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

Optimal. Leaf size=57 \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}-\frac{a x}{b^2}+\frac{\cosh (x)}{b} \]

[Out]

-((a*x)/b^2) - (2*a^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^2*Sqrt[a^2 + b^2]) + Cosh[x]/b

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Rubi [A]  time = 0.116113, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2746, 12, 2735, 2660, 618, 206} \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \sqrt{a^2+b^2}}-\frac{a x}{b^2}+\frac{\cosh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Sinh[x]),x]

[Out]

-((a*x)/b^2) - (2*a^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^2*Sqrt[a^2 + b^2]) + Cosh[x]/b

Rule 2746

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0930584, size = 61, normalized size = 1.07 \[ \frac{a \left (\frac{2 a \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-x\right )+b \cosh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(a + b*Sinh[x]),x]

[Out]

(a*(-x + (2*a*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2]) + b*Cosh[x])/b^2

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Maple [A]  time = 0.022, size = 92, normalized size = 1.6 \begin{align*}{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{2}}{{b}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b/(tanh(1/2*x)+1)-a/b^2*ln(tanh(1/2*x)+1)-1/b/(tanh(1/2*x)-1)+a/b^2*ln(tanh(1/2*x)-1)+2*a^2/b^2/(a^2+b^2)^(1
/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(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(sinh(x)^2/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14927, size = 640, normalized size = 11.23 \begin{align*} \frac{a^{2} b + b^{3} - 2 \,{\left (a^{3} + a b^{2}\right )} x \cosh \left (x\right ) +{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )^{2} +{\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a^{2} \sinh \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \,{\left ({\left (a^{3} + a b^{2}\right )} x -{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) +{\left (a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*b + b^3 - 2*(a^3 + a*b^2)*x*cosh(x) + (a^2*b + b^3)*cosh(x)^2 + (a^2*b + b^3)*sinh(x)^2 + 2*(a^2*cosh
(x) + a^2*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*c
osh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x
) + 2*(b*cosh(x) + a)*sinh(x) - b)) - 2*((a^3 + a*b^2)*x - (a^2*b + b^3)*cosh(x))*sinh(x))/((a^2*b^2 + b^4)*co
sh(x) + (a^2*b^2 + b^4)*sinh(x))

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

[Out]

Timed out

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Giac [A]  time = 1.48418, size = 116, normalized size = 2.04 \begin{align*} \frac{a^{2} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b^{2}} - \frac{a x}{b^{2}} + \frac{e^{\left (-x\right )}}{2 \, b} + \frac{e^{x}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^2) -
a*x/b^2 + 1/2*e^(-x)/b + 1/2*e^x/b