3.152 \(\int (a \sinh ^4(x))^{5/2} \, dx\)

Optimal. Leaf size=132 \[ -\frac {21}{128} a^2 \sinh (x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \sinh ^7(x) \cosh (x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \sinh ^5(x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {21}{160} a^2 \sinh ^3(x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {63}{256} a^2 \coth (x) \sqrt {a \sinh ^4(x)}-\frac {63}{256} a^2 x \text {csch}^2(x) \sqrt {a \sinh ^4(x)} \]

[Out]

63/256*a^2*coth(x)*(a*sinh(x)^4)^(1/2)-63/256*a^2*x*csch(x)^2*(a*sinh(x)^4)^(1/2)-21/128*a^2*cosh(x)*sinh(x)*(
a*sinh(x)^4)^(1/2)+21/160*a^2*cosh(x)*sinh(x)^3*(a*sinh(x)^4)^(1/2)-9/80*a^2*cosh(x)*sinh(x)^5*(a*sinh(x)^4)^(
1/2)+1/10*a^2*cosh(x)*sinh(x)^7*(a*sinh(x)^4)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2635, 8} \[ \frac {1}{10} a^2 \sinh ^7(x) \cosh (x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \sinh ^5(x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {21}{160} a^2 \sinh ^3(x) \cosh (x) \sqrt {a \sinh ^4(x)}-\frac {21}{128} a^2 \sinh (x) \cosh (x) \sqrt {a \sinh ^4(x)}+\frac {63}{256} a^2 \coth (x) \sqrt {a \sinh ^4(x)}-\frac {63}{256} a^2 x \text {csch}^2(x) \sqrt {a \sinh ^4(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^4)^(5/2),x]

[Out]

(63*a^2*Coth[x]*Sqrt[a*Sinh[x]^4])/256 - (63*a^2*x*Csch[x]^2*Sqrt[a*Sinh[x]^4])/256 - (21*a^2*Cosh[x]*Sinh[x]*
Sqrt[a*Sinh[x]^4])/128 + (21*a^2*Cosh[x]*Sinh[x]^3*Sqrt[a*Sinh[x]^4])/160 - (9*a^2*Cosh[x]*Sinh[x]^5*Sqrt[a*Si
nh[x]^4])/80 + (a^2*Cosh[x]*Sinh[x]^7*Sqrt[a*Sinh[x]^4])/10

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 )^{5/2} \, dx &=\left (a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^{10}(x) \, dx\\ &=\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}-\frac {1}{10} \left (9 a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^8(x) \, dx\\ &=-\frac {9}{80} a^2 \cosh (x) \sinh ^5(x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}+\frac {1}{80} \left (63 a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^6(x) \, dx\\ &=\frac {21}{160} a^2 \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \cosh (x) \sinh ^5(x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}-\frac {1}{32} \left (21 a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^4(x) \, dx\\ &=-\frac {21}{128} a^2 \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {21}{160} a^2 \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \cosh (x) \sinh ^5(x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}+\frac {1}{128} \left (63 a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int \sinh ^2(x) \, dx\\ &=\frac {63}{256} a^2 \coth (x) \sqrt {a \sinh ^4(x)}-\frac {21}{128} a^2 \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {21}{160} a^2 \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \cosh (x) \sinh ^5(x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}-\frac {1}{256} \left (63 a^2 \text {csch}^2(x) \sqrt {a \sinh ^4(x)}\right ) \int 1 \, dx\\ &=\frac {63}{256} a^2 \coth (x) \sqrt {a \sinh ^4(x)}-\frac {63}{256} a^2 x \text {csch}^2(x) \sqrt {a \sinh ^4(x)}-\frac {21}{128} a^2 \cosh (x) \sinh (x) \sqrt {a \sinh ^4(x)}+\frac {21}{160} a^2 \cosh (x) \sinh ^3(x) \sqrt {a \sinh ^4(x)}-\frac {9}{80} a^2 \cosh (x) \sinh ^5(x) \sqrt {a \sinh ^4(x)}+\frac {1}{10} a^2 \cosh (x) \sinh ^7(x) \sqrt {a \sinh ^4(x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 53, normalized size = 0.40 \[ \frac {a (-2520 x+2100 \sinh (2 x)-600 \sinh (4 x)+150 \sinh (6 x)-25 \sinh (8 x)+2 \sinh (10 x)) \text {csch}^6(x) \left (a \sinh ^4(x)\right )^{3/2}}{10240} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^4)^(5/2),x]

[Out]

(a*Csch[x]^6*(a*Sinh[x]^4)^(3/2)*(-2520*x + 2100*Sinh[2*x] - 600*Sinh[4*x] + 150*Sinh[6*x] - 25*Sinh[8*x] + 2*
Sinh[10*x]))/10240

________________________________________________________________________________________

fricas [B]  time = 0.58, size = 1597, normalized size = 12.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/20480*(40*a^2*cosh(x)*e^(2*x)*sinh(x)^19 + 2*a^2*e^(2*x)*sinh(x)^20 + 5*(76*a^2*cosh(x)^2 - 5*a^2)*e^(2*x)*s
inh(x)^18 + 30*(76*a^2*cosh(x)^3 - 15*a^2*cosh(x))*e^(2*x)*sinh(x)^17 + 15*(646*a^2*cosh(x)^4 - 255*a^2*cosh(x
)^2 + 10*a^2)*e^(2*x)*sinh(x)^16 + 48*(646*a^2*cosh(x)^5 - 425*a^2*cosh(x)^3 + 50*a^2*cosh(x))*e^(2*x)*sinh(x)
^15 + 60*(1292*a^2*cosh(x)^6 - 1275*a^2*cosh(x)^4 + 300*a^2*cosh(x)^2 - 10*a^2)*e^(2*x)*sinh(x)^14 + 120*(1292
*a^2*cosh(x)^7 - 1785*a^2*cosh(x)^5 + 700*a^2*cosh(x)^3 - 70*a^2*cosh(x))*e^(2*x)*sinh(x)^13 + 60*(4199*a^2*co
sh(x)^8 - 7735*a^2*cosh(x)^6 + 4550*a^2*cosh(x)^4 - 910*a^2*cosh(x)^2 + 35*a^2)*e^(2*x)*sinh(x)^12 + 80*(4199*
a^2*cosh(x)^9 - 9945*a^2*cosh(x)^7 + 8190*a^2*cosh(x)^5 - 2730*a^2*cosh(x)^3 + 315*a^2*cosh(x))*e^(2*x)*sinh(x
)^11 + 2*(184756*a^2*cosh(x)^10 - 546975*a^2*cosh(x)^8 + 600600*a^2*cosh(x)^6 - 300300*a^2*cosh(x)^4 + 69300*a
^2*cosh(x)^2 - 2520*a^2*x)*e^(2*x)*sinh(x)^10 + 20*(16796*a^2*cosh(x)^11 - 60775*a^2*cosh(x)^9 + 85800*a^2*cos
h(x)^7 - 60060*a^2*cosh(x)^5 + 23100*a^2*cosh(x)^3 - 2520*a^2*x*cosh(x))*e^(2*x)*sinh(x)^9 + 30*(8398*a^2*cosh
(x)^12 - 36465*a^2*cosh(x)^10 + 64350*a^2*cosh(x)^8 - 60060*a^2*cosh(x)^6 + 34650*a^2*cosh(x)^4 - 7560*a^2*x*c
osh(x)^2 - 70*a^2)*e^(2*x)*sinh(x)^8 + 240*(646*a^2*cosh(x)^13 - 3315*a^2*cosh(x)^11 + 7150*a^2*cosh(x)^9 - 85
80*a^2*cosh(x)^7 + 6930*a^2*cosh(x)^5 - 2520*a^2*x*cosh(x)^3 - 70*a^2*cosh(x))*e^(2*x)*sinh(x)^7 + 60*(1292*a^
2*cosh(x)^14 - 7735*a^2*cosh(x)^12 + 20020*a^2*cosh(x)^10 - 30030*a^2*cosh(x)^8 + 32340*a^2*cosh(x)^6 - 17640*
a^2*x*cosh(x)^4 - 980*a^2*cosh(x)^2 + 10*a^2)*e^(2*x)*sinh(x)^6 + 24*(1292*a^2*cosh(x)^15 - 8925*a^2*cosh(x)^1
3 + 27300*a^2*cosh(x)^11 - 50050*a^2*cosh(x)^9 + 69300*a^2*cosh(x)^7 - 52920*a^2*x*cosh(x)^5 - 4900*a^2*cosh(x
)^3 + 150*a^2*cosh(x))*e^(2*x)*sinh(x)^5 + 30*(323*a^2*cosh(x)^16 - 2550*a^2*cosh(x)^14 + 9100*a^2*cosh(x)^12
- 20020*a^2*cosh(x)^10 + 34650*a^2*cosh(x)^8 - 35280*a^2*x*cosh(x)^6 - 4900*a^2*cosh(x)^4 + 300*a^2*cosh(x)^2
- 5*a^2)*e^(2*x)*sinh(x)^4 + 120*(19*a^2*cosh(x)^17 - 170*a^2*cosh(x)^15 + 700*a^2*cosh(x)^13 - 1820*a^2*cosh(
x)^11 + 3850*a^2*cosh(x)^9 - 5040*a^2*x*cosh(x)^7 - 980*a^2*cosh(x)^5 + 100*a^2*cosh(x)^3 - 5*a^2*cosh(x))*e^(
2*x)*sinh(x)^3 + 5*(76*a^2*cosh(x)^18 - 765*a^2*cosh(x)^16 + 3600*a^2*cosh(x)^14 - 10920*a^2*cosh(x)^12 + 2772
0*a^2*cosh(x)^10 - 45360*a^2*x*cosh(x)^8 - 11760*a^2*cosh(x)^6 + 1800*a^2*cosh(x)^4 - 180*a^2*cosh(x)^2 + 5*a^
2)*e^(2*x)*sinh(x)^2 + 10*(4*a^2*cosh(x)^19 - 45*a^2*cosh(x)^17 + 240*a^2*cosh(x)^15 - 840*a^2*cosh(x)^13 + 25
20*a^2*cosh(x)^11 - 5040*a^2*x*cosh(x)^9 - 1680*a^2*cosh(x)^7 + 360*a^2*cosh(x)^5 - 60*a^2*cosh(x)^3 + 5*a^2*c
osh(x))*e^(2*x)*sinh(x) + (2*a^2*cosh(x)^20 - 25*a^2*cosh(x)^18 + 150*a^2*cosh(x)^16 - 600*a^2*cosh(x)^14 + 21
00*a^2*cosh(x)^12 - 5040*a^2*x*cosh(x)^10 - 2100*a^2*cosh(x)^8 + 600*a^2*cosh(x)^6 - 150*a^2*cosh(x)^4 + 25*a^
2*cosh(x)^2 - 2*a^2)*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)^
10*e^(4*x) - 2*cosh(x)^10*e^(2*x) + (e^(4*x) - 2*e^(2*x) + 1)*sinh(x)^10 + cosh(x)^10 + 10*(cosh(x)*e^(4*x) -
2*cosh(x)*e^(2*x) + cosh(x))*sinh(x)^9 + 45*(cosh(x)^2*e^(4*x) - 2*cosh(x)^2*e^(2*x) + cosh(x)^2)*sinh(x)^8 +
120*(cosh(x)^3*e^(4*x) - 2*cosh(x)^3*e^(2*x) + cosh(x)^3)*sinh(x)^7 + 210*(cosh(x)^4*e^(4*x) - 2*cosh(x)^4*e^(
2*x) + cosh(x)^4)*sinh(x)^6 + 252*(cosh(x)^5*e^(4*x) - 2*cosh(x)^5*e^(2*x) + cosh(x)^5)*sinh(x)^5 + 210*(cosh(
x)^6*e^(4*x) - 2*cosh(x)^6*e^(2*x) + cosh(x)^6)*sinh(x)^4 + 120*(cosh(x)^7*e^(4*x) - 2*cosh(x)^7*e^(2*x) + cos
h(x)^7)*sinh(x)^3 + 45*(cosh(x)^8*e^(4*x) - 2*cosh(x)^8*e^(2*x) + cosh(x)^8)*sinh(x)^2 + 10*(cosh(x)^9*e^(4*x)
- 2*cosh(x)^9*e^(2*x) + cosh(x)^9)*sinh(x))

________________________________________________________________________________________

giac [A]  time = 0.30, size = 114, normalized size = 0.86 \[ -\frac {1}{20480} \, {\left (5040 \, a^{2} x - 2 \, a^{2} e^{\left (10 \, x\right )} + 25 \, a^{2} e^{\left (8 \, x\right )} - 150 \, a^{2} e^{\left (6 \, x\right )} + 600 \, a^{2} e^{\left (4 \, x\right )} - 2100 \, a^{2} e^{\left (2 \, x\right )} - {\left (5754 \, a^{2} e^{\left (10 \, x\right )} - 2100 \, a^{2} e^{\left (8 \, x\right )} + 600 \, a^{2} e^{\left (6 \, x\right )} - 150 \, a^{2} e^{\left (4 \, x\right )} + 25 \, a^{2} e^{\left (2 \, x\right )} - 2 \, a^{2}\right )} e^{\left (-10 \, x\right )}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(5/2),x, algorithm="giac")

[Out]

-1/20480*(5040*a^2*x - 2*a^2*e^(10*x) + 25*a^2*e^(8*x) - 150*a^2*e^(6*x) + 600*a^2*e^(4*x) - 2100*a^2*e^(2*x)
- (5754*a^2*e^(10*x) - 2100*a^2*e^(8*x) + 600*a^2*e^(6*x) - 150*a^2*e^(4*x) + 25*a^2*e^(2*x) - 2*a^2)*e^(-10*x
))*sqrt(a)

________________________________________________________________________________________

maple [A]  time = 0.20, size = 171, normalized size = 1.30 \[ \frac {\left (-1+\cosh \left (2 x \right )\right ) \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (\cosh \left (2 x \right )+1\right )}\, a^{\frac {3}{2}} \left (8 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}\, \left (\sinh ^{4}\left (2 x \right )\right )-50 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}\, \cosh \left (2 x \right ) \left (\sinh ^{2}\left (2 x \right )\right )+160 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}\, \left (\sinh ^{2}\left (2 x \right )\right )-325 \cosh \left (2 x \right ) \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}+640 \sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\, \sqrt {a}-315 \ln \left (\sqrt {a}\, \cosh \left (2 x \right )+\sqrt {a \left (\sinh ^{2}\left (2 x \right )\right )}\right ) a \right )}{2560 \sinh \left (2 x \right ) \sqrt {\left (-1+\cosh \left (2 x \right )\right )^{2} a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^4)^(5/2),x)

[Out]

1/2560*(-1+cosh(2*x))*(a*(-1+cosh(2*x))*(cosh(2*x)+1))^(1/2)*a^(3/2)*(8*(a*sinh(2*x)^2)^(1/2)*a^(1/2)*sinh(2*x
)^4-50*(a*sinh(2*x)^2)^(1/2)*a^(1/2)*cosh(2*x)*sinh(2*x)^2+160*(a*sinh(2*x)^2)^(1/2)*a^(1/2)*sinh(2*x)^2-325*c
osh(2*x)*(a*sinh(2*x)^2)^(1/2)*a^(1/2)+640*(a*sinh(2*x)^2)^(1/2)*a^(1/2)-315*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)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 100, normalized size = 0.76 \[ -\frac {63}{256} \, a^{\frac {5}{2}} x - \frac {1}{20480} \, {\left (25 \, a^{\frac {5}{2}} e^{\left (-2 \, x\right )} - 150 \, a^{\frac {5}{2}} e^{\left (-4 \, x\right )} + 600 \, a^{\frac {5}{2}} e^{\left (-6 \, x\right )} - 2100 \, a^{\frac {5}{2}} e^{\left (-8 \, x\right )} + 2100 \, a^{\frac {5}{2}} e^{\left (-12 \, x\right )} - 600 \, a^{\frac {5}{2}} e^{\left (-14 \, x\right )} + 150 \, a^{\frac {5}{2}} e^{\left (-16 \, x\right )} - 25 \, a^{\frac {5}{2}} e^{\left (-18 \, x\right )} + 2 \, a^{\frac {5}{2}} e^{\left (-20 \, x\right )} - 2 \, a^{\frac {5}{2}}\right )} e^{\left (10 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)^4)^(5/2),x, algorithm="maxima")

[Out]

-63/256*a^(5/2)*x - 1/20480*(25*a^(5/2)*e^(-2*x) - 150*a^(5/2)*e^(-4*x) + 600*a^(5/2)*e^(-6*x) - 2100*a^(5/2)*
e^(-8*x) + 2100*a^(5/2)*e^(-12*x) - 600*a^(5/2)*e^(-14*x) + 150*a^(5/2)*e^(-16*x) - 25*a^(5/2)*e^(-18*x) + 2*a
^(5/2)*e^(-20*x) - 2*a^(5/2))*e^(10*x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {sinh}\relax (x)}^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^4)^(5/2),x)

[Out]

int((a*sinh(x)^4)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sinh ^{4}{\relax (x )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)**4)**(5/2),x)

[Out]

Integral((a*sinh(x)**4)**(5/2), x)

________________________________________________________________________________________