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

Optimal. Leaf size=38 \[ \frac{\left (a^2+b^2\right ) \log (a+b \sinh (x))}{b^3}-\frac{a \sinh (x)}{b^2}+\frac{\sinh ^2(x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0611395, antiderivative size = 38, 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} \[ \frac{\left (a^2+b^2\right ) \log (a+b \sinh (x))}{b^3}-\frac{a \sinh (x)}{b^2}+\frac{\sinh ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Sinh[x]),x]

[Out]

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

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

Mathematica [A]  time = 0.0348453, size = 38, normalized size = 1. \[ -\frac{-\left (a^2+b^2\right ) \log (a+b \sinh (x))+a b \sinh (x)-\frac{1}{2} b^2 \sinh ^2(x)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 185, normalized size = 4.9 \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}{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}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }+{\frac{1}{b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(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/b*ln(tanh(
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/
b*ln(tanh(1/2*x)-1)+1/b^3*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*a^2+1/b*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)

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Maxima [B]  time = 1.29438, size = 109, normalized size = 2.87 \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(cosh(x)^3/(a+b*sinh(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 = 1.80766, size = 629, normalized size = 16.55 \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 \sinh \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(cosh(x)^3/(a+b*sinh(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*sinh(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(cosh(x)**3/(a+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.15903, size = 82, normalized size = 2.16 \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(cosh(x)^3/(a+b*sinh(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