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+cosh(x)/b-2*a^2*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/b^2/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [B]  time = 0.58, size = 238, normalized size = 4.18 \[ \frac {a^{2} b + b^{3} - 2 \, {\left (a^{3} + a b^{2}\right )} x \cosh \relax (x) + {\left (a^{2} b + b^{3}\right )} \cosh \relax (x)^{2} + {\left (a^{2} b + b^{3}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a^{2} \sinh \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - 2 \, {\left ({\left (a^{3} + a b^{2}\right )} x - {\left (a^{2} b + b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) + {\left (a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)\right )}} \]

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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giac [A]  time = 0.19, size = 86, normalized size = 1.51 \[ \frac {a^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {a x}{b^{2}} + \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} \]

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

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maple [A]  time = 0.04, size = 92, normalized size = 1.61 \[ \frac {2 a^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}+\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))-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)

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maxima [A]  time = 0.50, size = 84, normalized size = 1.47 \[ \frac {a^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} - \frac {a x}{b^{2}} + \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.54, size = 129, normalized size = 2.26 \[ \frac {{\mathrm {e}}^{-x}}{2\,b}+\frac {{\mathrm {e}}^x}{2\,b}-\frac {a\,x}{b^2}-\frac {a^2\,\ln \left (-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}-\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}}+\frac {a^2\,\ln \left (\frac {2\,a^2\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}\right )}{b^2\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b*sinh(x)),x)

[Out]

exp(-x)/(2*b) + exp(x)/(2*b) - (a*x)/b^2 - (a^2*log(- (2*a^2*exp(x))/b^3 - (2*a^2*(b - a*exp(x)))/(b^3*(a^2 +
b^2)^(1/2))))/(b^2*(a^2 + b^2)^(1/2)) + (a^2*log((2*a^2*(b - a*exp(x)))/(b^3*(a^2 + b^2)^(1/2)) - (2*a^2*exp(x
))/b^3))/(b^2*(a^2 + b^2)^(1/2))

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

[Out]

Timed out

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