3.77 \(\int \frac {\sinh ^3(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=107 \[ -\frac {b \sinh (x) \cosh (x)}{2 a^2}+\frac {b x \left (a^2-2 b^2\right )}{2 a^4}-\frac {2 b^4 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^4 \sqrt {a^2+b^2}}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}+\frac {\sinh ^2(x) \cosh (x)}{3 a} \]

[Out]

1/2*b*(a^2-2*b^2)*x/a^4-1/3*(2*a^2-3*b^2)*cosh(x)/a^3-1/2*b*cosh(x)*sinh(x)/a^2+1/3*cosh(x)*sinh(x)^2/a-2*b^4*
arctanh((a-b*tanh(1/2*x))/(a^2+b^2)^(1/2))/a^4/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {b x \left (a^2-2 b^2\right )}{2 a^4}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {2 b^4 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a^4 \sqrt {a^2+b^2}}-\frac {b \sinh (x) \cosh (x)}{2 a^2}+\frac {\sinh ^2(x) \cosh (x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Csch[x]),x]

[Out]

(b*(a^2 - 2*b^2)*x)/(2*a^4) - (2*b^4*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^4*Sqrt[a^2 + b^2]) - ((2*a
^2 - 3*b^2)*Cosh[x])/(3*a^3) - (b*Cosh[x]*Sinh[x])/(2*a^2) + (Cosh[x]*Sinh[x]^2)/(3*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 ^3(x)}{a+b \text {csch}(x)} \, dx &=\frac {\cosh (x) \sinh ^2(x)}{3 a}-\frac {i \int \frac {\left (-3 i b-2 i a \text {csch}(x)-2 i b \text {csch}^2(x)\right ) \sinh ^2(x)}{a+b \text {csch}(x)} \, dx}{3 a}\\ &=-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}+\frac {\int \frac {\left (-2 \left (2 a^2-3 b^2\right )-a b \text {csch}(x)+3 b^2 \text {csch}^2(x)\right ) \sinh (x)}{a+b \text {csch}(x)} \, dx}{6 a^2}\\ &=-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}+\frac {i \int \frac {-3 i b \left (a^2-2 b^2\right )-3 i a b^2 \text {csch}(x)}{a+b \text {csch}(x)} \, dx}{6 a^3}\\ &=\frac {b \left (a^2-2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}+\frac {b^4 \int \frac {\text {csch}(x)}{a+b \text {csch}(x)} \, dx}{a^4}\\ &=\frac {b \left (a^2-2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}+\frac {b^3 \int \frac {1}{1+\frac {a \sinh (x)}{b}} \, dx}{a^4}\\ &=\frac {b \left (a^2-2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}+\frac {\left (2 b^3\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^4}\\ &=\frac {b \left (a^2-2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}-\frac {\left (4 b^3\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^4}\\ &=\frac {b \left (a^2-2 b^2\right ) x}{2 a^4}-\frac {2 b^4 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}-\tanh \left (\frac {x}{2}\right )\right )}{\sqrt {a^2+b^2}}\right )}{a^4 \sqrt {a^2+b^2}}-\frac {\left (2 a^2-3 b^2\right ) \cosh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh (x) \sinh ^2(x)}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 104, normalized size = 0.97 \[ \frac {\left (12 a b^2-9 a^3\right ) \cosh (x)+a^3 \cosh (3 x)+3 b \left (\frac {8 b^3 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+2 a^2 x-a^2 \sinh (2 x)-4 b^2 x\right )}{12 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(a + b*Csch[x]),x]

[Out]

((-9*a^3 + 12*a*b^2)*Cosh[x] + a^3*Cosh[3*x] + 3*b*(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] - a^2*Sinh[2*x]))/(12*a^4)

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fricas [B]  time = 0.43, size = 807, normalized size = 7.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+b*csch(x)),x, algorithm="fricas")

[Out]

1/24*((a^5 + a^3*b^2)*cosh(x)^6 + (a^5 + a^3*b^2)*sinh(x)^6 - 3*(a^4*b + a^2*b^3)*cosh(x)^5 - 3*(a^4*b + a^2*b
^3 - 2*(a^5 + a^3*b^2)*cosh(x))*sinh(x)^5 + a^5 + a^3*b^2 + 12*(a^4*b - a^2*b^3 - 2*b^5)*x*cosh(x)^3 - 3*(3*a^
5 - a^3*b^2 - 4*a*b^4)*cosh(x)^4 - 3*(3*a^5 - a^3*b^2 - 4*a*b^4 - 5*(a^5 + a^3*b^2)*cosh(x)^2 + 5*(a^4*b + a^2
*b^3)*cosh(x))*sinh(x)^4 + 2*(10*(a^5 + a^3*b^2)*cosh(x)^3 - 15*(a^4*b + a^2*b^3)*cosh(x)^2 + 6*(a^4*b - a^2*b
^3 - 2*b^5)*x - 6*(3*a^5 - a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x)^3 - 3*(3*a^5 - a^3*b^2 - 4*a*b^4)*cosh(x)^2 - 3
*(3*a^5 - a^3*b^2 - 4*a*b^4 - 5*(a^5 + a^3*b^2)*cosh(x)^4 + 10*(a^4*b + a^2*b^3)*cosh(x)^3 - 12*(a^4*b - a^2*b
^3 - 2*b^5)*x*cosh(x) + 6*(3*a^5 - a^3*b^2 - 4*a*b^4)*cosh(x)^2)*sinh(x)^2 + 24*(b^4*cosh(x)^3 + 3*b^4*cosh(x)
^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4*sinh(x)^3)*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*c
osh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a)) + 3*(a^4*b + a^2*b^3)*cosh(x) + 3*(2*(a
^5 + a^3*b^2)*cosh(x)^5 + a^4*b + a^2*b^3 - 5*(a^4*b + a^2*b^3)*cosh(x)^4 + 12*(a^4*b - a^2*b^3 - 2*b^5)*x*cos
h(x)^2 - 4*(3*a^5 - a^3*b^2 - 4*a*b^4)*cosh(x)^3 - 2*(3*a^5 - a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x))/((a^6 + a^4
*b^2)*cosh(x)^3 + 3*(a^6 + a^4*b^2)*cosh(x)^2*sinh(x) + 3*(a^6 + a^4*b^2)*cosh(x)*sinh(x)^2 + (a^6 + a^4*b^2)*
sinh(x)^3)

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giac [A]  time = 0.14, size = 155, normalized size = 1.45 \[ \frac {b^{4} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} + \frac {a^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} - 9 \, a^{2} e^{x} + 12 \, b^{2} e^{x}}{24 \, a^{3}} + \frac {{\left (a^{2} b - 2 \, b^{3}\right )} x}{2 \, a^{4}} + \frac {{\left (3 \, a^{2} b e^{x} + a^{3} - 3 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+b*csch(x)),x, algorithm="giac")

[Out]

b^4*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^4) +
1/24*(a^2*e^(3*x) - 3*a*b*e^(2*x) - 9*a^2*e^x + 12*b^2*e^x)/a^3 + 1/2*(a^2*b - 2*b^3)*x/a^4 + 1/24*(3*a^2*b*e
^x + a^3 - 3*(3*a^3 - 4*a*b^2)*e^(2*x))*e^(-3*x)/a^4

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maple [B]  time = 0.16, size = 262, normalized size = 2.45 \[ -\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{4}}+\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a^{2}}-\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{4}}+\frac {2 b^{4} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{4} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a+b*csch(x)),x)

[Out]

-1/3/a/(tanh(1/2*x)-1)^3-1/2/a/(tanh(1/2*x)-1)^2-1/2/a^2/(tanh(1/2*x)-1)^2*b+1/2/a/(tanh(1/2*x)-1)-1/2/a^2/(ta
nh(1/2*x)-1)*b-1/a^3/(tanh(1/2*x)-1)*b^2-1/2*b/a^2*ln(tanh(1/2*x)-1)+b^3/a^4*ln(tanh(1/2*x)-1)+1/3/a/(tanh(1/2
*x)+1)^3-1/2/a/(tanh(1/2*x)+1)^2+1/2/a^2/(tanh(1/2*x)+1)^2*b-1/2/a/(tanh(1/2*x)+1)-1/2/a^2/(tanh(1/2*x)+1)*b+1
/a^3/(tanh(1/2*x)+1)*b^2+1/2*b/a^2*ln(tanh(1/2*x)+1)-b^3/a^4*ln(tanh(1/2*x)+1)+2*b^4/a^4/(a^2+b^2)^(1/2)*arcta
nh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))

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maxima [A]  time = 0.41, size = 157, normalized size = 1.47 \[ \frac {b^{4} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} - \frac {{\left (3 \, a b e^{\left (-x\right )} - a^{2} + 3 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )} e^{\left (-2 \, x\right )}\right )} e^{\left (3 \, x\right )}}{24 \, a^{3}} + \frac {3 \, a b e^{\left (-2 \, x\right )} + a^{2} e^{\left (-3 \, x\right )} - 3 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )} e^{\left (-x\right )}}{24 \, a^{3}} + \frac {{\left (a^{2} b - 2 \, b^{3}\right )} x}{2 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+b*csch(x)),x, algorithm="maxima")

[Out]

b^4*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^4) - 1/24*(3*a*b
*e^(-x) - a^2 + 3*(3*a^2 - 4*b^2)*e^(-2*x))*e^(3*x)/a^3 + 1/24*(3*a*b*e^(-2*x) + a^2*e^(-3*x) - 3*(3*a^2 - 4*b
^2)*e^(-x))/a^3 + 1/2*(a^2*b - 2*b^3)*x/a^4

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mupad [B]  time = 1.83, size = 199, normalized size = 1.86 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,a}+\frac {{\mathrm {e}}^{3\,x}}{24\,a}+\frac {x\,\left (a^2\,b-2\,b^3\right )}{2\,a^4}-\frac {{\mathrm {e}}^x\,\left (3\,a^2-4\,b^2\right )}{8\,a^3}+\frac {b\,{\mathrm {e}}^{-2\,x}}{8\,a^2}-\frac {b\,{\mathrm {e}}^{2\,x}}{8\,a^2}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a^2-4\,b^2\right )}{8\,a^3}-\frac {b^4\,\ln \left (-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5}-\frac {2\,b^4\,\left (a-b\,{\mathrm {e}}^x\right )}{a^5\,\sqrt {a^2+b^2}}\right )}{a^4\,\sqrt {a^2+b^2}}+\frac {b^4\,\ln \left (\frac {2\,b^4\,\left (a-b\,{\mathrm {e}}^x\right )}{a^5\,\sqrt {a^2+b^2}}-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5}\right )}{a^4\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a + b/sinh(x)),x)

[Out]

exp(-3*x)/(24*a) + exp(3*x)/(24*a) + (x*(a^2*b - 2*b^3))/(2*a^4) - (exp(x)*(3*a^2 - 4*b^2))/(8*a^3) + (b*exp(-
2*x))/(8*a^2) - (b*exp(2*x))/(8*a^2) - (exp(-x)*(3*a^2 - 4*b^2))/(8*a^3) - (b^4*log(- (2*b^4*exp(x))/a^5 - (2*
b^4*(a - b*exp(x)))/(a^5*(a^2 + b^2)^(1/2))))/(a^4*(a^2 + b^2)^(1/2)) + (b^4*log((2*b^4*(a - b*exp(x)))/(a^5*(
a^2 + b^2)^(1/2)) - (2*b^4*exp(x))/a^5))/(a^4*(a^2 + b^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3/(a+b*csch(x)),x)

[Out]

Integral(sinh(x)**3/(a + b*csch(x)), x)

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