3.154 \(\int \sqrt{a \sinh ^4(x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{2} \coth (x) \sqrt{a \sinh ^4(x)}-\frac{1}{2} x \text{csch}^2(x) \sqrt{a \sinh ^4(x)} \]

[Out]

(Coth[x]*Sqrt[a*Sinh[x]^4])/2 - (x*Csch[x]^2*Sqrt[a*Sinh[x]^4])/2

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Rubi [A]  time = 0.0149813, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3207, 2635, 8} \[ \frac{1}{2} \coth (x) \sqrt{a \sinh ^4(x)}-\frac{1}{2} x \text{csch}^2(x) \sqrt{a \sinh ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Sinh[x]^4],x]

[Out]

(Coth[x]*Sqrt[a*Sinh[x]^4])/2 - (x*Csch[x]^2*Sqrt[a*Sinh[x]^4])/2

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sqrt{a \sinh ^4(x)} \, dx &=\left (\text{csch}^2(x) \sqrt{a \sinh ^4(x)}\right ) \int \sinh ^2(x) \, dx\\ &=\frac{1}{2} \coth (x) \sqrt{a \sinh ^4(x)}-\frac{1}{2} \left (\text{csch}^2(x) \sqrt{a \sinh ^4(x)}\right ) \int 1 \, dx\\ &=\frac{1}{2} \coth (x) \sqrt{a \sinh ^4(x)}-\frac{1}{2} x \text{csch}^2(x) \sqrt{a \sinh ^4(x)}\\ \end{align*}

Mathematica [A]  time = 0.0380094, size = 24, normalized size = 0.67 \[ \frac{1}{2} \sqrt{a \sinh ^4(x)} \left (\coth (x)-x \text{csch}^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Sinh[x]^4],x]

[Out]

((Coth[x] - x*Csch[x]^2)*Sqrt[a*Sinh[x]^4])/2

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Maple [B]  time = 0.095, size = 90, normalized size = 2.5 \begin{align*}{\frac{\sqrt{8} \left ( -1+\cosh \left ( 2\,x \right ) \right ) \sqrt{2}}{16\,\sinh \left ( 2\,x \right ) }\sqrt{a \left ( -1+\cosh \left ( 2\,x \right ) \right ) \left ( \cosh \left ( 2\,x \right ) +1 \right ) } \left ( \sqrt{a \left ( \sinh \left ( 2\,x \right ) \right ) ^{2}}\sqrt{a}-\ln \left ( \sqrt{a}\cosh \left ( 2\,x \right ) +\sqrt{a \left ( \sinh \left ( 2\,x \right ) \right ) ^{2}} \right ) a \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a \left ( -1+\cosh \left ( 2\,x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^4)^(1/2),x)

[Out]

1/16*8^(1/2)*(-1+cosh(2*x))*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^(1/2)*2^(1/2)*((a*sinh(2*x)^2)^(1/2)*a^(1/2)-ln(a
^(1/2)*cosh(2*x)+(a*sinh(2*x)^2)^(1/2))*a)/a^(1/2)/sinh(2*x)/(a*(-1+cosh(2*x))^2)^(1/2)

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Maxima [A]  time = 1.87253, size = 36, normalized size = 1. \begin{align*} -\frac{1}{8} \,{\left (\sqrt{a} e^{\left (-4 \, x\right )} - \sqrt{a}\right )} e^{\left (2 \, x\right )} - \frac{1}{2} \, \sqrt{a} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(1/2),x, algorithm="maxima")

[Out]

-1/8*(sqrt(a)*e^(-4*x) - sqrt(a))*e^(2*x) - 1/2*sqrt(a)*x

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Fricas [B]  time = 1.7905, size = 551, normalized size = 15.31 \begin{align*} \frac{{\left (4 \, \cosh \left (x\right ) e^{\left (2 \, x\right )} \sinh \left (x\right )^{3} + e^{\left (2 \, x\right )} \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 2 \, x\right )} e^{\left (2 \, x\right )} \sinh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - 2 \, x \cosh \left (x\right )\right )} e^{\left (2 \, x\right )} \sinh \left (x\right ) +{\left (\cosh \left (x\right )^{4} - 4 \, x \cosh \left (x\right )^{2} - 1\right )} e^{\left (2 \, x\right )}\right )} \sqrt{a e^{\left (8 \, x\right )} - 4 \, a e^{\left (6 \, x\right )} + 6 \, a e^{\left (4 \, x\right )} - 4 \, a e^{\left (2 \, x\right )} + a} e^{\left (-2 \, x\right )}}{8 \,{\left (\cosh \left (x\right )^{2} e^{\left (4 \, x\right )} - 2 \, \cosh \left (x\right )^{2} e^{\left (2 \, x\right )} +{\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) e^{\left (4 \, x\right )} - 2 \, \cosh \left (x\right ) e^{\left (2 \, x\right )} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(1/2),x, algorithm="fricas")

[Out]

1/8*(4*cosh(x)*e^(2*x)*sinh(x)^3 + e^(2*x)*sinh(x)^4 + 2*(3*cosh(x)^2 - 2*x)*e^(2*x)*sinh(x)^2 + 4*(cosh(x)^3
- 2*x*cosh(x))*e^(2*x)*sinh(x) + (cosh(x)^4 - 4*x*cosh(x)^2 - 1)*e^(2*x))*sqrt(a*e^(8*x) - 4*a*e^(6*x) + 6*a*e
^(4*x) - 4*a*e^(2*x) + a)*e^(-2*x)/(cosh(x)^2*e^(4*x) - 2*cosh(x)^2*e^(2*x) + (e^(4*x) - 2*e^(2*x) + 1)*sinh(x
)^2 + cosh(x)^2 + 2*(cosh(x)*e^(4*x) - 2*cosh(x)*e^(2*x) + cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sinh ^{4}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)**4)**(1/2),x)

[Out]

Integral(sqrt(a*sinh(x)**4), x)

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Giac [A]  time = 1.22562, size = 35, normalized size = 0.97 \begin{align*} \frac{1}{8} \,{\left ({\left (2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )} - 4 \, x + e^{\left (2 \, x\right )}\right )} \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(1/2),x, algorithm="giac")

[Out]

1/8*((2*e^(2*x) - 1)*e^(-2*x) - 4*x + e^(2*x))*sqrt(a)