\(\int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [354]

Optimal result

Integrand size = 32, antiderivative size = 648 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {(e+f x)^2}{b d}-\frac {a^2 (e+f x)^2}{b \left (a^2+b^2\right ) d}+\frac {4 a f (e+f x) \arctan \left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}+\frac {a b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {2 f (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{b d^2}+\frac {2 a^2 f (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {2 i a f^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}+\frac {2 i a f^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac {2 a b f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^2}+\frac {2 a b f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^2}-\frac {f^2 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{b d^3}+\frac {a^2 f^2 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{b \left (a^2+b^2\right ) d^3}+\frac {2 a b f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^3}-\frac {2 a b f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^3}-\frac {a (e+f x)^2 \text {sech}(c+d x)}{\left (a^2+b^2\right ) d}+\frac {(e+f x)^2 \tanh (c+d x)}{b d}-\frac {a^2 (e+f x)^2 \tanh (c+d x)}{b \left (a^2+b^2\right ) d} \] Output:

(f*x+e)^2/b/d-a^2*(f*x+e)^2/b/(a^2+b^2)/d+4*a*f*(f*x+e)*arctan(exp(d*x+c)) 
/(a^2+b^2)/d^2-a*b*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b 
^2)^(3/2)/d+a*b*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2) 
^(3/2)/d-2*f*(f*x+e)*ln(1+exp(2*d*x+2*c))/b/d^2+2*a^2*f*(f*x+e)*ln(1+exp(2 
*d*x+2*c))/b/(a^2+b^2)/d^2+2*I*a*f^2*polylog(2,I*exp(d*x+c))/(a^2+b^2)/d^3 
-2*I*a*f^2*polylog(2,-I*exp(d*x+c))/(a^2+b^2)/d^3-2*a*b*f*(f*x+e)*polylog( 
2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2)/d^2+2*a*b*f*(f*x+e)*p 
olylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2)/d^2-f^2*polylo 
g(2,-exp(2*d*x+2*c))/b/d^3+a^2*f^2*polylog(2,-exp(2*d*x+2*c))/b/(a^2+b^2)/ 
d^3+2*a*b*f^2*polylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2) 
/d^3-2*a*b*f^2*polylog(3,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)^(3/2 
)/d^3-a*(f*x+e)^2*sech(d*x+c)/(a^2+b^2)/d+(f*x+e)^2*tanh(d*x+c)/b/d-a^2*(f 
*x+e)^2*tanh(d*x+c)/b/(a^2+b^2)/d
 

Mathematica [A] (warning: unable to verify)

Time = 3.88 (sec) , antiderivative size = 633, normalized size of antiderivative = 0.98 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {f \left (4 b d^2 e e^{2 c} x-4 b d^2 e \left (1+e^{2 c}\right ) x+2 b d^2 e^{2 c} f x^2-2 b d^2 \left (1+e^{2 c}\right ) f x^2+4 a d e \left (1+e^{2 c}\right ) \arctan \left (e^{c+d x}\right )+2 b d e \left (1+e^{2 c}\right ) \left (2 d x-\log \left (1+e^{2 (c+d x)}\right )\right )+2 i a \left (1+e^{2 c}\right ) f \left (d x \left (\log \left (1-i e^{c+d x}\right )-\log \left (1+i e^{c+d x}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{c+d x}\right )+\operatorname {PolyLog}\left (2,i e^{c+d x}\right )\right )+b \left (1+e^{2 c}\right ) f \left (2 d x \left (d x-\log \left (1+e^{2 (c+d x)}\right )\right )-\operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )\right )\right )}{\left (a^2+b^2\right ) \left (1+e^{2 c}\right )}+\frac {a b \left (2 d^2 e^2 \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )+2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )+2 d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 f^2 \operatorname {PolyLog}\left (3,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-2 f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {d^2 (e+f x)^2 \text {sech}(c+d x) (-a+b \text {sech}(c) \sinh (d x))}{a^2+b^2}}{d^3} \] Input:

Integrate[((e + f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]), 
x]
 

Output:

((f*(4*b*d^2*e*E^(2*c)*x - 4*b*d^2*e*(1 + E^(2*c))*x + 2*b*d^2*E^(2*c)*f*x 
^2 - 2*b*d^2*(1 + E^(2*c))*f*x^2 + 4*a*d*e*(1 + E^(2*c))*ArcTan[E^(c + d*x 
)] + 2*b*d*e*(1 + E^(2*c))*(2*d*x - Log[1 + E^(2*(c + d*x))]) + (2*I)*a*(1 
 + E^(2*c))*f*(d*x*(Log[1 - I*E^(c + d*x)] - Log[1 + I*E^(c + d*x)]) - Pol 
yLog[2, (-I)*E^(c + d*x)] + PolyLog[2, I*E^(c + d*x)]) + b*(1 + E^(2*c))*f 
*(2*d*x*(d*x - Log[1 + E^(2*(c + d*x))]) - PolyLog[2, -E^(2*(c + d*x))]))) 
/((a^2 + b^2)*(1 + E^(2*c))) + (a*b*(2*d^2*e^2*ArcTanh[(a + b*E^(c + d*x)) 
/Sqrt[a^2 + b^2]] - 2*d^2*e*f*x*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^ 
2])] - d^2*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + 2*d^2* 
e*f*x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + d^2*f^2*x^2*Log[1 + 
 (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 2*d*f*(e + f*x)*PolyLog[2, (b*E^ 
(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 2*d*f*(e + f*x)*PolyLog[2, -((b*E^(c 
+ d*x))/(a + Sqrt[a^2 + b^2]))] + 2*f^2*PolyLog[3, (b*E^(c + d*x))/(-a + S 
qrt[a^2 + b^2])] - 2*f^2*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2] 
))]))/(a^2 + b^2)^(3/2) + (d^2*(e + f*x)^2*Sech[c + d*x]*(-a + b*Sech[c]*S 
inh[d*x]))/(a^2 + b^2))/d^3
 

Rubi [A] (verified)

Time = 3.85 (sec) , antiderivative size = 573, normalized size of antiderivative = 0.88, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {6117, 3042, 4672, 26, 3042, 26, 4201, 2620, 2715, 2838, 6107, 3042, 3803, 25, 2694, 27, 2620, 3011, 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.

\(\displaystyle \int \frac {(e+f x)^2 \tanh (c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 6117

\(\displaystyle \frac {\int (e+f x)^2 \text {sech}^2(c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\int (e+f x)^2 \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{b}\)

\(\Big \downarrow \) 4672

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 i f \int -i (e+f x) \tanh (c+d x)dx}{d}}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 f \int (e+f x) \tanh (c+d x)dx}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 f \int -i (e+f x) \tan (i c+i d x)dx}{d}}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \int (e+f x) \tan (i c+i d x)dx}{d}}{b}\)

\(\Big \downarrow \) 4201

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}}dx-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}-\frac {f \int \log \left (1+e^{2 (c+d x)}\right )dx}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}-\frac {f \int e^{-2 (c+d x)} \log \left (1+e^{2 (c+d x)}\right )de^{2 (c+d x)}}{4 d^2}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {a \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)}dx}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 6107

\(\displaystyle -\frac {a \left (\frac {b^2 \int \frac {(e+f x)^2}{a+b \sinh (c+d x)}dx}{a^2+b^2}+\frac {\int (e+f x)^2 \text {sech}^2(c+d x) (a-b \sinh (c+d x))dx}{a^2+b^2}\right )}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3803

\(\displaystyle -\frac {a \left (\frac {2 b^2 \int -\frac {e^{c+d x} (e+f x)^2}{-2 e^{c+d x} a-b e^{2 (c+d x)}+b}dx}{a^2+b^2}+\frac {\int (e+f x)^2 \text {sech}^2(c+d x) (a-b \sinh (c+d x))dx}{a^2+b^2}\right )}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a \left (\frac {\int (e+f x)^2 \text {sech}^2(c+d x) (a-b \sinh (c+d x))dx}{a^2+b^2}-\frac {2 b^2 \int \frac {e^{c+d x} (e+f x)^2}{-2 e^{c+d x} a-b e^{2 (c+d x)}+b}dx}{a^2+b^2}\right )}{b}+\frac {\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}}{b}\)

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[((e + f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

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 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 3803

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

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && 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 6107

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

rule 6117

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(p_.)*Tanh[(c_.) + 
 (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> S 
imp[1/b   Int[(e + f*x)^m*Sech[c + d*x]^(p + 1)*Tanh[c + d*x]^(n - 1), x], 
x] - Simp[a/b   Int[(e + f*x)^m*Sech[c + d*x]^(p + 1)*(Tanh[c + d*x]^(n - 1 
)/(a + b*Sinh[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 
 0] && IGtQ[n, 0] && IGtQ[p, 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 7293

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

Maple [F]

\[\int \frac {\left (f x +e \right )^{2} \operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}d x\]

Input:

int((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)
 

Output:

int((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(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. 3662 vs. \(2 (601) = 1202\).

Time = 0.21 (sec) , antiderivative size = 3662, normalized size of antiderivative = 5.65 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm 
="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\left (e + f x\right )^{2} \tanh {\left (c + d x \right )} \operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:

integrate((f*x+e)**2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)
 

Output:

Integral((e + f*x)**2*tanh(c + d*x)*sech(c + d*x)/(a + b*sinh(c + d*x)), x 
)
 

Maxima [F]

\[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \operatorname {sech}\left (d x + c\right ) \tanh \left (d x + c\right )}{b \sinh \left (d x + c\right ) + a} \,d x } \] Input:

integrate((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm 
="maxima")
 

Output:

2*b*e*f*(2*(d*x + c)/((a^2 + b^2)*d^2) - log(e^(2*d*x + 2*c) + 1)/((a^2 + 
b^2)*d^2)) + 4*a*f^2*integrate(x*e^(d*x + c)/(a^2*d*e^(2*d*x + 2*c) + b^2* 
d*e^(2*d*x + 2*c) + a^2*d + b^2*d), x) + 4*b*f^2*integrate(x/(a^2*d*e^(2*d 
*x + 2*c) + b^2*d*e^(2*d*x + 2*c) + a^2*d + b^2*d), x) - e^2*(a*b*log((b*e 
^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2))) 
/((a^2 + b^2)^(3/2)*d) + 2*(a*e^(-d*x - c) - b)/((a^2 + b^2 + (a^2 + b^2)* 
e^(-2*d*x - 2*c))*d)) + 4*a*e*f*arctan(e^(d*x + c))/((a^2 + b^2)*d^2) - 2* 
(b*f^2*x^2 + 2*b*e*f*x + (a*f^2*x^2*e^c + 2*a*e*f*x*e^c)*e^(d*x))/(a^2*d + 
 b^2*d + (a^2*d*e^(2*c) + b^2*d*e^(2*c))*e^(2*d*x)) - integrate(-2*(a*b*f^ 
2*x^2*e^c + 2*a*b*e*f*x*e^c)*e^(d*x)/(a^2*b + b^3 - (a^2*b*e^(2*c) + b^3*e 
^(2*c))*e^(2*d*x) - 2*(a^3*e^c + a*b^2*e^c)*e^(d*x)), x)
 

Giac [F(-1)]

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

integrate((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm 
="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\mathrm {tanh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \] Input:

int((tanh(c + d*x)*(e + f*x)^2)/(cosh(c + d*x)*(a + b*sinh(c + d*x))),x)
 

Output:

int((tanh(c + d*x)*(e + f*x)^2)/(cosh(c + d*x)*(a + b*sinh(c + d*x))), x)
 

Reduce [F]

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

int((f*x+e)^2*sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)
 

Output:

( - 2*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqr 
t(a**2 + b**2))*a*b**2*d**2*e**2*i - 2*sqrt(a**2 + b**2)*atan((e**(c + d*x 
)*b*i + a*i)/sqrt(a**2 + b**2))*a*b**2*d**2*e**2*i - 8*e**(5*c + 2*d*x)*in 
t((e**(3*d*x)*x**2)/(e**(6*c + 6*d*x)*b + 2*e**(5*c + 5*d*x)*a + e**(4*c + 
 4*d*x)*b + 4*e**(3*c + 3*d*x)*a - e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - 
 b),x)*a**5*d**3*f**2 - 16*e**(5*c + 2*d*x)*int((e**(3*d*x)*x**2)/(e**(6*c 
 + 6*d*x)*b + 2*e**(5*c + 5*d*x)*a + e**(4*c + 4*d*x)*b + 4*e**(3*c + 3*d* 
x)*a - e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b),x)*a**3*b**2*d**3*f**2 - 
 8*e**(5*c + 2*d*x)*int((e**(3*d*x)*x**2)/(e**(6*c + 6*d*x)*b + 2*e**(5*c 
+ 5*d*x)*a + e**(4*c + 4*d*x)*b + 4*e**(3*c + 3*d*x)*a - e**(2*c + 2*d*x)* 
b + 2*e**(c + d*x)*a - b),x)*a*b**4*d**3*f**2 - 16*e**(5*c + 2*d*x)*int((e 
**(3*d*x)*x)/(e**(6*c + 6*d*x)*b + 2*e**(5*c + 5*d*x)*a + e**(4*c + 4*d*x) 
*b + 4*e**(3*c + 3*d*x)*a - e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b),x)* 
a**5*d**3*e*f - 32*e**(5*c + 2*d*x)*int((e**(3*d*x)*x)/(e**(6*c + 6*d*x)*b 
 + 2*e**(5*c + 5*d*x)*a + e**(4*c + 4*d*x)*b + 4*e**(3*c + 3*d*x)*a - e**( 
2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b),x)*a**3*b**2*d**3*e*f - 16*e**(5*c 
+ 2*d*x)*int((e**(3*d*x)*x)/(e**(6*c + 6*d*x)*b + 2*e**(5*c + 5*d*x)*a + e 
**(4*c + 4*d*x)*b + 4*e**(3*c + 3*d*x)*a - e**(2*c + 2*d*x)*b + 2*e**(c + 
d*x)*a - b),x)*a*b**4*d**3*e*f + 4*e**(4*c + 2*d*x)*int((e**(2*d*x)*x)/(e* 
*(6*c + 6*d*x)*b + 2*e**(5*c + 5*d*x)*a + e**(4*c + 4*d*x)*b + 4*e**(3*...