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

Optimal. Leaf size=261 \[ -\frac {b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}-\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a d \sqrt {a^2+b^2}}-\frac {f \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {f \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x) \tanh ^{-1}\left (e^{c+d x}\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 261, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 1.82, size = 306, normalized size = 1.17 \[ \frac {\frac {b \left (2 d e \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-f (c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-2 c f \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )\right )}{\sqrt {a^2+b^2}}+d e \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+f \left (\text {Li}_2\left (-e^{-c-d x}\right )-\text {Li}_2\left (e^{-c-d x}\right )+(c+d x) \left (\log \left (1-e^{-c-d x}\right )-\log \left (e^{-c-d x}+1\right )\right )\right )-c f \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a d^2} \]

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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fricas [B]  time = 0.51, size = 649, normalized size = 2.49 \[ -\frac {b^{2} f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\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 ) - b^{2} f \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} {\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^{2} + b^{2}\right )} f {\rm Li}_2\left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (a^{2} + b^{2}\right )} f {\rm Li}_2\left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )\right ) - {\left (b^{2} d e - b^{2} c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \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 (b^{2} d e - b^{2} c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \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 (b^{2} d f x + b^{2} c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \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 (b^{2} d f x + b^{2} c f\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} \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 ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} d e\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) - {\left ({\left (a^{2} + b^{2}\right )} d e - {\left (a^{2} + b^{2}\right )} c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) - {\left ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} c f\right )} \log \left (-\cosh \left (d x + c\right ) - \sinh \left (d x + c\right ) + 1\right )}{{\left (a^{3} + a b^{2}\right )} d^{2}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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maple [B]  time = 0.18, size = 532, normalized size = 2.04 \[ -\frac {e \ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {2 e b \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d a \sqrt {a^{2}+b^{2}}}+\frac {e \ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) f x}{a d}-\frac {f \dilog \left ({\mathrm e}^{d x +c}+1\right )}{d^{2} a}-\frac {f b \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d a \sqrt {a^{2}+b^{2}}}-\frac {f b \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d a \sqrt {a^{2}+b^{2}}}+\frac {f b \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} a \sqrt {a^{2}+b^{2}}}-\frac {f b \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}+\frac {f b \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}-\frac {f \dilog \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}-\frac {2 f c b \arctanh \left (\frac {2 b \,{\mathrm e}^{d x +c}+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d^{2} a \sqrt {a^{2}+b^{2}}}-\frac {f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}} \]

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/a/d*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/a/d*e*
ln(exp(d*x+c)-1)-1/a/d*ln(exp(d*x+c)+1)*f*x-1/d^2*f/a*dilog(exp(d*x+c)+1)-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*e
xp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))-1/d^2*f*dilog(exp(d*x+c))/a-2/d^2*f*c*b/a/(a^2+b^2)^(1/2)*ar
ctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-1/a/d^2*f*c*ln(exp(d*x+c)-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e {\left (\frac {b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d}\right )} + 2 \, f \int \frac {2 \, x}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {e+f\,x}{\mathrm {sinh}\left (c+d\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right ) \operatorname {csch}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

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]

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

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