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]

1/2*coth(x)*(a*sinh(x)^4)^(1/2)-1/2*x*csch(x)^2*(a*sinh(x)^4)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 8

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

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 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
]])

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*}

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Mathematica [A]  time = 0.04, 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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fricas [B]  time = 0.54, size = 180, normalized size = 5.00 \[ \frac {{\left (4 \, \cosh \relax (x) e^{\left (2 \, x\right )} \sinh \relax (x)^{3} + e^{\left (2 \, x\right )} \sinh \relax (x)^{4} + 2 \, {\left (3 \, \cosh \relax (x)^{2} - 2 \, x\right )} e^{\left (2 \, x\right )} \sinh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} - 2 \, x \cosh \relax (x)\right )} e^{\left (2 \, x\right )} \sinh \relax (x) + {\left (\cosh \relax (x)^{4} - 4 \, x \cosh \relax (x)^{2} - 1\right )} e^{\left (2 \, x\right )}\right )} \sqrt {a e^{\left (8 \, x\right )} - 4 \, a e^{\left (6 \, x\right )} + 6 \, a e^{\left (4 \, x\right )} - 4 \, a e^{\left (2 \, x\right )} + a} e^{\left (-2 \, x\right )}}{8 \, {\left (\cosh \relax (x)^{2} e^{\left (4 \, x\right )} - 2 \, \cosh \relax (x)^{2} e^{\left (2 \, x\right )} + {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \sinh \relax (x)^{2} + \cosh \relax (x)^{2} + 2 \, {\left (\cosh \relax (x) e^{\left (4 \, x\right )} - 2 \, \cosh \relax (x) e^{\left (2 \, x\right )} + \cosh \relax (x)\right )} \sinh \relax (x)\right )}} \]

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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giac [A]  time = 0.30, size = 26, normalized size = 0.72 \[ \frac {1}{8} \, {\left ({\left (2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )} - 4 \, x + e^{\left (2 \, x\right )}\right )} \sqrt {a} \]

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)

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maple [B]  time = 0.18, size = 84, normalized size = 2.33 \[ \frac {\left (-1+\cosh \left (2 x \right )\right ) \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (\cosh \left (2 x \right )+1\right )}\, \left (\sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}-\ln \left (\sqrt {a}\, \cosh \left (2 x \right )+\sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\right ) a \right )}{4 \sqrt {a}\, \sinh \left (2 x \right ) \sqrt {\left (-1+\cosh \left (2 x \right )\right )^{2} a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-1+cosh(2*x))*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^(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)/((-1+cosh(2*x))^2*a)^(1/2)

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maxima [A]  time = 0.42, size = 27, normalized size = 0.75 \[ -\frac {1}{8} \, {\left (\sqrt {a} e^{\left (-4 \, x\right )} - \sqrt {a}\right )} e^{\left (2 \, x\right )} - \frac {1}{2} \, \sqrt {a} x \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {a\,{\mathrm {sinh}\relax (x)}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*sinh(x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sinh ^{4}{\relax (x )}}\, dx \]

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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