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

Optimal. Leaf size=80 \[ -\frac {b \cosh (x)}{a^2}-\frac {x \left (a^2-2 b^2\right )}{2 a^3}+\frac {2 b^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}+\frac {\sinh (x) \cosh (x)}{2 a} \]

[Out]

-1/2*(a^2-2*b^2)*x/a^3-b*cosh(x)/a^2+1/2*cosh(x)*sinh(x)/a+2*b^3*arctanh((a-b*tanh(1/2*x))/(a^2+b^2)^(1/2))/a^
3/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3853, 4104, 3919, 3831, 2660, 618, 206} \[ -\frac {x \left (a^2-2 b^2\right )}{2 a^3}+\frac {2 b^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}-\frac {b \cosh (x)}{a^2}+\frac {\sinh (x) \cosh (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Csch[x]),x]

[Out]

-((a^2 - 2*b^2)*x)/(2*a^3) + (2*b^3*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 + b^2]) - (b*Cos
h[x])/a^2 + (Cosh[x]*Sinh[x])/(2*a)

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 3831

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

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3919

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

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\sinh ^2(x)}{a+b \text {csch}(x)} \, dx &=\frac {\cosh (x) \sinh (x)}{2 a}-\frac {i \int \frac {\left (-2 i b-i a \text {csch}(x)-i b \text {csch}^2(x)\right ) \sinh (x)}{a+b \text {csch}(x)} \, dx}{2 a}\\ &=-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}+\frac {\int \frac {-a^2+2 b^2-a b \text {csch}(x)}{a+b \text {csch}(x)} \, dx}{2 a^2}\\ &=-\frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}-\frac {b^3 \int \frac {\text {csch}(x)}{a+b \text {csch}(x)} \, dx}{a^3}\\ &=-\frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}-\frac {b^2 \int \frac {1}{1+\frac {a \sinh (x)}{b}} \, dx}{a^3}\\ &=-\frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}-x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^3}\\ &=-\frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}+\frac {\left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (1+\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}-2 \tanh \left (\frac {x}{2}\right )\right )}{a^3}\\ &=-\frac {\left (a^2-2 b^2\right ) x}{2 a^3}+\frac {2 b^3 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}-\tanh \left (\frac {x}{2}\right )\right )}{\sqrt {a^2+b^2}}\right )}{a^3 \sqrt {a^2+b^2}}-\frac {b \cosh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 82, normalized size = 1.02 \[ \frac {-\frac {8 b^3 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-2 a^2 x+a^2 \sinh (2 x)-4 a b \cosh (x)+4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*x + 4*b^2*x - (8*b^3*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Cosh[x] + a^
2*Sinh[2*x])/(4*a^3)

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fricas [B]  time = 0.42, size = 456, normalized size = 5.70 \[ \frac {{\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x)^{4} + {\left (a^{4} + a^{2} b^{2}\right )} \sinh \relax (x)^{4} - a^{4} - a^{2} b^{2} - 4 \, {\left (a^{4} - a^{2} b^{2} - 2 \, b^{4}\right )} x \cosh \relax (x)^{2} - 4 \, {\left (a^{3} b + a b^{3}\right )} \cosh \relax (x)^{3} - 4 \, {\left (a^{3} b + a b^{3} - {\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 2 \, {\left (3 \, {\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - a^{2} b^{2} - 2 \, b^{4}\right )} x - 6 \, {\left (a^{3} b + a b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 8 \, {\left (b^{3} \cosh \relax (x)^{2} + 2 \, b^{3} \cosh \relax (x) \sinh \relax (x) + b^{3} \sinh \relax (x)^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) - 4 \, {\left (a^{3} b + a b^{3}\right )} \cosh \relax (x) - 4 \, {\left (a^{3} b + a b^{3} - {\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x)^{3} + 2 \, {\left (a^{4} - a^{2} b^{2} - 2 \, b^{4}\right )} x \cosh \relax (x) + 3 \, {\left (a^{3} b + a b^{3}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)}{8 \, {\left ({\left (a^{5} + a^{3} b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{5} + a^{3} b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{5} + a^{3} b^{2}\right )} \sinh \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a^4 + a^2*b^2)*cosh(x)^4 + (a^4 + a^2*b^2)*sinh(x)^4 - a^4 - a^2*b^2 - 4*(a^4 - a^2*b^2 - 2*b^4)*x*cosh(
x)^2 - 4*(a^3*b + a*b^3)*cosh(x)^3 - 4*(a^3*b + a*b^3 - (a^4 + a^2*b^2)*cosh(x))*sinh(x)^3 + 2*(3*(a^4 + a^2*b
^2)*cosh(x)^2 - 2*(a^4 - a^2*b^2 - 2*b^4)*x - 6*(a^3*b + a*b^3)*cosh(x))*sinh(x)^2 + 8*(b^3*cosh(x)^2 + 2*b^3*
cosh(x)*sinh(x) + b^3*sinh(x)^2)*sqrt(a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*
b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^
2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a)) - 4*(a^3*b + a*b^3)*cosh(x) - 4*(a^3*b + a*b^3 - (a^4 + a^2*
b^2)*cosh(x)^3 + 2*(a^4 - a^2*b^2 - 2*b^4)*x*cosh(x) + 3*(a^3*b + a*b^3)*cosh(x)^2)*sinh(x))/((a^5 + a^3*b^2)*
cosh(x)^2 + 2*(a^5 + a^3*b^2)*cosh(x)*sinh(x) + (a^5 + a^3*b^2)*sinh(x)^2)

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giac [A]  time = 0.13, size = 115, normalized size = 1.44 \[ -\frac {b^{3} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {{\left (4 \, a b e^{x} + a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3*log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3)
+ 1/8*(a*e^(2*x) - 4*b*e^x)/a^2 - 1/2*(a^2 - 2*b^2)*x/a^3 - 1/8*(4*a*b*e^x + a^2)*e^(-2*x)/a^3

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maple [B]  time = 0.17, size = 174, normalized size = 2.18 \[ \frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{a^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}}-\frac {2 b^{3} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/(tanh(1/2*x)-1)^2+1/2/a/(tanh(1/2*x)-1)+1/a^2/(tanh(1/2*x)-1)*b+1/2/a*ln(tanh(1/2*x)-1)-1/a^3*ln(tanh(1/
2*x)-1)*b^2-1/2/a/(tanh(1/2*x)+1)^2+1/2/a/(tanh(1/2*x)+1)-1/a^2/(tanh(1/2*x)+1)*b-1/2/a*ln(tanh(1/2*x)+1)+1/a^
3*ln(tanh(1/2*x)+1)*b^2-2*b^3/a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))

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maxima [A]  time = 0.40, size = 116, normalized size = 1.45 \[ -\frac {b^{3} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} - \frac {{\left (4 \, b e^{\left (-x\right )} - a\right )} e^{\left (2 \, x\right )}}{8 \, a^{2}} - \frac {4 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )}}{8 \, a^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3*log((a*e^(-x) - b - sqrt(a^2 + b^2))/(a*e^(-x) - b + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3) - 1/8*(4*b*e
^(-x) - a)*e^(2*x)/a^2 - 1/8*(4*b*e^(-x) + a*e^(-2*x))/a^2 - 1/2*(a^2 - 2*b^2)*x/a^3

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mupad [B]  time = 1.67, size = 157, normalized size = 1.96 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}-\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}-\frac {x\,\left (a^2-2\,b^2\right )}{2\,a^3}-\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}-\frac {2\,b^3\,\left (a-b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a^2+b^2}}\right )}{a^3\,\sqrt {a^2+b^2}}+\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}+\frac {2\,b^3\,\left (a-b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a^2+b^2}}\right )}{a^3\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b/sinh(x)),x)

[Out]

exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) - (b*exp(-x))/(2*a^2) - (x*(a^2 - 2*b^2))/(2*a^3) - (b^3
*log((2*b^3*exp(x))/a^4 - (2*b^3*(a - b*exp(x)))/(a^4*(a^2 + b^2)^(1/2))))/(a^3*(a^2 + b^2)^(1/2)) + (b^3*log(
(2*b^3*exp(x))/a^4 + (2*b^3*(a - b*exp(x)))/(a^4*(a^2 + b^2)^(1/2))))/(a^3*(a^2 + b^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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