3.17 \(\int (b \sinh (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=88 \[ \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{3 d \sqrt {b \sinh (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2635, 2642, 2641} \[ \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{3 d \sqrt {b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(3/2),x]

[Out]

(((2*I)/3)*b^2*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[I*Sinh[c + d*x]])/(d*Sqrt[b*Sinh[c + d*x]]) + (2*b*Co
sh[c + d*x]*Sqrt[b*Sinh[c + d*x]])/(3*d)

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 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

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

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Mathematica [C]  time = 0.13, size = 88, normalized size = 1.00 \[ \frac {b^2 \left (\sinh (2 (c+d x))-2 \sqrt {-\sinh (2 c+2 d x)-\cosh (2 c+2 d x)+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\cosh (2 (c+d x))+\sinh (2 (c+d x))\right )\right )}{3 d \sqrt {b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(3/2),x]

[Out]

(b^2*(Sinh[2*(c + d*x)] - 2*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]]*Sqrt[1 - C
osh[2*c + 2*d*x] - Sinh[2*c + 2*d*x]]))/(3*d*Sqrt[b*Sinh[c + d*x]])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sinh \left (d x + c\right )} b \sinh \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(d*x + c))*b*sinh(d*x + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sinh \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c))^(3/2), x)

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maple [A]  time = 0.07, size = 106, normalized size = 1.20 \[ -\frac {b^{2} \left (i \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cosh ^{2}\left (d x +c \right )\right ) \sinh \left (d x +c \right )\right )}{3 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sinh(d*x+c))^(3/2),x)

[Out]

-1/3*b^2*(I*(1-I*sinh(d*x+c))^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)*EllipticF((1-I*sinh(
d*x+c))^(1/2),1/2*2^(1/2))-2*cosh(d*x+c)^2*sinh(d*x+c))/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sinh \left (d x + c\right )\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sinh(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sinh(c + d*x))^(3/2),x)

[Out]

int((b*sinh(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sinh {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sinh(d*x+c))**(3/2),x)

[Out]

Integral((b*sinh(c + d*x))**(3/2), x)

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