\(\int \frac {(e+f x)^3 \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [306]

Optimal result

Integrand size = 26, antiderivative size = 937 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Output:

-6*I*a*f^3*polylog(4,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^4-b*(f*x+e)^3*ln(1-I*b 
*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)/d-b*(f*x+e)^3*ln(1-I*b*exp( 
I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)/d+b*(f*x+e)^3*ln(1+exp(2*I*(d*x+ 
c)))/(a^2-b^2)/d-3/2*I*b*f*(f*x+e)^2*polylog(2,-exp(2*I*(d*x+c)))/(a^2-b^2 
)/d^2+6*I*a*f^3*polylog(4,I*exp(I*(d*x+c)))/(a^2-b^2)/d^4+3*I*b*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)/d^2+3*I*a*f* 
(f*x+e)^2*polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^2+3/4*I*b*f^3*polylog(4 
,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^4-6*a*f^2*(f*x+e)*polylog(3,-I*exp(I*(d*x+ 
c)))/(a^2-b^2)/d^3+6*a*f^2*(f*x+e)*polylog(3,I*exp(I*(d*x+c)))/(a^2-b^2)/d 
^3-6*b*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)/d^3-6*b*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)/d^3+3/2*b*f^2*(f*x+e)*polylog(3,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^ 
3-6*I*b*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)/d^ 
4-6*I*b*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)/d^ 
4-3*I*a*f*(f*x+e)^2*polylog(2,I*exp(I*(d*x+c)))/(a^2-b^2)/d^2+3*I*b*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)/d^2-2*I* 
a*(f*x+e)^3*arctan(exp(I*(d*x+c)))/(a^2-b^2)/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. \(2506\) vs. \(2(937)=1874\).

Time = 11.80 (sec) , antiderivative size = 2506, normalized size of antiderivative = 2.67 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

((I*b*(e + f*x)^4)/f - (2*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^3*Log[1 - I/ 
E^(I*(c + d*x))])/d + (2*(a + b)*(1 + E^((2*I)*c))*(e + f*x)^3*Log[1 + I/E 
^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d^2*(e + f*x)^2*Pol 
yLog[2, (-I)/E^(I*(c + d*x))] + 2*f*(d*(e + f*x)*PolyLog[3, (-I)/E^(I*(c + 
 d*x))] - I*f*PolyLog[4, (-I)/E^(I*(c + d*x))])))/d^4 - ((6*I)*(a - b)*(1 
+ E^((2*I)*c))*f*(d^2*(e + f*x)^2*PolyLog[2, I/E^(I*(c + d*x))] - (2*I)*d* 
f*(e + f*x)*PolyLog[3, I/E^(I*(c + d*x))] - 2*f^2*PolyLog[4, I/E^(I*(c + d 
*x))]))/d^4)/(2*(a^2 - b^2)*(1 + E^((2*I)*c))) + (b*((-4*I)*d^4*e^3*E^((2* 
I)*c)*x - (6*I)*d^4*e^2*E^((2*I)*c)*f*x^2 - (4*I)*d^4*e*E^((2*I)*c)*f^2*x^ 
3 - I*d^4*E^((2*I)*c)*f^3*x^4 - (2*I)*d^3*e^3*ArcTan[(2*a*E^(I*(c + d*x))) 
/(b*(-1 + E^((2*I)*(c + d*x))))] + (2*I)*d^3*e^3*E^((2*I)*c)*ArcTan[(2*a*E 
^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] - d^3*e^3*Log[4*a^2*E^((2* 
I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + d^3*e^3*E^((2*I)*c)*Lo 
g[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - 6*d^3*e^ 
2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2 
*I)*c)])] + 6*d^3*e^2*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E 
^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e*f^2*x^2*Log[1 + (b*E^( 
I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e* 
E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^ 
2 + b^2)*E^((2*I)*c)])] - 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(...
 

Rubi [A] (verified)

Time = 3.07 (sec) , antiderivative size = 814, normalized size of antiderivative = 0.87, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5044, 5030, 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.

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

\(\Big \downarrow \) 5044

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 5030

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

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]

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

Input:

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

Output:

int((f*x+e)^3*sec(d*x+c)/(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. 3069 vs. \(2 (822) = 1644\).

Time = 0.38 (sec) , antiderivative size = 3069, normalized size of antiderivative = 3.28 \[ \int \frac {(e+f x)^3 \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((f*x+e)^3*sec(d*x+c)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

(int((sec(c + d*x)*x**3)/(sin(c + d*x)*b + a),x)*a**2*d*f**3 - int((sec(c 
+ d*x)*x**3)/(sin(c + d*x)*b + a),x)*b**2*d*f**3 + 3*int((sec(c + d*x)*x** 
2)/(sin(c + d*x)*b + a),x)*a**2*d*e*f**2 - 3*int((sec(c + d*x)*x**2)/(sin( 
c + d*x)*b + a),x)*b**2*d*e*f**2 + 3*int((sec(c + d*x)*x)/(sin(c + d*x)*b 
+ a),x)*a**2*d*e**2*f - 3*int((sec(c + d*x)*x)/(sin(c + d*x)*b + a),x)*b** 
2*d*e**2*f - log(tan((c + d*x)/2) - 1)*a*e**3 + log(tan((c + d*x)/2) - 1)* 
b*e**3 + log(tan((c + d*x)/2) + 1)*a*e**3 + log(tan((c + d*x)/2) + 1)*b*e* 
*3 - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b*e**3)/(d*(a** 
2 - b**2))