3.1.9 \(\int \sinh ^{\frac {3}{2}}(a+b x) \, dx\) [9]

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

[Out]

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

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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, 2720} \begin {gather*} \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}+\frac {2 i \sqrt {i \sinh (a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{3 b \sqrt {\sinh (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(3/2),x]

[Out]

(((2*I)/3)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b*Sqrt[Sinh[a + b*x]]) + (2*Cosh[a + b
*x]*Sqrt[Sinh[a + b*x]])/(3*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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 {3}{2}}(a+b x) \, dx &=\frac {2 \cosh (a+b x) \sqrt {\sinh (a+b x)}}{3 b}-\frac {1}{3} \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx\\ &=\frac {2 \cosh (a+b x) \sqrt {\sinh (a+b x)}}{3 b}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{3 \sqrt {\sinh (a+b x)}}\\ &=\frac {2 i F\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {i \sinh (a+b x)}}{3 b \sqrt {\sinh (a+b x)}}+\frac {2 \cosh (a+b x) \sqrt {\sinh (a+b x)}}{3 b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.07, size = 83, normalized size = 1.04 \begin {gather*} \frac {\sinh (2 (a+b x))-2 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (a+b x))+\sinh (2 (a+b x))\right ) \sqrt {1-\cosh (2 a+2 b x)-\sinh (2 a+2 b x)}}{3 b \sqrt {\sinh (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(3/2),x]

[Out]

(Sinh[2*(a + b*x)] - 2*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)]]*Sqrt[1 - Cosh[2
*a + 2*b*x] - Sinh[2*a + 2*b*x]])/(3*b*Sqrt[Sinh[a + b*x]])

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Maple [A]
time = 0.62, size = 100, normalized size = 1.25

method	result	size
default	\(\frac {-\frac {i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \left (\cosh ^{2}\left (b x +a \right )\right ) \sinh \left (b x +a \right )}{3}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(-1/3*I*(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)*EllipticF((1-I*sinh(b*x+
a))^(1/2),1/2*2^(1/2))+2/3*cosh(b*x+a)^2*sinh(b*x+a))/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)^(3/2),x, algorithm="maxima")

[Out]

integrate(sinh(b*x + a)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 103, normalized size = 1.29 \begin {gather*} -\frac {2 \, {\left (\sqrt {2} \cosh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \sqrt {\sinh \left (b x + a\right )}}{3 \, {\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*(sqrt(2)*cosh(b*x + a) + sqrt(2)*sinh(b*x + a))*weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a
)) - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*sqrt(sinh(b*x + a)))/(b*cosh(b*x
+ a) + b*sinh(b*x + a))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sinh ^{\frac {3}{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)**(3/2),x)

[Out]

Integral(sinh(a + b*x)**(3/2), 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)^(3/2),x, algorithm="giac")

[Out]

integrate(sinh(b*x + a)^(3/2), 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 )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^(3/2),x)

[Out]

int(sinh(a + b*x)^(3/2), x)

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