Integrand size = 10, antiderivative size = 53 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \coth (x) \sqrt {a \sinh ^2(x)}-\frac {4}{15} a \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2} \] Output:
8/15*a^2*coth(x)*(a*sinh(x)^2)^(1/2)-4/15*a*coth(x)*(a*sinh(x)^2)^(3/2)+1/ 5*coth(x)*(a*sinh(x)^2)^(5/2)
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {1}{240} a^2 (150 \cosh (x)-25 \cosh (3 x)+3 \cosh (5 x)) \text {csch}(x) \sqrt {a \sinh ^2(x)} \] Input:
Integrate[(a*Sinh[x]^2)^(5/2),x]
Output:
(a^2*(150*Cosh[x] - 25*Cosh[3*x] + 3*Cosh[5*x])*Csch[x]*Sqrt[a*Sinh[x]^2]) /240
Time = 0.42 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3682, 3042, 3682, 3042, 3686, 3042, 26, 3118}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a \sinh ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-a \sin (i x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \int \left (a \sinh ^2(x)\right )^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \int \left (-a \sin (i x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}-\frac {2}{3} a \int \sqrt {a \sinh ^2(x)}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}-\frac {2}{3} a \int \sqrt {-a \sin (i x)^2}dx\right )\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}-\frac {2}{3} a \text {csch}(x) \sqrt {a \sinh ^2(x)} \int \sinh (x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}-\frac {2}{3} a \text {csch}(x) \sqrt {a \sinh ^2(x)} \int -i \sin (i x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}+\frac {2}{3} i a \text {csch}(x) \sqrt {a \sinh ^2(x)} \int \sin (i x)dx\right )\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {1}{5} \coth (x) \left (a \sinh ^2(x)\right )^{5/2}-\frac {4}{5} a \left (\frac {1}{3} \coth (x) \left (a \sinh ^2(x)\right )^{3/2}-\frac {2}{3} a \coth (x) \sqrt {a \sinh ^2(x)}\right )\) |
Input:
Int[(a*Sinh[x]^2)^(5/2),x]
Output:
(Coth[x]*(a*Sinh[x]^2)^(5/2))/5 - (4*a*((-2*a*Coth[x]*Sqrt[a*Sinh[x]^2])/3 + (Coth[x]*(a*Sinh[x]^2)^(3/2))/3))/5
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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] + Simp[b*((2*p - 1)/(2*p)) Int[(b*S in[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] && !IntegerQ[p] && G tQ[p, 1]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[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]])
Time = 0.44 (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 | \(\frac {a^{2} {\mathrm e}^{6 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{160 \,{\mathrm e}^{2 x}-160}-\frac {5 a^{2} {\mathrm e}^{4 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{96 \left ({\mathrm e}^{2 x}-1\right )}+\frac {5 a^{2} {\mathrm e}^{2 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{16 \left ({\mathrm e}^{2 x}-1\right )}+\frac {5 \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}\, a^{2}}{16 \left ({\mathrm e}^{2 x}-1\right )}-\frac {5 a^{2} {\mathrm e}^{-2 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{96 \left ({\mathrm e}^{2 x}-1\right )}+\frac {a^{2} {\mathrm e}^{-4 x} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{160 \,{\mathrm e}^{2 x}-160}\) | \(196\) |
Input:
int((a*sinh(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/15*a^3*sinh(x)*cosh(x)*(3*sinh(x)^4-4*sinh(x)^2+8)/(a*sinh(x)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (41) = 82\).
Time = 0.12 (sec) , antiderivative size = 511, normalized size of antiderivative = 9.64 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx =\text {Too large to display} \] Input:
integrate((a*sinh(x)^2)^(5/2),x, algorithm="fricas")
Output:
1/480*(30*a^2*cosh(x)*e^x*sinh(x)^9 + 3*a^2*e^x*sinh(x)^10 + 5*(27*a^2*cos h(x)^2 - 5*a^2)*e^x*sinh(x)^8 + 40*(9*a^2*cosh(x)^3 - 5*a^2*cosh(x))*e^x*s inh(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)*sinh(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))
\[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\int \left (a \sinh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a*sinh(x)**2)**(5/2),x)
Output:
Integral((a*sinh(x)**2)**(5/2), x)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=-\frac {1}{160} \, a^{\frac {5}{2}} e^{\left (5 \, x\right )} + \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (3 \, x\right )} - \frac {5}{16} \, a^{\frac {5}{2}} e^{\left (-x\right )} + \frac {5}{96} \, a^{\frac {5}{2}} e^{\left (-3 \, x\right )} - \frac {1}{160} \, a^{\frac {5}{2}} e^{\left (-5 \, x\right )} - \frac {5}{16} \, a^{\frac {5}{2}} e^{x} \] Input:
integrate((a*sinh(x)^2)^(5/2),x, algorithm="maxima")
Output:
-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
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (41) = 82\).
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.92 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {1}{480} \, {\left ({\left (150 \, e^{\left (4 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, e^{\left (2 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 3 \, \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} e^{\left (-5 \, x\right )} + 3 \, e^{\left (5 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) - 25 \, e^{\left (3 \, x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + 150 \, e^{x} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} a^{\frac {5}{2}} \] Input:
integrate((a*sinh(x)^2)^(5/2),x, algorithm="giac")
Output:
1/480*((150*e^(4*x)*sgn(e^(3*x) - e^x) - 25*e^(2*x)*sgn(e^(3*x) - e^x) + 3 *sgn(e^(3*x) - e^x))*e^(-5*x) + 3*e^(5*x)*sgn(e^(3*x) - e^x) - 25*e^(3*x)* sgn(e^(3*x) - e^x) + 150*e^x*sgn(e^(3*x) - e^x))*a^(5/2)
Timed out. \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\int {\left (a\,{\mathrm {sinh}\left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int((a*sinh(x)^2)^(5/2),x)
Output:
int((a*sinh(x)^2)^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \left (a \sinh ^2(x)\right )^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (3 e^{10 x}-25 e^{8 x}+150 e^{6 x}+150 e^{4 x}-25 e^{2 x}+3\right )}{480 e^{5 x}} \] Input:
int((a*sinh(x)^2)^(5/2),x)
Output:
(sqrt(a)*a**2*(3*e**(10*x) - 25*e**(8*x) + 150*e**(6*x) + 150*e**(4*x) - 2 5*e**(2*x) + 3))/(480*e**(5*x))