3.81 \(\int \frac{\sinh ^3(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=115 \[ \frac{\left (2 a^2+b^2\right ) \cosh (x)}{b^2 \left (a^2+b^2\right )}-\frac{2 a^2 \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^3 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \sinh (x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{2 a x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.238455, antiderivative size = 115, 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 = {2792, 3023, 2735, 2660, 618, 206} \[ \frac{\left (2 a^2+b^2\right ) \cosh (x)}{b^2 \left (a^2+b^2\right )}-\frac{2 a^2 \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^3 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \sinh (x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Sinh[x])^2,x]

[Out]

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

Rule 2792

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

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

Mathematica [A]  time = 0.35159, size = 95, normalized size = 0.83 \[ \frac{-\frac{2 a^2 \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\cosh (x) \left (\frac{a^3 b}{\left (a^2+b^2\right ) (a+b \sinh (x))}+b\right )-2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(a + b*Sinh[x])^2,x]

[Out]

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

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Maple [A]  time = 0.04, size = 213, normalized size = 1.9 \begin{align*}{\frac{1}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{a\ln \left ( \tanh \left ( x/2 \right ) +1 \right ) }{{b}^{3}}}-{\frac{1}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{a\ln \left ( \tanh \left ( x/2 \right ) -1 \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}\tanh \left ( x/2 \right ) }{b \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}-2\,{\frac{{a}^{3}}{{b}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}+4\,{\frac{{a}^{4}}{{b}^{3} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+6\,{\frac{{a}^{2}}{b \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\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))^2,x)

[Out]

1/b^2/(tanh(1/2*x)+1)-2*a/b^3*ln(tanh(1/2*x)+1)-1/b^2/(tanh(1/2*x)-1)+2*a/b^3*ln(tanh(1/2*x)-1)-2/b*a^2/(a*tan
h(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)*tanh(1/2*x)-2/b^2*a^3/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)+4/
b^3*a^4/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+6/b*a^2/(a^2+b^2)^(3/2)*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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.48933, size = 2426, normalized size = 21.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

-1/2*(a^4*b^2 + 2*a^2*b^4 + b^6 - (a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^4 - (a^4*b^2 + 2*a^2*b^4 + b^6)*sinh(x)^
4 - 2*(a^5*b + 2*a^3*b^3 + a*b^5 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*x)*cosh(x)^3 - 2*(a^5*b + 2*a^3*b^3 + a*b^5 -
2*(a^5*b + 2*a^3*b^3 + a*b^5)*x + 2*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 + 4*(a^6 + a^4*b^2 + 2*(a^
6 + 2*a^4*b^2 + a^2*b^4)*x)*cosh(x)^2 + 2*(2*a^6 + 2*a^4*b^2 - 3*(a^4*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^2 + 4*(a^
6 + 2*a^4*b^2 + a^2*b^4)*x - 3*(a^5*b + 2*a^3*b^3 + a*b^5 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*x)*cosh(x))*sinh(x)^
2 - 2*((2*a^4*b + 3*a^2*b^3)*cosh(x)^3 + (2*a^4*b + 3*a^2*b^3)*sinh(x)^3 + 2*(2*a^5 + 3*a^3*b^2)*cosh(x)^2 + (
4*a^5 + 6*a^3*b^2 + 3*(2*a^4*b + 3*a^2*b^3)*cosh(x))*sinh(x)^2 - (2*a^4*b + 3*a^2*b^3)*cosh(x) - (2*a^4*b + 3*
a^2*b^3 - 3*(2*a^4*b + 3*a^2*b^3)*cosh(x)^2 - 4*(2*a^5 + 3*a^3*b^2)*cosh(x))*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*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)) - 2*(3*
a^5*b + 4*a^3*b^3 + a*b^5 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*x)*cosh(x) - 2*(3*a^5*b + 4*a^3*b^3 + a*b^5 + 2*(a^4
*b^2 + 2*a^2*b^4 + b^6)*cosh(x)^3 + 3*(a^5*b + 2*a^3*b^3 + a*b^5 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*x)*cosh(x)^2
+ 2*(a^5*b + 2*a^3*b^3 + a*b^5)*x - 4*(a^6 + a^4*b^2 + 2*(a^6 + 2*a^4*b^2 + a^2*b^4)*x)*cosh(x))*sinh(x))/((a^
4*b^4 + 2*a^2*b^6 + b^8)*cosh(x)^3 + (a^4*b^4 + 2*a^2*b^6 + b^8)*sinh(x)^3 + 2*(a^5*b^3 + 2*a^3*b^5 + a*b^7)*c
osh(x)^2 + (2*a^5*b^3 + 4*a^3*b^5 + 2*a*b^7 + 3*(a^4*b^4 + 2*a^2*b^6 + b^8)*cosh(x))*sinh(x)^2 - (a^4*b^4 + 2*
a^2*b^6 + b^8)*cosh(x) - (a^4*b^4 + 2*a^2*b^6 + b^8 - 3*(a^4*b^4 + 2*a^2*b^6 + b^8)*cosh(x)^2 - 4*(a^5*b^3 + 2
*a^3*b^5 + a*b^7)*cosh(x))*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)**3/(a+b*sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.35389, size = 248, normalized size = 2.16 \begin{align*} \frac{{\left (2 \, a^{4} + 3 \, a^{2} b^{2}\right )} \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 )}{{\left (a^{2} b^{3} + b^{5}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \, a x}{b^{3}} + \frac{e^{x}}{2 \, b^{2}} - \frac{{\left (a^{2} b^{2} + b^{4} +{\left (4 \, a^{4} - a^{2} b^{2} - b^{4}\right )} e^{\left (2 \, x\right )} - 2 \,{\left (3 \, a^{3} b + a b^{3}\right )} e^{x}\right )} e^{\left (-x\right )}}{2 \,{\left (a^{2} + b^{2}\right )}{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

(2*a^4 + 3*a^2*b^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)))/((a^2*b
^3 + b^5)*sqrt(a^2 + b^2)) - 2*a*x/b^3 + 1/2*e^x/b^2 - 1/2*(a^2*b^2 + b^4 + (4*a^4 - a^2*b^2 - b^4)*e^(2*x) -
2*(3*a^3*b + a*b^3)*e^x)*e^(-x)/((a^2 + b^2)*(b*e^(2*x) + 2*a*e^x - b)*b^3)