\(\int (a \sinh ^2(x))^{5/2} \, dx\) [140]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 53 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \coth (x) \sqrt {a \sinh ^2(x)}-\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2} \]

[Out]

-4/15*a*coth(x)*(a*sinh(x)^2)^(3/2)+1/5*coth(x)*(a*sinh(x)^2)^(5/2)+8/15*a^2*coth(x)*(a*sinh(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3282, 3286, 2718} \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \coth (x) \sqrt {a \sinh ^2(x)}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2} \]

[In]

Int[(a*Sinh[x]^2)^(5/2),x]

[Out]

(8*a^2*Coth[x]*Sqrt[a*Sinh[x]^2])/15 - (4*a*Coth[x]*(a*Sinh[x]^2)^(3/2))/15 + (Coth[x]*(a*Sinh[x]^2)^(5/2))/5

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3282

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-Cot[e + f*x])*((b*Sin[e + f*x]^2)^p/(2*f*p)),
x] + Dist[b*((2*p - 1)/(2*p)), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p]
&& GtQ[p, 1]

Rule 3286

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*} \text {integral}& = \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {1}{5} (4 a) \int \left (a \sinh ^2(x)\right )^{3/2} \, dx \\ & = -\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}+\frac {1}{15} \left (8 a^2\right ) \int \sqrt {a \sinh ^2(x)} \, dx \\ & = -\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}+\frac {1}{15} \left (8 a^2 \text {csch}(x) \sqrt {a \sinh ^2(x)}\right ) \int \sinh (x) \, dx \\ & = \frac {8}{15} a^2 \coth (x) \sqrt {a \sinh ^2(x)}-\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {1}{240} a^2 (150 \cosh (x)-25 \cosh (3 x)+3 \cosh (5 x)) \text {csch}(x) \sqrt {a \sinh ^2(x)} \]

[In]

Integrate[(a*Sinh[x]^2)^(5/2),x]

[Out]

(a^2*(150*Cosh[x] - 25*Cosh[3*x] + 3*Cosh[5*x])*Csch[x]*Sqrt[a*Sinh[x]^2])/240

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60

method result size
default \(\frac {a^{3} \sinh \left (x \right ) \cosh \left (x \right ) \left (3 \sinh \left (x \right )^{4}-4 \sinh \left (x \right )^{2}+8\right )}{15 \sqrt {a \sinh \left (x \right )^{2}}}\) \(32\)
risch \(\frac {a^{2} {\mathrm e}^{6 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{160 \,{\mathrm e}^{2 x}-160}-\frac {5 a^{2} {\mathrm e}^{4 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{96 \left ({\mathrm e}^{2 x}-1\right )}+\frac {5 a^{2} {\mathrm e}^{2 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{16 \left ({\mathrm e}^{2 x}-1\right )}+\frac {5 \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}\, a^{2}}{16 \left ({\mathrm e}^{2 x}-1\right )}-\frac {5 a^{2} {\mathrm e}^{-2 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{96 \left ({\mathrm e}^{2 x}-1\right )}+\frac {a^{2} {\mathrm e}^{-4 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{160 \,{\mathrm e}^{2 x}-160}\) \(196\)

[In]

int((a*sinh(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15*a^3*sinh(x)*cosh(x)*(3*sinh(x)^4-4*sinh(x)^2+8)/(a*sinh(x)^2)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (41) = 82\).

Time = 0.28 (sec) , antiderivative size = 511, normalized size of antiderivative = 9.64 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {{\left (30 \, a^{2} \cosh \left (x\right ) e^{x} \sinh \left (x\right )^{9} + 3 \, a^{2} e^{x} \sinh \left (x\right )^{10} + 5 \, {\left (27 \, a^{2} \cosh \left (x\right )^{2} - 5 \, a^{2}\right )} e^{x} \sinh \left (x\right )^{8} + 40 \, {\left (9 \, a^{2} \cosh \left (x\right )^{3} - 5 \, a^{2} \cosh \left (x\right )\right )} e^{x} \sinh \left (x\right )^{7} + 10 \, {\left (63 \, a^{2} \cosh \left (x\right )^{4} - 70 \, a^{2} \cosh \left (x\right )^{2} + 15 \, a^{2}\right )} e^{x} \sinh \left (x\right )^{6} + 4 \, {\left (189 \, a^{2} \cosh \left (x\right )^{5} - 350 \, a^{2} \cosh \left (x\right )^{3} + 225 \, a^{2} \cosh \left (x\right )\right )} e^{x} \sinh \left (x\right )^{5} + 10 \, {\left (63 \, a^{2} \cosh \left (x\right )^{6} - 175 \, a^{2} \cosh \left (x\right )^{4} + 225 \, a^{2} \cosh \left (x\right )^{2} + 15 \, a^{2}\right )} e^{x} \sinh \left (x\right )^{4} + 40 \, {\left (9 \, a^{2} \cosh \left (x\right )^{7} - 35 \, a^{2} \cosh \left (x\right )^{5} + 75 \, a^{2} \cosh \left (x\right )^{3} + 15 \, a^{2} \cosh \left (x\right )\right )} e^{x} \sinh \left (x\right )^{3} + 5 \, {\left (27 \, a^{2} \cosh \left (x\right )^{8} - 140 \, a^{2} \cosh \left (x\right )^{6} + 450 \, a^{2} \cosh \left (x\right )^{4} + 180 \, a^{2} \cosh \left (x\right )^{2} - 5 \, a^{2}\right )} e^{x} \sinh \left (x\right )^{2} + 10 \, {\left (3 \, a^{2} \cosh \left (x\right )^{9} - 20 \, a^{2} \cosh \left (x\right )^{7} + 90 \, a^{2} \cosh \left (x\right )^{5} + 60 \, a^{2} \cosh \left (x\right )^{3} - 5 \, a^{2} \cosh \left (x\right )\right )} e^{x} \sinh \left (x\right ) + {\left (3 \, a^{2} \cosh \left (x\right )^{10} - 25 \, a^{2} \cosh \left (x\right )^{8} + 150 \, a^{2} \cosh \left (x\right )^{6} + 150 \, a^{2} \cosh \left (x\right )^{4} - 25 \, a^{2} \cosh \left (x\right )^{2} + 3 \, a^{2}\right )} e^{x}\right )} \sqrt {a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a} e^{\left (-x\right )}}{480 \, {\left (\cosh \left (x\right )^{5} e^{\left (2 \, x\right )} + {\left (e^{\left (2 \, x\right )} - 1\right )} \sinh \left (x\right )^{5} - \cosh \left (x\right )^{5} + 5 \, {\left (\cosh \left (x\right ) e^{\left (2 \, x\right )} - \cosh \left (x\right )\right )} \sinh \left (x\right )^{4} + 10 \, {\left (\cosh \left (x\right )^{2} e^{\left (2 \, x\right )} - \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{3} + 10 \, {\left (\cosh \left (x\right )^{3} e^{\left (2 \, x\right )} - \cosh \left (x\right )^{3}\right )} \sinh \left (x\right )^{2} + 5 \, {\left (\cosh \left (x\right )^{4} e^{\left (2 \, x\right )} - \cosh \left (x\right )^{4}\right )} \sinh \left (x\right )\right )}} \]

[In]

integrate((a*sinh(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/480*(30*a^2*cosh(x)*e^x*sinh(x)^9 + 3*a^2*e^x*sinh(x)^10 + 5*(27*a^2*cosh(x)^2 - 5*a^2)*e^x*sinh(x)^8 + 40*(
9*a^2*cosh(x)^3 - 5*a^2*cosh(x))*e^x*sinh(x)^7 + 10*(63*a^2*cosh(x)^4 - 70*a^2*cosh(x)^2 + 15*a^2)*e^x*sinh(x)
^6 + 4*(189*a^2*cosh(x)^5 - 350*a^2*cosh(x)^3 + 225*a^2*cosh(x))*e^x*sinh(x)^5 + 10*(63*a^2*cosh(x)^6 - 175*a^
2*cosh(x)^4 + 225*a^2*cosh(x)^2 + 15*a^2)*e^x*sinh(x)^4 + 40*(9*a^2*cosh(x)^7 - 35*a^2*cosh(x)^5 + 75*a^2*cosh
(x)^3 + 15*a^2*cosh(x))*e^x*sinh(x)^3 + 5*(27*a^2*cosh(x)^8 - 140*a^2*cosh(x)^6 + 450*a^2*cosh(x)^4 + 180*a^2*
cosh(x)^2 - 5*a^2)*e^x*sinh(x)^2 + 10*(3*a^2*cosh(x)^9 - 20*a^2*cosh(x)^7 + 90*a^2*cosh(x)^5 + 60*a^2*cosh(x)^
3 - 5*a^2*cosh(x))*e^x*sinh(x) + (3*a^2*cosh(x)^10 - 25*a^2*cosh(x)^8 + 150*a^2*cosh(x)^6 + 150*a^2*cosh(x)^4
- 25*a^2*cosh(x)^2 + 3*a^2)*e^x)*sqrt(a*e^(4*x) - 2*a*e^(2*x) + a)*e^(-x)/(cosh(x)^5*e^(2*x) + (e^(2*x) - 1)*s
inh(x)^5 - cosh(x)^5 + 5*(cosh(x)*e^(2*x) - cosh(x))*sinh(x)^4 + 10*(cosh(x)^2*e^(2*x) - cosh(x)^2)*sinh(x)^3
+ 10*(cosh(x)^3*e^(2*x) - cosh(x)^3)*sinh(x)^2 + 5*(cosh(x)^4*e^(2*x) - cosh(x)^4)*sinh(x))

Sympy [F]

\[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\int \left (a \sinh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate((a*sinh(x)**2)**(5/2),x)

[Out]

Integral((a*sinh(x)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=-\frac {1}{160} \, a^{\frac {5}{2}} e^{\left (5 \, x\right )} + \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} - \frac {5}{16} \, a^{\frac {5}{2}} e^{\left (-x\right )} + \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (-3 \, x\right )} - \frac {1}{160} \, a^{\frac {5}{2}} e^{\left (-5 \, x\right )} - \frac {5}{16} \, a^{\frac {5}{2}} e^{x} \]

[In]

integrate((a*sinh(x)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/160*a^(5/2)*e^(5*x) + 5/96*a^(5/2)*e^(3*x) - 5/16*a^(5/2)*e^(-x) + 5/96*a^(5/2)*e^(-3*x) - 1/160*a^(5/2)*e^
(-5*x) - 5/16*a^(5/2)*e^x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (41) = 82\).

Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.26 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {1}{480} \, {\left (3 \, a^{2} e^{\left (5 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, a^{2} e^{\left (3 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 150 \, a^{2} e^{x} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + {\left (150 \, a^{2} e^{\left (4 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, a^{2} e^{\left (2 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 3 \, a^{2} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} e^{\left (-5 \, x\right )}\right )} \sqrt {a} \]

[In]

integrate((a*sinh(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/480*(3*a^2*e^(5*x)*sgn(e^(3*x) - e^x) - 25*a^2*e^(3*x)*sgn(e^(3*x) - e^x) + 150*a^2*e^x*sgn(e^(3*x) - e^x) +
 (150*a^2*e^(4*x)*sgn(e^(3*x) - e^x) - 25*a^2*e^(2*x)*sgn(e^(3*x) - e^x) + 3*a^2*sgn(e^(3*x) - e^x))*e^(-5*x))
*sqrt(a)

Mupad [F(-1)]

Timed out. \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\int {\left (a\,{\mathrm {sinh}\left (x\right )}^2\right )}^{5/2} \,d x \]

[In]

int((a*sinh(x)^2)^(5/2),x)

[Out]

int((a*sinh(x)^2)^(5/2), x)