3.240 \(\int \frac{(e+f x) \text{csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=261 \[ -\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2 \sqrt{a^2+b^2}}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2 \sqrt{a^2+b^2}}-\frac{f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d \sqrt{a^2+b^2}}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d \sqrt{a^2+b^2}}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d} \]

[Out]

(-2*(e + f*x)*ArcTanh[E^(c + d*x)])/(a*d) - (b*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*Sq
rt[a^2 + b^2]*d) + (b*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (f*Pol
yLog[2, -E^(c + d*x)])/(a*d^2) + (f*PolyLog[2, E^(c + d*x)])/(a*d^2) - (b*f*PolyLog[2, -((b*E^(c + d*x))/(a -
Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (b*f*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(a*Sq
rt[a^2 + b^2]*d^2)

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Rubi [A]  time = 0.459786, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {5575, 4182, 2279, 2391, 3322, 2264, 2190} \[ -\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2 \sqrt{a^2+b^2}}+\frac{b f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2 \sqrt{a^2+b^2}}-\frac{f \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{f \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d \sqrt{a^2+b^2}}+\frac{b (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d \sqrt{a^2+b^2}}-\frac{2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e + f*x)*ArcTanh[E^(c + d*x)])/(a*d) - (b*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(a*Sq
rt[a^2 + b^2]*d) + (b*(e + f*x)*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (f*Pol
yLog[2, -E^(c + d*x)])/(a*d^2) + (f*PolyLog[2, E^(c + d*x)])/(a*d^2) - (b*f*PolyLog[2, -((b*E^(c + d*x))/(a -
Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (b*f*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(a*Sq
rt[a^2 + b^2]*d^2)

Rule 5575

Int[(Csch[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Csch[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csch[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 4182

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

Rule 2279

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 2391

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 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[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 2264

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]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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]

Rubi steps

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

Mathematica [A]  time = 1.95028, size = 306, normalized size = 1.17 \[ \frac{\frac{b \left (-f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+2 d e \tanh ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{a^2+b^2}}\right )-f (c+d x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )-2 c f \tanh ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{a^2+b^2}}\right )\right )}{\sqrt{a^2+b^2}}+f \left (\text{PolyLog}\left (2,-e^{-c-d x}\right )-\text{PolyLog}\left (2,e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right )+d e \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )-c f \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{a d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.112, size = 532, normalized size = 2. \begin{align*}{\frac{e\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{da}}+2\,{\frac{eb}{da\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{dx+c}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{e\ln \left ({{\rm e}^{dx+c}}+1 \right ) }{da}}-{\frac{f{\it dilog} \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}-{\frac{bfx}{da}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{bfc}{a{d}^{2}}\ln \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{bfx}{da}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{bfc}{a{d}^{2}}\ln \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{bf}{a{d}^{2}}{\it dilog} \left ({ \left ( -b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}-a \right ) \left ( -a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{bf}{a{d}^{2}}{\it dilog} \left ({ \left ( b{{\rm e}^{dx+c}}+\sqrt{{a}^{2}+{b}^{2}}+a \right ) \left ( a+\sqrt{{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{\ln \left ({{\rm e}^{dx+c}}+1 \right ) fx}{da}}-{\frac{f{\it dilog} \left ({{\rm e}^{dx+c}}+1 \right ) }{a{d}^{2}}}-{\frac{fc\ln \left ({{\rm e}^{dx+c}}-1 \right ) }{a{d}^{2}}}-2\,{\frac{bfc}{a{d}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,b{{\rm e}^{dx+c}}+2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*e*ln(exp(d*x+c)-1)+2/d*e*b/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-1/d/a*e*l
n(exp(d*x+c)+1)-1/d^2*f/a*dilog(exp(d*x+c))-1/d*f*b/a/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a
+(a^2+b^2)^(1/2)))*x-1/d^2*f*b/a/(a^2+b^2)^(1/2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*c+
1/d*f*b/a/(a^2+b^2)^(1/2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*x+1/d^2*f*b/a/(a^2+b^2)^(1/
2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*c-1/d^2*f*b/a/(a^2+b^2)^(1/2)*dilog((-b*exp(d*x+c)
+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))+1/d^2*f*b/a/(a^2+b^2)^(1/2)*dilog((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(
a+(a^2+b^2)^(1/2)))-1/d/a*ln(exp(d*x+c)+1)*f*x-1/d^2*f/a*dilog(exp(d*x+c)+1)-1/d^2/a*f*c*ln(exp(d*x+c)-1)-2/d^
2*f*c*b/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.54721, size = 1621, normalized size = 6.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*f*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*s
qrt((a^2 + b^2)/b^2) - b)/b + 1) - b^2*f*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*c
osh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - (a^2 + b^2)*f*dilog(cosh(d*x + c) + sinh(d
*x + c)) + (a^2 + b^2)*f*dilog(-cosh(d*x + c) - sinh(d*x + c)) - (b^2*d*e - b^2*c*f)*sqrt((a^2 + b^2)/b^2)*log
(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + (b^2*d*e - b^2*c*f)*sqrt((a^2 + b^
2)/b^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + (b^2*d*f*x + b^2*c*f)*s
qrt((a^2 + b^2)/b^2)*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) - (b^2*d*f*x + b^2*c*f)*sqrt((a^2 + b^2)/b^2)*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^2 + b^2)*d*f*x + (a^2 + b^2)*d*e)*log(cos
h(d*x + c) + sinh(d*x + c) + 1) - ((a^2 + b^2)*d*e - (a^2 + b^2)*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - 1) -
((a^2 + b^2)*d*f*x + (a^2 + b^2)*c*f)*log(-cosh(d*x + c) - sinh(d*x + c) + 1))/((a^3 + a*b^2)*d^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out