3.1.8 \(\int \sinh ^{\frac {5}{2}}(a+b x) \, dx\) [8]

Optimal. Leaf size=80 \[ \frac {6 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{5 b \sqrt {i \sinh (a+b x)}}+\frac {2 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{5 b} \]

[Out]

2/5*cosh(b*x+a)*sinh(b*x+a)^(3/2)/b-6/5*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*x
)*EllipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*sinh(b*x+a)^(1/2)/b/(I*sinh(b*x+a))^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 80, 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 = {2715, 2721, 2719} \begin {gather*} \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}+\frac {6 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{5 b \sqrt {i \sinh (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(5/2),x]

[Out]

(((6*I)/5)*EllipticE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/(b*Sqrt[I*Sinh[a + b*x]]) + (2*Cosh[a + b
*x]*Sinh[a + b*x]^(3/2))/(5*b)

Rule 2715

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] && Integ
erQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \sinh ^{\frac {5}{2}}(a+b x) \, dx &=\frac {2 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{5 b}-\frac {3}{5} \int \sqrt {\sinh (a+b x)} \, dx\\ &=\frac {2 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{5 b}-\frac {\left (3 \sqrt {\sinh (a+b x)}\right ) \int \sqrt {i \sinh (a+b x)} \, dx}{5 \sqrt {i \sinh (a+b x)}}\\ &=\frac {6 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{5 b \sqrt {i \sinh (a+b x)}}+\frac {2 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 68, normalized size = 0.85 \begin {gather*} \frac {-6 E\left (\left .\frac {1}{4} (-2 i a+\pi -2 i b x)\right |2\right ) \sqrt {i \sinh (a+b x)}+\sinh (a+b x) \sinh (2 (a+b x))}{5 b \sqrt {\sinh (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(5/2),x]

[Out]

(-6*EllipticE[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*Sqrt[I*Sinh[a + b*x]] + Sinh[a + b*x]*Sinh[2*(a + b*x)])/(5*b*
Sqrt[Sinh[a + b*x]])

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Maple [A]
time = 0.66, size = 164, normalized size = 2.05

method	result	size
default	\(\frac {-\frac {6 \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {2 \left (\cosh ^{4}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cosh ^{2}\left (b x +a \right )\right )}{5}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-6/5*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b*x+a))^(1/2)*EllipticE((1-I*sinh(b*x+a)
)^(1/2),1/2*2^(1/2))+3/5*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b*x+a))^(1/2)*Ellipti
cF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))+2/5*cosh(b*x+a)^4-2/5*cosh(b*x+a)^2)/cosh(b*x+a)/sinh(b*x+a)^(1/2)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 202, normalized size = 2.52 \begin {gather*} \frac {12 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right ) + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 12 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1\right )} \sqrt {\sinh \left (b x + a\right )}}{10 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(12*(sqrt(2)*cosh(b*x + a)^2 + 2*sqrt(2)*cosh(b*x + a)*sinh(b*x + a) + sqrt(2)*sinh(b*x + a)^2)*weierstra
ssZeta(4, 0, weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a))) + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*si
nh(b*x + a)^3 + sinh(b*x + a)^4 + 6*(cosh(b*x + a)^2 + 2)*sinh(b*x + a)^2 + 12*cosh(b*x + a)^2 + 4*(cosh(b*x +
a)^3 + 6*cosh(b*x + a))*sinh(b*x + a) - 1)*sqrt(sinh(b*x + a)))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b
*x + a) + b*sinh(b*x + a)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh ^{\frac {5}{2}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {sinh}\left (a+b\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^(5/2),x)

[Out]

int(sinh(a + b*x)^(5/2), x)

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