3.3.98 \(\int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx\) [298]

3.3.98.1 Optimal result

Integrand size = 22, antiderivative size = 337 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=-\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b} \]

output

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

3.3.98.2 Mathematica [A] (verified)

Time = 3.75 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.57 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {2 i b^3 c^3 \arctan \left (e^{i (a+b x)}\right )-12 i b c d^2 \arctan \left (e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1-i e^{i (a+b x)}\right )+6 b d^3 x \log \left (1-i e^{i (a+b x)}\right )-3 b^3 c d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-b^3 d^3 x^3 \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1+i e^{i (a+b x)}\right )-6 b d^3 x \log \left (1+i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1+i e^{i (a+b x)}\right )-3 i d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+3 i d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )-6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )+b^2 (c+d x)^2 \sec (a+b x) (-3 d+b (c+d x) \tan (a+b x))}{2 b^4} \]

input

Integrate[(c + d*x)^3*Sec[a + b*x]*Tan[a + b*x]^2,x]
 

output

((2*I)*b^3*c^3*ArcTan[E^(I*(a + b*x))] - (12*I)*b*c*d^2*ArcTan[E^(I*(a + b 
*x))] - 3*b^3*c^2*d*x*Log[1 - I*E^(I*(a + b*x))] + 6*b*d^3*x*Log[1 - I*E^( 
I*(a + b*x))] - 3*b^3*c*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - b^3*d^3*x^3*L 
og[1 - I*E^(I*(a + b*x))] + 3*b^3*c^2*d*x*Log[1 + I*E^(I*(a + b*x))] - 6*b 
*d^3*x*Log[1 + I*E^(I*(a + b*x))] + 3*b^3*c*d^2*x^2*Log[1 + I*E^(I*(a + b* 
x))] + b^3*d^3*x^3*Log[1 + I*E^(I*(a + b*x))] - (3*I)*d*(-2*d^2 + b^2*(c + 
 d*x)^2)*PolyLog[2, (-I)*E^(I*(a + b*x))] + (3*I)*d*(-2*d^2 + b^2*(c + d*x 
)^2)*PolyLog[2, I*E^(I*(a + b*x))] + 6*b*c*d^2*PolyLog[3, (-I)*E^(I*(a + b 
*x))] + 6*b*d^3*x*PolyLog[3, (-I)*E^(I*(a + b*x))] - 6*b*c*d^2*PolyLog[3, 
I*E^(I*(a + b*x))] - 6*b*d^3*x*PolyLog[3, I*E^(I*(a + b*x))] + (6*I)*d^3*P 
olyLog[4, (-I)*E^(I*(a + b*x))] - (6*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))] 
+ b^2*(c + d*x)^2*Sec[a + b*x]*(-3*d + b*(c + d*x)*Tan[a + b*x]))/(2*b^4)
 

3.3.98.3 Rubi [A] (verified)

Time = 2.07 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.72, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {4913, 3042, 4669, 3011, 4674, 3042, 4669, 2715, 2838, 3011, 7163, 2720, 7143}

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

\(\displaystyle \int (c+d x)^3 \tan ^2(a+b x) \sec (a+b x) \, dx\)

\(\Big \downarrow \) 4913

\(\displaystyle \int (c+d x)^3 \sec ^3(a+b x)dx-\int (c+d x)^3 \sec (a+b x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

Int[(c + d*x)^3*Sec[a + b*x]*Tan[a + b*x]^2,x]
 

output

((2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + (3*d^2*(((-2*I)*(c + d*x)* 
ArcTan[E^(I*(a + b*x))])/b + (I*d*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 - 
(I*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2))/b^2 - (3*d*((I*(c + d*x)^2*PolyL 
og[2, (-I)*E^(I*(a + b*x))])/b - ((2*I)*d*(((-I)*(c + d*x)*PolyLog[3, (-I) 
*E^(I*(a + b*x))])/b + (d*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^2))/b))/b + 
(3*d*((I*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b - ((2*I)*d*(((-I)*(c 
 + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b + (d*PolyLog[4, I*E^(I*(a + b*x)) 
])/b^2))/b))/b + (((-2*I)*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + (3*d*(( 
I*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b - ((2*I)*d*(((-I)*(c + d 
*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b + (d*PolyLog[4, (-I)*E^(I*(a + b*x 
))])/b^2))/b))/b - (3*d*((I*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b - 
 ((2*I)*d*(((-I)*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b + (d*PolyLog[4 
, I*E^(I*(a + b*x))])/b^2))/b))/b)/2 - (3*d*(c + d*x)^2*Sec[a + b*x])/(2*b 
^2) + ((c + d*x)^3*Sec[a + b*x]*Tan[a + b*x])/(2*b)
 

3.3.98.3.1 Defintions of rubi rules used

rule 2715

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

rule 2720

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

rule 2838

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

rule 3011

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

rule 3042

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

rule 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 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 4913

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]*Tan[(a_.) + (b_.)*(x 
_)]^(p_), x_Symbol] :> -Int[(c + d*x)^m*Sec[a + b*x]*Tan[a + b*x]^(p - 2), 
x] + Int[(c + d*x)^m*Sec[a + b*x]^3*Tan[a + b*x]^(p - 2), x] /; FreeQ[{a, b 
, c, d, m}, x] && IGtQ[p/2, 0]
 

rule 7143

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

rule 7163

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

3.3.98.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1126 vs. \(2 (293 ) = 586\).

Time = 2.01 (sec) , antiderivative size = 1127, normalized size of antiderivative = 3.34

method	result	size
risch	\(\text {Expression too large to display}\)	\(1127\)

input

int((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x,method=_RETURNVERBOSE)
 

output

-3*I*d^3*polylog(2,I*exp(I*(b*x+a)))/b^4-3*I*d^3*polylog(4,I*exp(I*(b*x+a) 
))/b^4-3/2/b*c*d^2*ln(1-I*exp(I*(b*x+a)))*x^2-3/2/b^3*a^2*c*d^2*ln(1+I*exp 
(I*(b*x+a)))-3/2/b*c^2*d*ln(1-I*exp(I*(b*x+a)))*x-3/2/b^2*c^2*d*ln(1-I*exp 
(I*(b*x+a)))*a+3*I/b^3*c*d^2*a^2*arctan(exp(I*(b*x+a)))+3*I*d^3*polylog(2, 
-I*exp(I*(b*x+a)))/b^4+3*I*d^3*polylog(4,-I*exp(I*(b*x+a)))/b^4+3*I/b^2*c* 
d^2*polylog(2,I*exp(I*(b*x+a)))*x-3*I/b^2*c*d^2*polylog(2,-I*exp(I*(b*x+a) 
))*x-3*I/b^2*c^2*d*a*arctan(exp(I*(b*x+a)))+3/b^3*d^3*polylog(3,-I*exp(I*( 
b*x+a)))*x+1/2/b*d^3*ln(1+I*exp(I*(b*x+a)))*x^3-1/2/b*d^3*ln(1-I*exp(I*(b* 
x+a)))*x^3-3/b^3*d^3*ln(1+I*exp(I*(b*x+a)))*x+3/b^4*d^3*ln(1-I*exp(I*(b*x+ 
a)))*a+3/b^3*d^3*ln(1-I*exp(I*(b*x+a)))*x-3/b^4*d^3*ln(1+I*exp(I*(b*x+a))) 
*a-3/b^3*d^3*polylog(3,I*exp(I*(b*x+a)))*x+3/b^3*c*d^2*polylog(3,-I*exp(I* 
(b*x+a)))-1/2/b^4*a^3*d^3*ln(1-I*exp(I*(b*x+a)))-3/b^3*c*d^2*polylog(3,I*e 
xp(I*(b*x+a)))+I/b*c^3*arctan(exp(I*(b*x+a)))+3/2/b*c*d^2*ln(1+I*exp(I*(b* 
x+a)))*x^2+3/2/b^3*a^2*c*d^2*ln(1-I*exp(I*(b*x+a)))+3/2/b*c^2*d*ln(1+I*exp 
(I*(b*x+a)))*x+3/2/b^2*c^2*d*ln(1+I*exp(I*(b*x+a)))*a-I/b^4*d^3*a^3*arctan 
(exp(I*(b*x+a)))+6*I/b^4*d^3*a*arctan(exp(I*(b*x+a)))+3/2*I/b^2*c^2*d*poly 
log(2,I*exp(I*(b*x+a)))-6*I/b^3*c*d^2*arctan(exp(I*(b*x+a)))+3/2*I/b^2*d^3 
*polylog(2,I*exp(I*(b*x+a)))*x^2-3/2*I/b^2*d^3*polylog(2,-I*exp(I*(b*x+a)) 
)*x^2-3/2*I/b^2*c^2*d*polylog(2,-I*exp(I*(b*x+a)))-I/b^2/(exp(2*I*(b*x+a)) 
+1)^2*(d^3*x^3*b*exp(3*I*(b*x+a))+3*c*d^2*x^2*b*exp(3*I*(b*x+a))+3*c^2*...
 

3.3.98.5 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. 1315 vs. \(2 (273) = 546\).

Time = 0.33 (sec) , antiderivative size = 1315, normalized size of antiderivative = 3.90 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="fricas")
 

output

1/4*(-6*I*d^3*cos(b*x + a)^2*polylog(4, I*cos(b*x + a) + sin(b*x + a)) - 6 
*I*d^3*cos(b*x + a)^2*polylog(4, I*cos(b*x + a) - sin(b*x + a)) + 6*I*d^3* 
cos(b*x + a)^2*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*cos(b* 
x + a)^2*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 
2*I*b^2*c*d^2*x - I*b^2*c^2*d + 2*I*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + 
a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d + 2 
*I*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(I*b^2*d^3 
*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*co 
s(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^ 
2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a)) - (b^3 
*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a) 
^2*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*( 
a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a)^2*log(cos(b*x + a) - I*si 
n(b*x + a) + I) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b 
*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^2*log(I 
*cos(b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b 
^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*co 
s(b*x + a)^2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3 
*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2* 
d - 2*b*d^3)*x)*cos(b*x + a)^2*log(-I*cos(b*x + a) + sin(b*x + a) + 1) ...
 

3.3.98.6 Sympy [F]

\[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \tan ^{2}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

input

integrate((d*x+c)**3*sec(b*x+a)*tan(b*x+a)**2,x)
 

output

Integral((c + d*x)**3*tan(a + b*x)**2*sec(a + b*x), x)
 

3.3.98.7 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. 3831 vs. \(2 (273) = 546\).

Time = 1.37 (sec) , antiderivative size = 3831, normalized size of antiderivative = 11.37 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="maxima")
 

output

-1/4*(c^3*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - l 
og(sin(b*x + a) - 1)) - 3*a*c^2*d*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + l 
og(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b + 3*a^2*c*d^2*(2*sin(b*x + 
 a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/ 
b^2 - a^3*d^3*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) 
 - log(sin(b*x + a) - 1))/b^3 - 4*(2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^ 
3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 
2)*d^3)*(b*x + a) + ((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - 
a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a) 
)*cos(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 
 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + 
 a))*cos(2*b*x + 2*a) + (I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d^3 + 3*( 
I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 
 - 2*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 2*(I*(b*x + a)^3*d^3 - 6*I*b*c* 
d^2 + 6*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2 
*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos 
(b*x + a), sin(b*x + a) + 1) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 
3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d 
^3)*(b*x + a) + ((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^ 
3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))...
 

3.3.98.8 Giac [F]

\[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right ) \tan \left (b x + a\right )^{2} \,d x } \]

input

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="giac")
 

output

integrate((d*x + c)^3*sec(b*x + a)*tan(b*x + a)^2, x)
 

3.3.98.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Hanged} \]

input

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

output

\text{Hanged}