3.259 \(\int \frac{x}{a+b \sinh ^2(x)} \, dx\)

Optimal. Leaf size=215 \[ \frac{\text{PolyLog}\left (2,-\frac{b e^{2 x}}{-2 \sqrt{a} \sqrt{a-b}+2 a-b}\right )}{4 \sqrt{a} \sqrt{a-b}}-\frac{\text{PolyLog}\left (2,-\frac{b e^{2 x}}{2 \sqrt{a} \sqrt{a-b}+2 a-b}\right )}{4 \sqrt{a} \sqrt{a-b}}+\frac{x \log \left (\frac{b e^{2 x}}{-2 \sqrt{a} \sqrt{a-b}+2 a-b}+1\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{x \log \left (\frac{b e^{2 x}}{2 \sqrt{a} \sqrt{a-b}+2 a-b}+1\right )}{2 \sqrt{a} \sqrt{a-b}} \]

[Out]

(x*Log[1 + (b*E^(2*x))/(2*a - 2*Sqrt[a]*Sqrt[a - b] - b)])/(2*Sqrt[a]*Sqrt[a - b]) - (x*Log[1 + (b*E^(2*x))/(2
*a + 2*Sqrt[a]*Sqrt[a - b] - b)])/(2*Sqrt[a]*Sqrt[a - b]) + PolyLog[2, -((b*E^(2*x))/(2*a - 2*Sqrt[a]*Sqrt[a -
b] - b))]/(4*Sqrt[a]*Sqrt[a - b]) - PolyLog[2, -((b*E^(2*x))/(2*a + 2*Sqrt[a]*Sqrt[a - b] - b))]/(4*Sqrt[a]*S
qrt[a - b])

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Rubi [A]  time = 0.329364, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5629, 3320, 2264, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{b e^{2 x}}{-2 \sqrt{a} \sqrt{a-b}+2 a-b}\right )}{4 \sqrt{a} \sqrt{a-b}}-\frac{\text{PolyLog}\left (2,-\frac{b e^{2 x}}{2 \sqrt{a} \sqrt{a-b}+2 a-b}\right )}{4 \sqrt{a} \sqrt{a-b}}+\frac{x \log \left (\frac{b e^{2 x}}{-2 \sqrt{a} \sqrt{a-b}+2 a-b}+1\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{x \log \left (\frac{b e^{2 x}}{2 \sqrt{a} \sqrt{a-b}+2 a-b}+1\right )}{2 \sqrt{a} \sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sinh[x]^2),x]

[Out]

(x*Log[1 + (b*E^(2*x))/(2*a - 2*Sqrt[a]*Sqrt[a - b] - b)])/(2*Sqrt[a]*Sqrt[a - b]) - (x*Log[1 + (b*E^(2*x))/(2
*a + 2*Sqrt[a]*Sqrt[a - b] - b)])/(2*Sqrt[a]*Sqrt[a - b]) + PolyLog[2, -((b*E^(2*x))/(2*a - 2*Sqrt[a]*Sqrt[a -
b] - b))]/(4*Sqrt[a]*Sqrt[a - b]) - PolyLog[2, -((b*E^(2*x))/(2*a + 2*Sqrt[a]*Sqrt[a - b] - b))]/(4*Sqrt[a]*S
qrt[a - b])

Rule 5629

Int[(x_)^(m_.)*((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]^2)^(n_), x_Symbol] :> Dist[1/2^n, Int[x^m*(2*a - b + b*C
osh[2*c + 2*d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a - b, 0] && IGtQ[m, 0] && ILtQ[n, 0] && (EqQ[n,
-1] || (EqQ[m, 1] && EqQ[n, -2]))

Rule 3320

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

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]

Rubi steps

\begin{align*} \int \frac{x}{a+b \sinh ^2(x)} \, dx &=2 \int \frac{x}{2 a-b+b \cosh (2 x)} \, dx\\ &=4 \int \frac{e^{2 x} x}{b+2 (2 a-b) e^{2 x}+b e^{4 x}} \, dx\\ &=\frac{(2 b) \int \frac{e^{2 x} x}{-4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)+2 b e^{2 x}} \, dx}{\sqrt{a} \sqrt{a-b}}-\frac{(2 b) \int \frac{e^{2 x} x}{4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)+2 b e^{2 x}} \, dx}{\sqrt{a} \sqrt{a-b}}\\ &=\frac{x \log \left (1+\frac{b e^{2 x}}{2 a-2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{x \log \left (1+\frac{b e^{2 x}}{2 a+2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{\int \log \left (1+\frac{2 b e^{2 x}}{-4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)}\right ) \, dx}{2 \sqrt{a} \sqrt{a-b}}+\frac{\int \log \left (1+\frac{2 b e^{2 x}}{4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)}\right ) \, dx}{2 \sqrt{a} \sqrt{a-b}}\\ &=\frac{x \log \left (1+\frac{b e^{2 x}}{2 a-2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{x \log \left (1+\frac{b e^{2 x}}{2 a+2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{-4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)}\right )}{x} \, dx,x,e^{2 x}\right )}{4 \sqrt{a} \sqrt{a-b}}+\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 b x}{4 \sqrt{a} \sqrt{a-b}+2 (2 a-b)}\right )}{x} \, dx,x,e^{2 x}\right )}{4 \sqrt{a} \sqrt{a-b}}\\ &=\frac{x \log \left (1+\frac{b e^{2 x}}{2 a-2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}-\frac{x \log \left (1+\frac{b e^{2 x}}{2 a+2 \sqrt{a} \sqrt{a-b}-b}\right )}{2 \sqrt{a} \sqrt{a-b}}+\frac{\text{Li}_2\left (-\frac{b e^{2 x}}{2 a-2 \sqrt{a} \sqrt{a-b}-b}\right )}{4 \sqrt{a} \sqrt{a-b}}-\frac{\text{Li}_2\left (-\frac{b e^{2 x}}{2 a+2 \sqrt{a} \sqrt{a-b}-b}\right )}{4 \sqrt{a} \sqrt{a-b}}\\ \end{align*}

Mathematica [C]  time = 0.771916, size = 576, normalized size = 2.68 \[ -\frac{i \left (\text{PolyLog}\left (2,\frac{\left (2 i \sqrt{a (b-a)}-2 a+b\right ) \left (\sqrt{a (b-a)} \tanh (x)+i a\right )}{b \sqrt{a (b-a)} \tanh (x)-i a b}\right )-\text{PolyLog}\left (2,\frac{\left (-2 i \sqrt{a (b-a)}-2 a+b\right ) \left (\sqrt{a (b-a)} \tanh (x)+i a\right )}{b \sqrt{a (b-a)} \tanh (x)-i a b}\right )\right )-2 i \cos ^{-1}\left (1-\frac{2 a}{b}\right ) \tan ^{-1}\left (\frac{\sqrt{a b-a^2} \tanh (x)}{a}\right )-\log \left (\frac{2 a \left (\sqrt{a (b-a)}-i a+i b\right ) (\tanh (x)-1)}{b \sqrt{a (b-a)} \tanh (x)-i a b}\right ) \left (2 \tan ^{-1}\left (\frac{\sqrt{a b-a^2} \tanh (x)}{a}\right )+\cos ^{-1}\left (1-\frac{2 a}{b}\right )\right )-\log \left (\frac{2 a \left (\sqrt{a (b-a)}+i a-i b\right ) (\tanh (x)+1)}{b \sqrt{a (b-a)} \tanh (x)-i a b}\right ) \left (\cos ^{-1}\left (1-\frac{2 a}{b}\right )-2 \tan ^{-1}\left (\frac{\sqrt{a b-a^2} \tanh (x)}{a}\right )\right )+\log \left (\frac{\sqrt{2} e^{-x} \sqrt{a (b-a)}}{\sqrt{b} \sqrt{2 a+b \cosh (2 x)-b}}\right ) \left (2 \left (\tan ^{-1}\left (\frac{\sqrt{a b-a^2} \tanh (x)}{a}\right )+\tan ^{-1}\left (\frac{a \coth (x)}{\sqrt{-a (a-b)}}\right )\right )+\cos ^{-1}\left (1-\frac{2 a}{b}\right )\right )+\log \left (\frac{\sqrt{2} e^x \sqrt{a (b-a)}}{\sqrt{b} \sqrt{2 a+b \cosh (2 x)-b}}\right ) \left (\cos ^{-1}\left (1-\frac{2 a}{b}\right )-2 \left (\tan ^{-1}\left (\frac{\sqrt{a b-a^2} \tanh (x)}{a}\right )+\tan ^{-1}\left (\frac{a \coth (x)}{\sqrt{-a (a-b)}}\right )\right )\right )+4 x \tan ^{-1}\left (\frac{a \coth (x)}{\sqrt{-a (a-b)}}\right )}{4 \sqrt{a (b-a)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sinh[x]^2),x]

[Out]

-(4*x*ArcTan[(a*Coth[x])/Sqrt[-(a*(a - b))]] - (2*I)*ArcCos[1 - (2*a)/b]*ArcTan[(Sqrt[-a^2 + a*b]*Tanh[x])/a]
+ (ArcCos[1 - (2*a)/b] + 2*(ArcTan[(a*Coth[x])/Sqrt[-(a*(a - b))]] + ArcTan[(Sqrt[-a^2 + a*b]*Tanh[x])/a]))*Lo
g[(Sqrt[2]*Sqrt[a*(-a + b)])/(Sqrt[b]*E^x*Sqrt[2*a - b + b*Cosh[2*x]])] + (ArcCos[1 - (2*a)/b] - 2*(ArcTan[(a*
Coth[x])/Sqrt[-(a*(a - b))]] + ArcTan[(Sqrt[-a^2 + a*b]*Tanh[x])/a]))*Log[(Sqrt[2]*Sqrt[a*(-a + b)]*E^x)/(Sqrt
[b]*Sqrt[2*a - b + b*Cosh[2*x]])] - (ArcCos[1 - (2*a)/b] + 2*ArcTan[(Sqrt[-a^2 + a*b]*Tanh[x])/a])*Log[(2*a*((
-I)*a + I*b + Sqrt[a*(-a + b)])*(-1 + Tanh[x]))/((-I)*a*b + b*Sqrt[a*(-a + b)]*Tanh[x])] - (ArcCos[1 - (2*a)/b
] - 2*ArcTan[(Sqrt[-a^2 + a*b]*Tanh[x])/a])*Log[(2*a*(I*a - I*b + Sqrt[a*(-a + b)])*(1 + Tanh[x]))/((-I)*a*b +
b*Sqrt[a*(-a + b)]*Tanh[x])] + I*(-PolyLog[2, ((-2*a + b - (2*I)*Sqrt[a*(-a + b)])*(I*a + Sqrt[a*(-a + b)]*Ta
nh[x]))/((-I)*a*b + b*Sqrt[a*(-a + b)]*Tanh[x])] + PolyLog[2, ((-2*a + b + (2*I)*Sqrt[a*(-a + b)])*(I*a + Sqrt
[a*(-a + b)]*Tanh[x]))/((-I)*a*b + b*Sqrt[a*(-a + b)]*Tanh[x])]))/(4*Sqrt[a*(-a + b)])

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Maple [B]  time = 0.047, size = 505, normalized size = 2.4 \begin{align*}{x\ln \left ( 1-{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ) \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}+{ax\ln \left ( 1-{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}-{\frac{bx}{2}\ln \left ( 1-{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}-{{x}^{2} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}-{a{x}^{2}{\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}+{\frac{b{x}^{2}}{2}{\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}+{\frac{1}{2}{\it polylog} \left ( 2,{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ) \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}+{\frac{a}{2}{\it polylog} \left ( 2,{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}-{\frac{b}{4}{\it polylog} \left ( 2,{b{{\rm e}^{2\,x}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}} \left ( -2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}}+{\frac{x}{2}\ln \left ( 1-{b{{\rm e}^{2\,x}} \left ( 2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}}-{\frac{{x}^{2}}{2}{\frac{1}{\sqrt{a \left ( a-b \right ) }}}}+{\frac{1}{4}{\it polylog} \left ( 2,{b{{\rm e}^{2\,x}} \left ( 2\,\sqrt{a \left ( a-b \right ) }-2\,a+b \right ) ^{-1}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*sinh(x)^2),x)

[Out]

1/(-2*(a*(a-b))^(1/2)-2*a+b)*ln(1-b*exp(2*x)/(-2*(a*(a-b))^(1/2)-2*a+b))*x+1/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1/
2)-2*a+b)*ln(1-b*exp(2*x)/(-2*(a*(a-b))^(1/2)-2*a+b))*a*x-1/2/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1/2)-2*a+b)*ln(1-
b*exp(2*x)/(-2*(a*(a-b))^(1/2)-2*a+b))*b*x-1/(-2*(a*(a-b))^(1/2)-2*a+b)*x^2-1/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1
/2)-2*a+b)*a*x^2+1/2/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1/2)-2*a+b)*b*x^2+1/2/(-2*(a*(a-b))^(1/2)-2*a+b)*polylog(2
,b*exp(2*x)/(-2*(a*(a-b))^(1/2)-2*a+b))+1/2/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1/2)-2*a+b)*polylog(2,b*exp(2*x)/(-
2*(a*(a-b))^(1/2)-2*a+b))*a-1/4/(a*(a-b))^(1/2)/(-2*(a*(a-b))^(1/2)-2*a+b)*polylog(2,b*exp(2*x)/(-2*(a*(a-b))^
(1/2)-2*a+b))*b+1/2/(a*(a-b))^(1/2)*x*ln(1-b*exp(2*x)/(2*(a*(a-b))^(1/2)-2*a+b))-1/2/(a*(a-b))^(1/2)*x^2+1/4/(
a*(a-b))^(1/2)*polylog(2,b*exp(2*x)/(2*(a*(a-b))^(1/2)-2*a+b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \sinh \left (x\right )^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*sinh(x)^2),x, algorithm="maxima")

[Out]

integrate(x/(b*sinh(x)^2 + a), x)

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Fricas [B]  time = 2.25311, size = 1968, normalized size = 9.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*sinh(x)^2),x, algorithm="fricas")

[Out]

-1/2*(b*x*sqrt((a^2 - a*b)/b^2)*log((((2*a - b)*cosh(x) + (2*a - b)*sinh(x) - 2*(b*cosh(x) + b*sinh(x))*sqrt((
a^2 - a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 - a*b)/b^2) + 2*a - b)/b) + b)/b) + b*x*sqrt((a^2 - a*b)/b^2)*log(-(((2*
a - b)*cosh(x) + (2*a - b)*sinh(x) - 2*(b*cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 - a
*b)/b^2) + 2*a - b)/b) - b)/b) - b*x*sqrt((a^2 - a*b)/b^2)*log((((2*a - b)*cosh(x) + (2*a - b)*sinh(x) + 2*(b*
cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b) + b)/b) - b*x*sqrt((
a^2 - a*b)/b^2)*log(-(((2*a - b)*cosh(x) + (2*a - b)*sinh(x) + 2*(b*cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2)
)*sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b) - b)/b) + b*sqrt((a^2 - a*b)/b^2)*dilog(-(((2*a - b)*cosh(x) +
(2*a - b)*sinh(x) - 2*(b*cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 - a*b)/b^2) + 2*a -
b)/b) + b)/b + 1) + b*sqrt((a^2 - a*b)/b^2)*dilog((((2*a - b)*cosh(x) + (2*a - b)*sinh(x) - 2*(b*cosh(x) + b*
sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt(-(2*b*sqrt((a^2 - a*b)/b^2) + 2*a - b)/b) - b)/b + 1) - b*sqrt((a^2 - a*b
)/b^2)*dilog(-(((2*a - b)*cosh(x) + (2*a - b)*sinh(x) + 2*(b*cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt(
(2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b) + b)/b + 1) - b*sqrt((a^2 - a*b)/b^2)*dilog((((2*a - b)*cosh(x) + (2*
a - b)*sinh(x) + 2*(b*cosh(x) + b*sinh(x))*sqrt((a^2 - a*b)/b^2))*sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b
) - b)/b + 1))/(a^2 - a*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{a + b \sinh ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*sinh(x)**2),x)

[Out]

Integral(x/(a + b*sinh(x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b \sinh \left (x\right )^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*sinh(x)^2),x, algorithm="giac")

[Out]

integrate(x/(b*sinh(x)^2 + a), x)