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

Optimal. Leaf size=421 \[ \frac {2 (e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {i f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac {i a^2 f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {i f \text {PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}-\frac {i a^2 f \text {PolyLog}\left (2,i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {a f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {a f \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2} \]

[Out]

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+exp(2*d*x+2*c))
/(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-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+1/2*a*f*polylog(2,-exp(2*d*x+
2*c))/(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

________________________________________________________________________________________

Rubi [A]
time = 0.44, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5686, 4265, 2317, 2438, 5692, 5680, 2221, 6874, 3799} \begin {gather*} -\frac {2 a^2 (e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{b d \left (a^2+b^2\right )}+\frac {i a^2 f \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac {i a^2 f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac {a f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}-\frac {a (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}-\frac {a (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}+\frac {a (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac {2 (e+f x) \text {ArcTan}\left (e^{c+d x}\right )}{b d}-\frac {i f \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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]
- Dist[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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 3799

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] + Dist[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 4265

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] + (-Dist[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] + Dist[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 5680

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(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 5686

Int[(((e_.) + (f_.)*(x_))^(m_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]*Tanh[c + d*x]^(n - 1), x], x] - Dist[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 5692

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

Rule 6874

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\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}\\ &=\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {a \int (e+f x) \text {sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{b \left (a^2+b^2\right )}-\frac {(a b) \int \frac {(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}-\frac {(i f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{b d}+\frac {(i f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{b d}\\ &=\frac {a (e+f x)^2}{2 \left (a^2+b^2\right ) f}+\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {a \int (a (e+f x) \text {sech}(c+d x)-b (e+f x) \tanh (c+d x)) \, dx}{b \left (a^2+b^2\right )}-\frac {(a b) \int \frac {e^{c+d x} (e+f x)}{a-\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}-\frac {(a b) \int \frac {e^{c+d x} (e+f x)}{a+\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}-\frac {(i f) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^2}+\frac {(i f) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^2}\\ &=\frac {a (e+f x)^2}{2 \left (a^2+b^2\right ) f}+\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b 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 {i f \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}+\frac {a \int (e+f x) \tanh (c+d x) \, dx}{a^2+b^2}-\frac {a^2 \int (e+f x) \text {sech}(c+d x) \, dx}{b \left (a^2+b^2\right )}+\frac {(a f) \int \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac {(a f) \int \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}\\ &=\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \tan ^{-1}\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 {i f \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}+\frac {(2 a) \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}+\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a-\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {(a f) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {\left (i a^2 f\right ) \int \log \left (1-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}-\frac {\left (i a^2 f\right ) \int \log \left (1+i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}\\ &=\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \tan ^{-1}\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 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {\left (i a^2 f\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {\left (i a^2 f\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {(a f) \int \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d}\\ &=\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \tan ^{-1}\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 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i a^2 f \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {i a^2 f \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {(a f) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}\\ &=\frac {2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x) \tan ^{-1}\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 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {i a^2 f \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {i f \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {i a^2 f \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {a f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.98, size = 438, normalized size = 1.04 \begin {gather*} \frac {-2 a c d e+2 a c^2 f-2 a d^2 e x+2 a c d f x+4 b d e \text {ArcTan}(\cosh (c+d x)+\sinh (c+d x))+4 b d f x \text {ArcTan}(\cosh (c+d x)+\sinh (c+d x))-2 a c f \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-2 a d f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-2 a c f \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 a d f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 a d e \log (a+b \sinh (c+d x))+2 a c f \log (a+b \sinh (c+d x))+2 a d e \log (1+\cosh (2 (c+d x))+\sinh (2 (c+d x)))+2 a d f x \log (1+\cosh (2 (c+d x))+\sinh (2 (c+d x)))-2 a f \text {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-2 a f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 i b f \text {PolyLog}(2,-i (\cosh (c+d x)+\sinh (c+d x)))+2 i b f \text {PolyLog}(2,i (\cosh (c+d x)+\sinh (c+d x)))+a f \text {PolyLog}(2,-\cosh (2 (c+d x))-\sinh (2 (c+d x)))}{2 \left (a^2+b^2\right ) d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*c*d*e + 2*a*c^2*f - 2*a*d^2*e*x + 2*a*c*d*f*x + 4*b*d*e*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] + 4*b*d*f*
x*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] - 2*a*c*f*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*a*d*f*x*L
og[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*a*c*f*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 2*a*d
*f*x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 2*a*d*e*Log[a + b*Sinh[c + d*x]] + 2*a*c*f*Log[a + b*Sin
h[c + d*x]] + 2*a*d*e*Log[1 + Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]] + 2*a*d*f*x*Log[1 + Cosh[2*(c + d*x)] + S
inh[2*(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]))] - (2*I)*b*f*PolyLog[2, (-I)*(Cosh[c + d*x] + Sinh[c + d*x])] + (2*I)*b*f*PolyLog[2
, I*(Cosh[c + d*x] + Sinh[c + d*x])] + a*f*PolyLog[2, -Cosh[2*(c + d*x)] - Sinh[2*(c + d*x)]])/(2*(a^2 + b^2)*
d^2)

________________________________________________________________________________________

Maple [B] 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 = 3.09, size = 1287, normalized size = 3.06

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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))*e + 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)

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 631, normalized size = 1.50 \begin {gather*} -\frac {a f {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + a f {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) - {\left (a f + i \, b f\right )} {\rm Li}_2\left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right )\right ) - {\left (a f - i \, b f\right )} {\rm Li}_2\left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right )\right ) - {\left (a c f - a d \cosh \left (1\right ) - a d \sinh \left (1\right )\right )} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - {\left (a c f - a d \cosh \left (1\right ) - a d \sinh \left (1\right )\right )} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + {\left (a d f x + a c f\right )} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + {\left (a d f x + a c f\right )} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + {\left (a c f + i \, b c f - a d \cosh \left (1\right ) - i \, b d \cosh \left (1\right ) - a d \sinh \left (1\right ) - i \, b d \sinh \left (1\right )\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + i\right ) + {\left (a c f - i \, b c f - a d \cosh \left (1\right ) + i \, b d \cosh \left (1\right ) - a d \sinh \left (1\right ) + i \, b d \sinh \left (1\right )\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - i\right ) - {\left (a d f x - i \, b d f x + a c f - i \, b c f\right )} \log \left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right ) + 1\right ) - {\left (a d f x + i \, b d f x + a c f + i \, b c f\right )} \log \left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right ) + 1\right )}{{\left (a^{2} + b^{2}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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*c*f - a*d*cosh(1) - a*d*sinh(1))*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*s
qrt((a^2 + b^2)/b^2) + 2*a) - (a*c*f - a*d*cosh(1) - a*d*sinh(1))*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*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*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*c*f + I*b*c*f - a*d*cosh(1) - I*b*d*
cosh(1) - a*d*sinh(1) - I*b*d*sinh(1))*log(cosh(d*x + c) + sinh(d*x + c) + I) + (a*c*f - I*b*c*f - a*d*cosh(1)
+ I*b*d*cosh(1) - a*d*sinh(1) + I*b*d*sinh(1))*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right ) \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________