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

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

[Out]

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

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Rubi [A]  time = 0.059733, 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} \[ -\frac{a^2+b^2}{b^3 (a+b \sinh (x))}-\frac{2 a \log (a+b \sinh (x))}{b^3}+\frac{\sinh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Sinh[x])^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0702072, size = 37, normalized size = 0.92 \[ -\frac{\frac{a^2+b^2}{a+b \sinh (x)}+2 a \log (a+b \sinh (x))-b \sinh (x)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.052, size = 141, normalized size = 3.5 \begin{align*} 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}}}-{\frac{1}{{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{a\tanh \left ( x/2 \right ) }{{b}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-2\,{\frac{\tanh \left ( x/2 \right ) }{ \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) a}}-2\,{\frac{a\ln \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a+b*sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.1072, size = 138, normalized size = 3.45 \begin{align*} \frac{2 \, a b e^{\left (-x\right )} + b^{2} -{\left (4 \, a^{2} + 5 \, b^{2}\right )} e^{\left (-2 \, x\right )}}{2 \,{\left (b^{4} e^{\left (-x\right )} + 2 \, a b^{3} e^{\left (-2 \, x\right )} - b^{4} e^{\left (-3 \, x\right )}\right )}} - \frac{2 \, a x}{b^{3}} - \frac{e^{\left (-x\right )}}{2 \, b^{2}} - \frac{2 \, a \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(cosh(x)^3/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.16806, size = 999, normalized size = 24.98 \begin{align*} \frac{b^{2} \cosh \left (x\right )^{4} + b^{2} \sinh \left (x\right )^{4} + 2 \,{\left (2 \, a b x + a b\right )} \cosh \left (x\right )^{3} + 2 \,{\left (2 \, a b x + 2 \, b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )^{3} + 2 \,{\left (4 \, a^{2} x - 2 \, a^{2} - 3 \, b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \, b^{2} \cosh \left (x\right )^{2} + 4 \, a^{2} x - 2 \, a^{2} - 3 \, b^{2} + 3 \,{\left (2 \, a b x + a b\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + b^{2} - 2 \,{\left (2 \, a b x + a b\right )} \cosh \left (x\right ) - 4 \,{\left (a b \cosh \left (x\right )^{3} + a b \sinh \left (x\right )^{3} + 2 \, a^{2} \cosh \left (x\right )^{2} - a b \cosh \left (x\right ) +{\left (3 \, a b \cosh \left (x\right ) + 2 \, a^{2}\right )} \sinh \left (x\right )^{2} +{\left (3 \, a b \cosh \left (x\right )^{2} + 4 \, a^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \,{\left (2 \, b^{2} \cosh \left (x\right )^{3} - 2 \, a b x + 3 \,{\left (2 \, a b x + a b\right )} \cosh \left (x\right )^{2} - a b + 2 \,{\left (4 \, a^{2} x - 2 \, a^{2} - 3 \, b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left (b^{4} \cosh \left (x\right )^{3} + b^{4} \sinh \left (x\right )^{3} + 2 \, a b^{3} \cosh \left (x\right )^{2} - b^{4} \cosh \left (x\right ) +{\left (3 \, b^{4} \cosh \left (x\right ) + 2 \, a b^{3}\right )} \sinh \left (x\right )^{2} +{\left (3 \, b^{4} \cosh \left (x\right )^{2} + 4 \, a b^{3} \cosh \left (x\right ) - b^{4}\right )} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*cosh(x)^4 + b^2*sinh(x)^4 + 2*(2*a*b*x + a*b)*cosh(x)^3 + 2*(2*a*b*x + 2*b^2*cosh(x) + a*b)*sinh(x)^3
+ 2*(4*a^2*x - 2*a^2 - 3*b^2)*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + 4*a^2*x - 2*a^2 - 3*b^2 + 3*(2*a*b*x + a*b)*co
sh(x))*sinh(x)^2 + b^2 - 2*(2*a*b*x + a*b)*cosh(x) - 4*(a*b*cosh(x)^3 + a*b*sinh(x)^3 + 2*a^2*cosh(x)^2 - a*b*
cosh(x) + (3*a*b*cosh(x) + 2*a^2)*sinh(x)^2 + (3*a*b*cosh(x)^2 + 4*a^2*cosh(x) - a*b)*sinh(x))*log(2*(b*sinh(x
) + a)/(cosh(x) - sinh(x))) + 2*(2*b^2*cosh(x)^3 - 2*a*b*x + 3*(2*a*b*x + a*b)*cosh(x)^2 - a*b + 2*(4*a^2*x -
2*a^2 - 3*b^2)*cosh(x))*sinh(x))/(b^4*cosh(x)^3 + b^4*sinh(x)^3 + 2*a*b^3*cosh(x)^2 - b^4*cosh(x) + (3*b^4*cos
h(x) + 2*a*b^3)*sinh(x)^2 + (3*b^4*cosh(x)^2 + 4*a*b^3*cosh(x) - b^4)*sinh(x))

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Sympy [A]  time = 2.02208, size = 209, normalized size = 5.22 \begin{align*} \begin{cases} \tilde{\infty } \left (2 \sinh{\left (x \right )} - \frac{\cosh ^{2}{\left (x \right )}}{\sinh{\left (x \right )}}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{2 \sinh{\left (x \right )} - \frac{\cosh ^{2}{\left (x \right )}}{\sinh{\left (x \right )}}}{b^{2}} & \text{for}\: a = 0 \\\frac{- \frac{2 \sinh ^{3}{\left (x \right )}}{3} + \sinh{\left (x \right )} \cosh ^{2}{\left (x \right )}}{a^{2}} & \text{for}\: b = 0 \\- \frac{2 a^{3} \log{\left (\frac{a}{b} + \sinh{\left (x \right )} \right )}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} - \frac{2 a^{3}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} - \frac{2 a^{2} b \log{\left (\frac{a}{b} + \sinh{\left (x \right )} \right )} \sinh{\left (x \right )}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} + \frac{a b^{2} \sinh ^{2}{\left (x \right )}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} - \frac{b^{3} \sinh ^{3}{\left (x \right )}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} + \frac{b^{3} \sinh{\left (x \right )} \cosh ^{2}{\left (x \right )}}{a^{2} b^{3} + a b^{4} \sinh{\left (x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(a+b*sinh(x))**2,x)

[Out]

Piecewise((zoo*(2*sinh(x) - cosh(x)**2/sinh(x)), Eq(a, 0) & Eq(b, 0)), ((2*sinh(x) - cosh(x)**2/sinh(x))/b**2,
Eq(a, 0)), ((-2*sinh(x)**3/3 + sinh(x)*cosh(x)**2)/a**2, Eq(b, 0)), (-2*a**3*log(a/b + sinh(x))/(a**2*b**3 +
a*b**4*sinh(x)) - 2*a**3/(a**2*b**3 + a*b**4*sinh(x)) - 2*a**2*b*log(a/b + sinh(x))*sinh(x)/(a**2*b**3 + a*b**
4*sinh(x)) + a*b**2*sinh(x)**2/(a**2*b**3 + a*b**4*sinh(x)) - b**3*sinh(x)**3/(a**2*b**3 + a*b**4*sinh(x)) + b
**3*sinh(x)*cosh(x)**2/(a**2*b**3 + a*b**4*sinh(x)), True))

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Giac [B]  time = 1.21533, size = 111, normalized size = 2.78 \begin{align*} -\frac{e^{\left (-x\right )} - e^{x}}{2 \, b^{2}} - \frac{2 \, a \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b^{3}} + \frac{2 \,{\left (a b{\left (e^{\left (-x\right )} - e^{x}\right )} - a^{2} + b^{2}\right )}}{{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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