3.153 \(\int (a \sinh ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=78 \[ -\frac {5}{24} a \sinh (x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {1}{6} a \sinh ^3(x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {5}{16} a \coth (x) \sqrt {a \sinh ^4(x)}-\frac {5}{16} a x \text {csch}^2(x) \sqrt {a \sinh ^4(x)} \]

[Out]

5/16*a*coth(x)*(a*sinh(x)^4)^(1/2)-5/16*a*x*csch(x)^2*(a*sinh(x)^4)^(1/2)-5/24*a*cosh(x)*sinh(x)*(a*sinh(x)^4)
^(1/2)+1/6*a*cosh(x)*sinh(x)^3*(a*sinh(x)^4)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ \frac {1}{6} a \sinh ^3(x) \cosh (x) \sqrt {a \sinh ^4(x)}-\frac {5}{24} a \sinh (x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {5}{16} a \coth (x) \sqrt {a \sinh ^4(x)}-\frac {5}{16} a x \text {csch}^2(x) \sqrt {a \sinh ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^4)^(3/2),x]

[Out]

(5*a*Coth[x]*Sqrt[a*Sinh[x]^4])/16 - (5*a*x*Csch[x]^2*Sqrt[a*Sinh[x]^4])/16 - (5*a*Cosh[x]*Sinh[x]*Sqrt[a*Sinh
[x]^4])/24 + (a*Cosh[x]*Sinh[x]^3*Sqrt[a*Sinh[x]^4])/6

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 \left (a \sinh ^4(x)\right )^{3/2} \, dx &=\left (a \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^6(x) \, dx\\ &=\frac {1}{6} a \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {1}{6} \left (5 a \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^4(x) \, dx\\ &=-\frac {5}{24} a \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {1}{6} a \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}+\frac {1}{8} \left (5 a \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^2(x) \, dx\\ &=\frac {5}{16} a \coth (x) \sqrt {a \sinh ^4(x)}-\frac {5}{24} a \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {1}{6} a \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {1}{16} \left (5 a \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int 1 \, dx\\ &=\frac {5}{16} a \coth (x) \sqrt {a \sinh ^4(x)}-\frac {5}{16} a x \text {csch}^2(x) \sqrt {a \sinh ^4(x)}-\frac {5}{24} a \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {1}{6} a \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 38, normalized size = 0.49 \[ \frac {1}{192} (-60 x+45 \sinh (2 x)-9 \sinh (4 x)+\sinh (6 x)) \text {csch}^6(x) \left (a \sinh ^4(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^4)^(3/2),x]

[Out]

(Csch[x]^6*(a*Sinh[x]^4)^(3/2)*(-60*x + 45*Sinh[2*x] - 9*Sinh[4*x] + Sinh[6*x]))/192

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fricas [B]  time = 0.64, size = 659, normalized size = 8.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/384*(12*a*cosh(x)*e^(2*x)*sinh(x)^11 + a*e^(2*x)*sinh(x)^12 + 3*(22*a*cosh(x)^2 - 3*a)*e^(2*x)*sinh(x)^10 +
10*(22*a*cosh(x)^3 - 9*a*cosh(x))*e^(2*x)*sinh(x)^9 + 45*(11*a*cosh(x)^4 - 9*a*cosh(x)^2 + a)*e^(2*x)*sinh(x)^
8 + 72*(11*a*cosh(x)^5 - 15*a*cosh(x)^3 + 5*a*cosh(x))*e^(2*x)*sinh(x)^7 + 6*(154*a*cosh(x)^6 - 315*a*cosh(x)^
4 + 210*a*cosh(x)^2 - 20*a*x)*e^(2*x)*sinh(x)^6 + 36*(22*a*cosh(x)^7 - 63*a*cosh(x)^5 + 70*a*cosh(x)^3 - 20*a*
x*cosh(x))*e^(2*x)*sinh(x)^5 + 45*(11*a*cosh(x)^8 - 42*a*cosh(x)^6 + 70*a*cosh(x)^4 - 40*a*x*cosh(x)^2 - a)*e^
(2*x)*sinh(x)^4 + 20*(11*a*cosh(x)^9 - 54*a*cosh(x)^7 + 126*a*cosh(x)^5 - 120*a*x*cosh(x)^3 - 9*a*cosh(x))*e^(
2*x)*sinh(x)^3 + 3*(22*a*cosh(x)^10 - 135*a*cosh(x)^8 + 420*a*cosh(x)^6 - 600*a*x*cosh(x)^4 - 90*a*cosh(x)^2 +
3*a)*e^(2*x)*sinh(x)^2 + 6*(2*a*cosh(x)^11 - 15*a*cosh(x)^9 + 60*a*cosh(x)^7 - 120*a*x*cosh(x)^5 - 30*a*cosh(
x)^3 + 3*a*cosh(x))*e^(2*x)*sinh(x) + (a*cosh(x)^12 - 9*a*cosh(x)^10 + 45*a*cosh(x)^8 - 120*a*x*cosh(x)^6 - 45
*a*cosh(x)^4 + 9*a*cosh(x)^2 - a)*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)^6*e^(4*x) - 2*cosh(x)^6*e^(2*x) + (e^(4*x) - 2*e^(2*x) + 1)*sinh(x)^6 + cosh(x)^6 + 6*(cosh(x)*e^
(4*x) - 2*cosh(x)*e^(2*x) + cosh(x))*sinh(x)^5 + 15*(cosh(x)^2*e^(4*x) - 2*cosh(x)^2*e^(2*x) + cosh(x)^2)*sinh
(x)^4 + 20*(cosh(x)^3*e^(4*x) - 2*cosh(x)^3*e^(2*x) + cosh(x)^3)*sinh(x)^3 + 15*(cosh(x)^4*e^(4*x) - 2*cosh(x)
^4*e^(2*x) + cosh(x)^4)*sinh(x)^2 + 6*(cosh(x)^5*e^(4*x) - 2*cosh(x)^5*e^(2*x) + cosh(x)^5)*sinh(x))

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giac [A]  time = 0.17, size = 50, normalized size = 0.64 \[ \frac {1}{384} \, {\left ({\left (110 \, e^{\left (6 \, x\right )} - 45 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-6 \, x\right )} - 120 \, x + e^{\left (6 \, x\right )} - 9 \, e^{\left (4 \, x\right )} + 45 \, e^{\left (2 \, x\right )}\right )} a^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/384*((110*e^(6*x) - 45*e^(4*x) + 9*e^(2*x) - 1)*e^(-6*x) - 120*x + e^(6*x) - 9*e^(4*x) + 45*e^(2*x))*a^(3/2)

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maple [A]  time = 0.15, size = 125, normalized size = 1.60 \[ \frac {\left (-1+\cosh \left (2 x \right )\right ) \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (\cosh \left (2 x \right )+1\right )}\, \sqrt {a}\, \left (2 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}\, \left (\sinh ^{2}\left (2 x \right )\right )-9 \cosh \left (2 x \right ) \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}+24 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}-15 \ln \left (\sqrt {a}\, \cosh \left (2 x \right )+\sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\right ) a \right )}{96 \sinh \left (2 x \right ) \sqrt {\left (-1+\cosh \left (2 x \right )\right )^{2} a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^4)^(3/2),x)

[Out]

1/96*(-1+cosh(2*x))*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^(1/2)*a^(1/2)*(2*(a*sinh(2*x)^2)^(1/2)*a^(1/2)*sinh(2*x)^
2-9*cosh(2*x)*(a*sinh(2*x)^2)^(1/2)*a^(1/2)+24*(a*sinh(2*x)^2)^(1/2)*a^(1/2)-15*ln(a^(1/2)*cosh(2*x)+(a*sinh(2
*x)^2)^(1/2))*a)/sinh(2*x)/((-1+cosh(2*x))^2*a)^(1/2)

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maxima [A]  time = 0.42, size = 63, normalized size = 0.81 \[ -\frac {5}{16} \, a^{\frac {3}{2}} x - \frac {1}{384} \, {\left (9 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} - 45 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 45 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} - 9 \, a^{\frac {3}{2}} e^{\left (-10 \, x\right )} + a^{\frac {3}{2}} e^{\left (-12 \, x\right )} - a^{\frac {3}{2}}\right )} e^{\left (6 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(3/2),x, algorithm="maxima")

[Out]

-5/16*a^(3/2)*x - 1/384*(9*a^(3/2)*e^(-2*x) - 45*a^(3/2)*e^(-4*x) + 45*a^(3/2)*e^(-8*x) - 9*a^(3/2)*e^(-10*x)
+ a^(3/2)*e^(-12*x) - a^(3/2))*e^(6*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {sinh}\relax (x)}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^4)^(3/2),x)

[Out]

int((a*sinh(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sinh ^{4}{\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)**4)**(3/2),x)

[Out]

Integral((a*sinh(x)**4)**(3/2), x)

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