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

Optimal. Leaf size=40 \[ \frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac{a \cosh (x)}{b^2}+\frac{\cosh ^2(x)}{2 b} \]

[Out]

-((a*Cosh[x])/b^2) + Cosh[x]^2/(2*b) + ((a^2 - b^2)*Log[a + b*Cosh[x]])/b^3

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Rubi [A]  time = 0.0689581, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2668, 697} \[ \frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac{a \cosh (x)}{b^2}+\frac{\cosh ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Cosh[x]),x]

[Out]

-((a*Cosh[x])/b^2) + Cosh[x]^2/(2*b) + ((a^2 - b^2)*Log[a + b*Cosh[x]])/b^3

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0528166, size = 40, normalized size = 1. \[ \frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{b^3}-\frac{a \cosh (x)}{b^2}+\frac{\cosh (2 x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.02, size = 283, normalized size = 7.1 \begin{align*}{\frac{{a}^{3}}{{b}^{3} \left ( a-b \right ) }\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }-{\frac{{a}^{2}}{{b}^{2} \left ( a-b \right ) }\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }-{\frac{a}{ \left ( a-b \right ) b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }+{\frac{1}{a-b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,b} \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}{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{a}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,b} \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}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)*a^3-1/b^2/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)*a^
2-1/b/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)*a+1/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)+1/2/b/(t
anh(1/2*x)+1)^2-1/b^2/(tanh(1/2*x)+1)*a-1/2/b/(tanh(1/2*x)+1)-1/b^3*ln(tanh(1/2*x)+1)*a^2+1/b*ln(tanh(1/2*x)+1
)+1/2/b/(tanh(1/2*x)-1)^2+1/b^2/(tanh(1/2*x)-1)*a+1/2/b/(tanh(1/2*x)-1)-1/b^3*ln(tanh(1/2*x)-1)*a^2+1/b*ln(tan
h(1/2*x)-1)

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Maxima [B]  time = 1.02369, size = 113, normalized size = 2.82 \begin{align*} -\frac{{\left (4 \, a e^{\left (-x\right )} - b\right )} e^{\left (2 \, x\right )}}{8 \, b^{2}} - \frac{4 \, a e^{\left (-x\right )} - b e^{\left (-2 \, x\right )}}{8 \, b^{2}} + \frac{{\left (a^{2} - b^{2}\right )} x}{b^{3}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.94193, size = 629, normalized size = 15.72 \begin{align*} \frac{b^{2} \cosh \left (x\right )^{4} + b^{2} \sinh \left (x\right )^{4} - 4 \, a b \cosh \left (x\right )^{3} - 8 \,{\left (a^{2} - b^{2}\right )} x \cosh \left (x\right )^{2} + 4 \,{\left (b^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )^{3} - 4 \, a b \cosh \left (x\right ) + 2 \,{\left (3 \, b^{2} \cosh \left (x\right )^{2} - 6 \, a b \cosh \left (x\right ) - 4 \,{\left (a^{2} - b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + b^{2} + 8 \,{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \,{\left (b^{2} \cosh \left (x\right )^{3} - 3 \, a b \cosh \left (x\right )^{2} - 4 \,{\left (a^{2} - b^{2}\right )} x \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )}{8 \,{\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \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*cosh(x)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.18362, size = 76, normalized size = 1.9 \begin{align*} \frac{b{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4 \, a{\left (e^{\left (-x\right )} + e^{x}\right )}}{8 \, b^{2}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left ({\left | b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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