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

Optimal. Leaf size=82 \[ \frac{x \left (2 a^2-b^2\right )}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^3 \sqrt{a^2+b^2}}-\frac{a \cosh (x)}{b^2}+\frac{\sinh (x) \cosh (x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.184749, antiderivative size = 82, 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 = {2793, 3023, 2735, 2660, 618, 206} \[ \frac{x \left (2 a^2-b^2\right )}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^3 \sqrt{a^2+b^2}}-\frac{a \cosh (x)}{b^2}+\frac{\sinh (x) \cosh (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Sinh[x]),x]

[Out]

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

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A]  time = 0.136388, size = 82, normalized size = 1. \[ \frac{-\frac{8 a^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+4 a^2 x-4 a b \cosh (x)-2 b^2 x+b^2 \sinh (2 x)}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 174, normalized size = 2.1 \begin{align*} -{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{1}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-2\,{\frac{{a}^{3}}{{b}^{3}\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)^3/(a+b*sinh(x)),x)

[Out]

-1/2/b/(tanh(1/2*x)+1)^2+1/2/b/(tanh(1/2*x)+1)-1/b^2/(tanh(1/2*x)+1)*a+1/b^3*ln(tanh(1/2*x)+1)*a^2-1/2/b*ln(ta
nh(1/2*x)+1)+1/2/b/(tanh(1/2*x)-1)^2+1/2/b/(tanh(1/2*x)-1)+1/b^2/(tanh(1/2*x)-1)*a-1/b^3*ln(tanh(1/2*x)-1)*a^2
+1/2/b*ln(tanh(1/2*x)-1)-2*a^3/b^3/(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)^3/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.09283, size = 1142, normalized size = 13.93 \begin{align*} \frac{{\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{4} +{\left (a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{4} - a^{2} b^{2} - b^{4} + 4 \,{\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x \cosh \left (x\right )^{2} - 4 \,{\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )^{3} - 4 \,{\left (a^{3} b + a b^{3} -{\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 2 \,{\left (3 \,{\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x - 6 \,{\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 8 \,{\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2}\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 ) - 4 \,{\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right ) - 4 \,{\left (a^{3} b + a b^{3} -{\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{3} - 2 \,{\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x \cosh \left (x\right ) + 3 \,{\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )}{8 \,{\left ({\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{2} b^{3} + b^{5}\right )} \sinh \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.4004, size = 158, normalized size = 1.93 \begin{align*} -\frac{a^{3} \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^{3}} + \frac{b e^{\left (2 \, x\right )} - 4 \, a e^{x}}{8 \, b^{2}} + \frac{{\left (2 \, a^{2} - b^{2}\right )} x}{2 \, b^{3}} - \frac{{\left (4 \, a b e^{x} + b^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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