\(\int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [350]

Optimal result

Integrand size = 24, antiderivative size = 421 \[ \int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \arctan \left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac {i a^2 f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}-\frac {i a^2 f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {a f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.68 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.24 \[ \int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {4 b d e \arctan \left (e^{c+d x}\right )-4 b c f \arctan \left (e^{c+d x}\right )+\frac {4 a^2 \left (a^2+b^2\right )^{5/2} d e \arctan \left (\frac {a+b e^{c+d x}}{\sqrt {-a^2-b^2}}\right )}{\left (-\left (a^2+b^2\right )^2\right )^{3/2}}+\frac {4 a^2 \sqrt {-\left (a^2+b^2\right )^2} d e \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+2 i b f (c+d x) \log \left (1-i e^{c+d x}\right )-2 i b f (c+d x) \log \left (1+i e^{c+d x}\right )-2 a f (c+d x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-2 a f (c+d x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 a d e \log \left (1+e^{2 (c+d x)}\right )-2 a c f \log \left (1+e^{2 (c+d x)}\right )+2 a f (c+d x) \log \left (1+e^{2 (c+d x)}\right )+2 a c f \log \left (b-2 a e^{c+d x}-b e^{2 (c+d x)}\right )-2 a d e \log \left (2 a e^{c+d x}+b \left (-1+e^{2 (c+d x)}\right )\right )-2 i b f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )+2 i b f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )-2 a f \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-2 a f \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+a f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2} \] Input:

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

Output:

(4*b*d*e*ArcTan[E^(c + d*x)] - 4*b*c*f*ArcTan[E^(c + d*x)] + (4*a^2*(a^2 + 
 b^2)^(5/2)*d*e*ArcTan[(a + b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/(-(a^2 + b^2 
)^2)^(3/2) + (4*a^2*Sqrt[-(a^2 + b^2)^2]*d*e*ArcTanh[(a + b*E^(c + d*x))/S 
qrt[a^2 + b^2]])/(-a^2 - b^2)^(3/2) + (2*I)*b*f*(c + d*x)*Log[1 - I*E^(c + 
 d*x)] - (2*I)*b*f*(c + d*x)*Log[1 + I*E^(c + d*x)] - 2*a*f*(c + d*x)*Log[ 
1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*a*f*(c + d*x)*Log[1 + (b*E^ 
(c + d*x))/(a + Sqrt[a^2 + b^2])] + 2*a*d*e*Log[1 + E^(2*(c + d*x))] - 2*a 
*c*f*Log[1 + E^(2*(c + d*x))] + 2*a*f*(c + d*x)*Log[1 + E^(2*(c + d*x))] + 
 2*a*c*f*Log[b - 2*a*E^(c + d*x) - b*E^(2*(c + d*x))] - 2*a*d*e*Log[2*a*E^ 
(c + d*x) + b*(-1 + E^(2*(c + d*x)))] - (2*I)*b*f*PolyLog[2, (-I)*E^(c + d 
*x)] + (2*I)*b*f*PolyLog[2, I*E^(c + d*x)] - 2*a*f*PolyLog[2, (b*E^(c + d* 
x))/(-a + Sqrt[a^2 + b^2])] - 2*a*f*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt 
[a^2 + b^2]))] + a*f*PolyLog[2, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^2)
 

Rubi [A] (verified)

Time = 2.03 (sec) , antiderivative size = 390, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6101, 3042, 4668, 2715, 2838, 6107, 6095, 2620, 2715, 2838, 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.

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

\(\Big \downarrow \) 6101

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4668

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 6107

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

\(\Big \downarrow \) 6095

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 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 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 3042

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

rule 4668

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 6095

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

rule 6101

Int[(((e_.) + (f_.)*(x_))^(m_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_ 
.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b   Int[(e + f*x)^m*Sech[ 
c + d*x]*Tanh[c + d*x]^(n - 1), x], x] - Simp[a/b   Int[(e + f*x)^m*Sech[c 
+ d*x]*(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]
 

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 7293

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1286 vs. \(2 (393 ) = 786\).

Time = 0.68 (sec) , antiderivative size = 1287, normalized size of antiderivative = 3.06

method	result	size
risch	\(\text {Expression too large to display}\)	\(1287\)

Input:

int((f*x+e)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

2/d^2*f/(2*a^2+2*b^2)*dilog(1+I*exp(d*x+c))*a+2/d^2*f/(2*a^2+2*b^2)*dilog( 
1-I*exp(d*x+c))*a-2/d^2*f/(2*a^2+2*b^2)*dilog((-b*exp(d*x+c)+(a^2+b^2)^(1/ 
2)-a)/(-a+(a^2+b^2)^(1/2)))*a-2/d^2*f/(2*a^2+2*b^2)*dilog((b*exp(d*x+c)+(a 
^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a+2/d*e/(2*a^2+2*b^2)*a*ln(1+exp(2*d 
*x+2*c))+4/d*e/(2*a^2+2*b^2)*b*arctan(exp(d*x+c))-2/d*e/(2*a^2+2*b^2)*a*ln 
(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)+2/d*e/(2*a^2+2*b^2)*(a^2+b^2)^(1/2)*ar 
ctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-2/d^2*c*f/(2*a^2+2*b^2)*(a 
^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-2/d*e*b^2/ 
(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^( 
1/2))-2/d*e/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a) 
/(a^2+b^2)^(1/2))*a^2-2/d*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2 
)-a)/(-a+(a^2+b^2)^(1/2)))*a*x-2/d*f/(2*a^2+2*b^2)*ln((b*exp(d*x+c)+(a^2+b 
^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a*x+2/d^2*f/(2*a^2+2*b^2)*ln(1+I*exp(d*x 
+c))*a*c+2/d^2*f/(2*a^2+2*b^2)*ln(1-I*exp(d*x+c))*a*c-2/d^2*f/(2*a^2+2*b^2 
)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a*c+2/d*f/(2*a^ 
2+2*b^2)*ln(1+I*exp(d*x+c))*a*x+2/d*f/(2*a^2+2*b^2)*ln(1-I*exp(d*x+c))*a*x 
-2/d^2*c*f/(2*a^2+2*b^2)*a*ln(1+exp(2*d*x+2*c))-4/d^2*c*f/(2*a^2+2*b^2)*b* 
arctan(exp(d*x+c))+2/d^2*c*f/(2*a^2+2*b^2)*a*ln(b*exp(2*d*x+2*c)+2*a*exp(d 
*x+c)-b)-2/d^2*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a 
^2+b^2)^(1/2)))*a*c-2*I/d^2*f/(2*a^2+2*b^2)*dilog(1+I*exp(d*x+c))*b+2*I...
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 589, normalized size of antiderivative = 1.40 \[ \int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-(a*f*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh 
(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + a*f*dilog((a*cosh(d*x + c) 
+ a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b 
^2) - b)/b + 1) - (a*f + I*b*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) - 
 (a*f - I*b*f)*dilog(-I*cosh(d*x + c) - I*sinh(d*x + c)) + (a*d*e - a*c*f) 
*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2) + 2 
*a) + (a*d*e - a*c*f)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt 
((a^2 + b^2)/b^2) + 2*a) + (a*d*f*x + a*c*f)*log(-(a*cosh(d*x + c) + a*sin 
h(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b 
)/b) + (a*d*f*x + a*c*f)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh 
(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - (a*d*e + I*b* 
d*e - a*c*f - I*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) + I) - (a*d*e - I 
*b*d*e - a*c*f + I*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - I) - (a*d*f* 
x - I*b*d*f*x + a*c*f - I*b*c*f)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1 
) - (a*d*f*x + I*b*d*f*x + a*c*f + I*b*c*f)*log(-I*cosh(d*x + c) - I*sinh( 
d*x + c) + 1))/((a^2 + b^2)*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 \mathit {atan} \left (e^{d x +c}\right ) b e +2 e^{3 c} \left (\int \frac {e^{3 d x} x}{e^{4 d x +4 c} b +2 e^{3 d x +3 c} a +2 e^{d x +c} a -b}d x \right ) a^{2} d f +2 e^{3 c} \left (\int \frac {e^{3 d x} x}{e^{4 d x +4 c} b +2 e^{3 d x +3 c} a +2 e^{d x +c} a -b}d x \right ) b^{2} d f -2 e^{c} \left (\int \frac {e^{d x} x}{e^{4 d x +4 c} b +2 e^{3 d x +3 c} a +2 e^{d x +c} a -b}d x \right ) a^{2} d f -2 e^{c} \left (\int \frac {e^{d x} x}{e^{4 d x +4 c} b +2 e^{3 d x +3 c} a +2 e^{d x +c} a -b}d x \right ) b^{2} d f +\mathrm {log}\left (e^{2 d x +2 c}+1\right ) a e -\mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) a e}{d \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

(2*atan(e**(c + d*x))*b*e + 2*e**(3*c)*int((e**(3*d*x)*x)/(e**(4*c + 4*d*x 
)*b + 2*e**(3*c + 3*d*x)*a + 2*e**(c + d*x)*a - b),x)*a**2*d*f + 2*e**(3*c 
)*int((e**(3*d*x)*x)/(e**(4*c + 4*d*x)*b + 2*e**(3*c + 3*d*x)*a + 2*e**(c 
+ d*x)*a - b),x)*b**2*d*f - 2*e**c*int((e**(d*x)*x)/(e**(4*c + 4*d*x)*b + 
2*e**(3*c + 3*d*x)*a + 2*e**(c + d*x)*a - b),x)*a**2*d*f - 2*e**c*int((e** 
(d*x)*x)/(e**(4*c + 4*d*x)*b + 2*e**(3*c + 3*d*x)*a + 2*e**(c + d*x)*a - b 
),x)*b**2*d*f + log(e**(2*c + 2*d*x) + 1)*a*e - log(e**(2*c + 2*d*x)*b + 2 
*e**(c + d*x)*a - b)*a*e)/(d*(a**2 + b**2))