x)/2)*b + a),x)*a**2*b**5*d**4*f**3 - 32*cos(c + d*x)*int((tan((c + 
d*x)/2)**3*x**3)/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c 
 + d*x)/2)**4*a - 4*tan((c + d*x)/2)**3*b - tan((c + d*x)/2)**2*a + 2*tan( 
(c + d*x)/2)*b + a),x)*b**7*d**4*f**3 - 48*cos(c + d*x)*int((tan((c + d*x) 
/2)**3*x**2)/(tan((c + d*x)/2)**6*a + 2*tan((c + d*x)/2)**5*b - tan((c ...