3.167 \(\int \frac{\sinh ^5(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=83 \[ \frac{\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac{a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac{\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}-\frac{a \cosh ^3(x)}{3 b^2}+\frac{\cosh ^4(x)}{4 b} \]

[Out]

-((a*(a^2 - 2*b^2)*Cosh[x])/b^4) + ((a^2 - 2*b^2)*Cosh[x]^2)/(2*b^3) - (a*Cosh[x]^3)/(3*b^2) + Cosh[x]^4/(4*b)
+ ((a^2 - b^2)^2*Log[a + b*Cosh[x]])/b^5

________________________________________________________________________________________

Rubi [A]  time = 0.107238, antiderivative size = 83, 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} \[ \frac{\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac{a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac{\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}-\frac{a \cosh ^3(x)}{3 b^2}+\frac{\cosh ^4(x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^5/(a + b*Cosh[x]),x]

[Out]

-((a*(a^2 - 2*b^2)*Cosh[x])/b^4) + ((a^2 - 2*b^2)*Cosh[x]^2)/(2*b^3) - (a*Cosh[x]^3)/(3*b^2) + Cosh[x]^4/(4*b)
+ ((a^2 - b^2)^2*Log[a + b*Cosh[x]])/b^5

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 ^5(x)}{a+b \cosh (x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{a+x} \, dx,x,b \cosh (x)\right )}{b^5}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a^3 \left (1-\frac{2 b^2}{a^2}\right )+\left (a^2-2 b^2\right ) x-a x^2+x^3+\frac{\left (a^2-b^2\right )^2}{a+x}\right ) \, dx,x,b \cosh (x)\right )}{b^5}\\ &=-\frac{a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac{\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac{a \cosh ^3(x)}{3 b^2}+\frac{\cosh ^4(x)}{4 b}+\frac{\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.107378, size = 84, normalized size = 1.01 \[ \frac{-12 b^2 \left (3 b^2-2 a^2\right ) \cosh (2 x)-24 a b \left (4 a^2-7 b^2\right ) \cosh (x)+96 \left (a^2-b^2\right )^2 \log (a+b \cosh (x))-8 a b^3 \cosh (3 x)+3 b^4 \cosh (4 x)}{96 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^5/(a + b*Cosh[x]),x]

[Out]

(-24*a*b*(4*a^2 - 7*b^2)*Cosh[x] - 12*b^2*(-2*a^2 + 3*b^2)*Cosh[2*x] - 8*a*b^3*Cosh[3*x] + 3*b^4*Cosh[4*x] + 9
6*(a^2 - b^2)^2*Log[a + b*Cosh[x]])/(96*b^5)

________________________________________________________________________________________

Maple [B]  time = 0.026, size = 599, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^5/(a+b*cosh(x)),x)

[Out]

-1/2/b/(tanh(1/2*x)+1)^3-3/8/b/(tanh(1/2*x)+1)^2+5/8/b/(tanh(1/2*x)+1)+1/2/b/(tanh(1/2*x)-1)^3-3/8/b/(tanh(1/2
*x)-1)^2-5/8/b/(tanh(1/2*x)-1)-1/b*ln(tanh(1/2*x)+1)-1/b*ln(tanh(1/2*x)-1)+2/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/b^5/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1
/2*x)^2*b-a-b)*a^5-1/b^4/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)*a^4-2/b^3/(a-b)*ln(a*tanh(1/2*x)^2-tanh
(1/2*x)^2*b-a-b)*a^3+1/4/b/(tanh(1/2*x)-1)^4-1/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)+1/4/b/(tanh(1/2*x
)+1)^4+2/b^3*ln(tanh(1/2*x)+1)*a^2+2/b^3*ln(tanh(1/2*x)-1)*a^2+1/2/b^2/(tanh(1/2*x)+1)^2*a-1/2/b^3/(tanh(1/2*x
)+1)*a^2+3/2/b^2/(tanh(1/2*x)+1)*a+1/2/b^2/(tanh(1/2*x)-1)^2*a+1/2/b^3/(tanh(1/2*x)-1)*a^2-3/2/b^2/(tanh(1/2*x
)-1)*a-1/3/b^2/(tanh(1/2*x)+1)^3*a-1/b^5*ln(tanh(1/2*x)+1)*a^4+1/2/b^3/(tanh(1/2*x)+1)^2*a^2-1/b^4/(tanh(1/2*x
)+1)*a^3+1/3/b^2/(tanh(1/2*x)-1)^3*a-1/b^5*ln(tanh(1/2*x)-1)*a^4+1/2/b^3/(tanh(1/2*x)-1)^2*a^2+1/b^4/(tanh(1/2
*x)-1)*a^3

________________________________________________________________________________________

Maxima [B]  time = 1.0499, size = 240, normalized size = 2.89 \begin{align*} -\frac{{\left (8 \, a b^{2} e^{\left (-x\right )} - 3 \, b^{3} - 12 \,{\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 24 \,{\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, b^{4}} - \frac{8 \, a b^{2} e^{\left (-3 \, x\right )} - 3 \, b^{3} e^{\left (-4 \, x\right )} + 24 \,{\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-x\right )} - 12 \,{\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )}}{192 \, b^{4}} + \frac{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x}{b^{5}} + \frac{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

-1/192*(8*a*b^2*e^(-x) - 3*b^3 - 12*(2*a^2*b - 3*b^3)*e^(-2*x) + 24*(4*a^3 - 7*a*b^2)*e^(-3*x))*e^(4*x)/b^4 -
1/192*(8*a*b^2*e^(-3*x) - 3*b^3*e^(-4*x) + 24*(4*a^3 - 7*a*b^2)*e^(-x) - 12*(2*a^2*b - 3*b^3)*e^(-2*x))/b^4 +
(a^4 - 2*a^2*b^2 + b^4)*x/b^5 + (a^4 - 2*a^2*b^2 + b^4)*log(2*a*e^(-x) + b*e^(-2*x) + b)/b^5

________________________________________________________________________________________

Fricas [B]  time = 1.93527, size = 2228, normalized size = 26.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

1/192*(3*b^4*cosh(x)^8 + 3*b^4*sinh(x)^8 - 8*a*b^3*cosh(x)^7 + 8*(3*b^4*cosh(x) - a*b^3)*sinh(x)^7 + 12*(2*a^2
*b^2 - 3*b^4)*cosh(x)^6 + 4*(21*b^4*cosh(x)^2 - 14*a*b^3*cosh(x) + 6*a^2*b^2 - 9*b^4)*sinh(x)^6 - 192*(a^4 - 2
*a^2*b^2 + b^4)*x*cosh(x)^4 - 24*(4*a^3*b - 7*a*b^3)*cosh(x)^5 + 24*(7*b^4*cosh(x)^3 - 7*a*b^3*cosh(x)^2 - 4*a
^3*b + 7*a*b^3 + 3*(2*a^2*b^2 - 3*b^4)*cosh(x))*sinh(x)^5 - 8*a*b^3*cosh(x) + 2*(105*b^4*cosh(x)^4 - 140*a*b^3
*cosh(x)^3 + 90*(2*a^2*b^2 - 3*b^4)*cosh(x)^2 - 96*(a^4 - 2*a^2*b^2 + b^4)*x - 60*(4*a^3*b - 7*a*b^3)*cosh(x))
*sinh(x)^4 + 3*b^4 - 24*(4*a^3*b - 7*a*b^3)*cosh(x)^3 + 8*(21*b^4*cosh(x)^5 - 35*a*b^3*cosh(x)^4 - 12*a^3*b +
21*a*b^3 + 30*(2*a^2*b^2 - 3*b^4)*cosh(x)^3 - 96*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x) - 30*(4*a^3*b - 7*a*b^3)*co
sh(x)^2)*sinh(x)^3 + 12*(2*a^2*b^2 - 3*b^4)*cosh(x)^2 + 12*(7*b^4*cosh(x)^6 - 14*a*b^3*cosh(x)^5 + 15*(2*a^2*b
^2 - 3*b^4)*cosh(x)^4 + 2*a^2*b^2 - 3*b^4 - 96*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^2 - 20*(4*a^3*b - 7*a*b^3)*co
sh(x)^3 - 6*(4*a^3*b - 7*a*b^3)*cosh(x))*sinh(x)^2 + 192*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + 4*(a^4 - 2*a^2*b
^2 + b^4)*cosh(x)^3*sinh(x) + 6*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2*sinh(x)^2 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x
)*sinh(x)^3 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^4)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + 8*(3*b^4*cosh(x)
^7 - 7*a*b^3*cosh(x)^6 + 9*(2*a^2*b^2 - 3*b^4)*cosh(x)^5 - 96*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^3 - 15*(4*a^3*
b - 7*a*b^3)*cosh(x)^4 - a*b^3 - 9*(4*a^3*b - 7*a*b^3)*cosh(x)^2 + 3*(2*a^2*b^2 - 3*b^4)*cosh(x))*sinh(x))/(b^
5*cosh(x)^4 + 4*b^5*cosh(x)^3*sinh(x) + 6*b^5*cosh(x)^2*sinh(x)^2 + 4*b^5*cosh(x)*sinh(x)^3 + b^5*sinh(x)^4)

________________________________________________________________________________________

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)**5/(a+b*cosh(x)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.22426, size = 167, normalized size = 2.01 \begin{align*} \frac{3 \, b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 8 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 24 \, a^{2} b{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 48 \, b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 96 \, a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )} + 192 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}}{192 \, b^{4}} + \frac{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^5/(a+b*cosh(x)),x, algorithm="giac")

[Out]

1/192*(3*b^3*(e^(-x) + e^x)^4 - 8*a*b^2*(e^(-x) + e^x)^3 + 24*a^2*b*(e^(-x) + e^x)^2 - 48*b^3*(e^(-x) + e^x)^2
- 96*a^3*(e^(-x) + e^x) + 192*a*b^2*(e^(-x) + e^x))/b^4 + (a^4 - 2*a^2*b^2 + b^4)*log(abs(b*(e^(-x) + e^x) +
2*a))/b^5