\(\int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) [310]

Optimal result

Integrand size = 28, antiderivative size = 548 \[ \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a (e+f x)^2}{\left (a^2+b^2\right ) d}-\frac {4 b f (e+f x) \arctan \left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {b^2 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {b^2 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {2 a f (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d^2}+\frac {2 i b f^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}+\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^2}-\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^2}-\frac {a f^2 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d^3}-\frac {2 b^2 f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^3}+\frac {2 b^2 f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d^3}+\frac {b (e+f x)^2 \text {sech}(c+d x)}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^2 \tanh (c+d x)}{\left (a^2+b^2\right ) d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.88 (sec) , antiderivative size = 633, normalized size of antiderivative = 1.16 \[ \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {f \left (4 a d^2 e e^{2 c} x-4 a d^2 e \left (1+e^{2 c}\right ) x+2 a d^2 e^{2 c} f x^2-2 a d^2 \left (1+e^{2 c}\right ) f x^2-4 b d e \left (1+e^{2 c}\right ) \arctan \left (e^{c+d x}\right )+2 a d e \left (1+e^{2 c}\right ) \left (2 d x-\log \left (1+e^{2 (c+d x)}\right )\right )+2 i b \left (1+e^{2 c}\right ) f \left (d x \left (-\log \left (1-i e^{c+d x}\right )+\log \left (1+i e^{c+d x}\right )\right )+\operatorname {PolyLog}\left (2,-i e^{c+d x}\right )-\operatorname {PolyLog}\left (2,i e^{c+d x}\right )\right )+a \left (1+e^{2 c}\right ) f \left (2 d x \left (d x-\log \left (1+e^{2 (c+d x)}\right )\right )-\operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )\right )\right )}{\left (a^2+b^2\right ) \left (1+e^{2 c}\right )}-\frac {b^2 \left (2 d^2 e^2 \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )+2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )+2 d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 f^2 \operatorname {PolyLog}\left (3,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-2 f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {d^2 (e+f x)^2 \text {sech}(c+d x) (b+a \text {sech}(c) \sinh (d x))}{a^2+b^2}}{d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 2.53 (sec) , antiderivative size = 469, normalized size of antiderivative = 0.86, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {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.

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

\(\Big \downarrow \) 6107

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3803

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-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 + Sq 
rt[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*T 
anh[c + d*x])/d)/(a^2 + b^2)
 

Defintions of rubi rules used

rule 25

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

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 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 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 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]

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

Input:

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

Output:

int((f*x+e)^2*sech(d*x+c)^2/(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. 3582 vs. \(2 (502) = 1004\).

Time = 0.21 (sec) , antiderivative size = 3582, normalized size of antiderivative = 6.54 \[ \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

2*a*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*b*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*a*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*(b^2*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*(b*e^(-d*x - c) + a)/((a^2 + b^2 + (a^2 + b^2)* 
e^(-2*d*x - 2*c))*d)) - 4*b*e*f*arctan(e^(d*x + c))/((a^2 + b^2)*d^2) - 2* 
(a*f^2*x^2 + 2*a*e*f*x - (b*f^2*x^2*e^c + 2*b*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*(b^2*f^ 
2*x^2*e^c + 2*b^2*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. \[ \int \frac {(e+f x)^2 \text {sech}^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((f*x+e)^2*sech(d*x+c)^2/(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)/sqrt( 
a**2 + b**2))*b**2*e**2*i + sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i 
)/sqrt(a**2 + b**2))*b**2*e**2*i + 4*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**4*d*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**2*b**2*d*f**2 + 4*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)*b**4*d*f**2 + 8*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**4*d*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**2*b**2*d*e*f + 8*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)*b**4*d*e*f + e**(2*c + 
2*d*x)*a**3*e**2 + e**(2*c + 2*d*x)*a*b**2*e**2 + e**(c + d*x)*a**2*b*e**2 
 + e**(c + d*x)*b**3*e**2 + 4*e**(3*c)*int((e**(3*d*x)*x**2)/(e**(6*c +...