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

Optimal. Leaf size=33 \[ \frac{(a-a \cosh (x))^4}{4 a^5}-\frac{2 (a-a \cosh (x))^3}{3 a^4} \]

[Out]

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

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Rubi [A]  time = 0.0556596, antiderivative size = 33, 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 = {2667, 43} \[ \frac{(a-a \cosh (x))^4}{4 a^5}-\frac{2 (a-a \cosh (x))^3}{3 a^4} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^5/(a + a*Cosh[x]),x]

[Out]

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

Rule 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0203131, size = 21, normalized size = 0.64 \[ \frac{2 \sinh ^6\left (\frac{x}{2}\right ) (3 \cosh (x)+5)}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5 + 3*Cosh[x])*Sinh[x/2]^6)/(3*a)

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Maple [B]  time = 0.027, size = 87, normalized size = 2.6 \begin{align*} 32\,{\frac{1}{a} \left ({\frac{1}{128\, \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{4}}}-{\frac{5}{192\, \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{3}}}+{\frac{5}{256\, \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{2}}}+{\frac{5}{256\,\tanh \left ( x/2 \right ) +256}}+{\frac{1}{128\, \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{4}}}+{\frac{5}{192\, \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{3}}}+{\frac{5}{256\, \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{2}}}-{\frac{5}{256\,\tanh \left ( x/2 \right ) -256}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/a*(1/128/(tanh(1/2*x)+1)^4-5/192/(tanh(1/2*x)+1)^3+5/256/(tanh(1/2*x)+1)^2+5/256/(tanh(1/2*x)+1)+1/128/(tan
h(1/2*x)-1)^4+5/192/(tanh(1/2*x)-1)^3+5/256/(tanh(1/2*x)-1)^2-5/256/(tanh(1/2*x)-1))

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Maxima [A]  time = 1.14559, size = 81, normalized size = 2.45 \begin{align*} -\frac{{\left (8 \, e^{\left (-x\right )} + 12 \, e^{\left (-2 \, x\right )} - 72 \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )}}{192 \, a} + \frac{72 \, e^{\left (-x\right )} - 12 \, e^{\left (-2 \, x\right )} - 8 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )}}{192 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*e^(-x) + 12*e^(-2*x) - 72*e^(-3*x) - 3)*e^(4*x)/a + 1/192*(72*e^(-x) - 12*e^(-2*x) - 8*e^(-3*x) + 3*
e^(-4*x))/a

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Fricas [A]  time = 1.79853, size = 165, normalized size = 5. \begin{align*} \frac{3 \, \cosh \left (x\right )^{4} + 3 \, \sinh \left (x\right )^{4} - 8 \, \cosh \left (x\right )^{3} + 6 \,{\left (3 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) - 2\right )} \sinh \left (x\right )^{2} - 12 \, \cosh \left (x\right )^{2} + 72 \, \cosh \left (x\right )}{96 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*cosh(x)^4 + 3*sinh(x)^4 - 8*cosh(x)^3 + 6*(3*cosh(x)^2 - 4*cosh(x) - 2)*sinh(x)^2 - 12*cosh(x)^2 + 72*
cosh(x))/a

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Sympy [B]  time = 4.05777, size = 104, normalized size = 3.15 \begin{align*} - \frac{4 \tanh ^{8}{\left (\frac{x}{2} \right )}}{3 a \tanh ^{8}{\left (\frac{x}{2} \right )} - 12 a \tanh ^{6}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} - 12 a \tanh ^{2}{\left (\frac{x}{2} \right )} + 3 a} + \frac{16 \tanh ^{6}{\left (\frac{x}{2} \right )}}{3 a \tanh ^{8}{\left (\frac{x}{2} \right )} - 12 a \tanh ^{6}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} - 12 a \tanh ^{2}{\left (\frac{x}{2} \right )} + 3 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**5/(a+a*cosh(x)),x)

[Out]

-4*tanh(x/2)**8/(3*a*tanh(x/2)**8 - 12*a*tanh(x/2)**6 + 18*a*tanh(x/2)**4 - 12*a*tanh(x/2)**2 + 3*a) + 16*tanh
(x/2)**6/(3*a*tanh(x/2)**8 - 12*a*tanh(x/2)**6 + 18*a*tanh(x/2)**4 - 12*a*tanh(x/2)**2 + 3*a)

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Giac [A]  time = 1.20462, size = 69, normalized size = 2.09 \begin{align*} \frac{{\left (72 \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 8 \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} + 3 \, e^{\left (4 \, x\right )} - 8 \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} + 72 \, e^{x}}{192 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*((72*e^(3*x) - 12*e^(2*x) - 8*e^x + 3)*e^(-4*x) + 3*e^(4*x) - 8*e^(3*x) - 12*e^(2*x) + 72*e^x)/a