\(\int (b \sinh (c+d x))^{5/2} \, dx\) [16]

Optimal result

Integrand size = 12, antiderivative size = 88 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {6 i b^2 E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 d \sqrt {i \sinh (c+d x)}}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2721, 2719} \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}+\frac {6 i b^2 E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 d \sqrt {i \sinh (c+d x)}} \]

[In]

Int[(b*Sinh[c + d*x])^(5/2),x]

[Out]

(((6*I)/5)*b^2*EllipticE[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[b*Sinh[c + d*x]])/(d*Sqrt[I*Sinh[c + d*x]]) + (2*b*Co
sh[c + d*x]*(b*Sinh[c + d*x])^(3/2))/(5*d)

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {b^2 \sqrt {b \sinh (c+d x)} \left (-\frac {6 i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right )}{\sqrt {i \sinh (c+d x)}}+\sinh (2 (c+d x))\right )}{5 d} \]

[In]

Integrate[(b*Sinh[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.93

method	result	size
default	\(-\frac {b^{3} \left (6 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \cosh \left (d x +c \right )^{4}+2 \cosh \left (d x +c \right )^{2}\right )}{5 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(170\)

[In]

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

[Out]

-1/5*b^3*(6*(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)*EllipticE((1-I*sinh(
d*x+c))^(1/2),1/2*2^(1/2))-3*(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)*Ell
ipticF((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))-2*cosh(d*x+c)^4+2*cosh(d*x+c)^2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)
/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.83 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {12 \, {\left (\sqrt {2} b^{2} \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {2} b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {2} b^{2} \sinh \left (d x + c\right )^{2}\right )} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + {\left (b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 12 \, b^{2} \cosh \left (d x + c\right )^{2} + 6 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 6 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}}{10 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/10*(12*(sqrt(2)*b^2*cosh(d*x + c)^2 + 2*sqrt(2)*b^2*cosh(d*x + c)*sinh(d*x + c) + sqrt(2)*b^2*sinh(d*x + c)^
2)*sqrt(b)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c))) + (b^2*cosh(d*x + c
)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 12*b^2*cosh(d*x + c)^2 + 6*(b^2*cosh(d*x + c
)^2 + 2*b^2)*sinh(d*x + c)^2 - b^2 + 4*(b^2*cosh(d*x + c)^3 + 6*b^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(
d*x + c)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2)

Sympy [F]

\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int \left (b \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

Integral((b*sinh(c + d*x))**(5/2), x)

Maxima [F]

\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(5/2), x)

Giac [F]

\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (b \sinh (c+d x))^{5/2} \, dx=\int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

int((b*sinh(c + d*x))^(5/2),x)

[Out]

int((b*sinh(c + d*x))^(5/2), x)