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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.036614, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2635, 2640, 2639} \[ \frac{2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}+\frac{6 i b^2 E\left (\left .\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \sinh (c+d x)}}{5 d \sqrt{i \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[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 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 2640

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

Rule 2639

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

Rubi steps

\begin{align*} \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{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]  time = 0.119512, size = 68, normalized size = 0.77 \[ \frac{b^2 \sqrt{b \sinh (c+d x)} \left (\sinh (2 (c+d x))-\frac{6 i E\left (\left .\frac{1}{4} (-2 i c-2 i d x+\pi )\right |2\right )}{\sqrt{i \sinh (c+d x)}}\right )}{5 d} \]

Antiderivative was successfully verified.

[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]  time = 0.047, size = 170, normalized size = 1.9 \begin{align*} -{\frac{{b}^{3}}{5\,d\cosh \left ( dx+c \right ) } \left ( 6\,\sqrt{1-i\sinh \left ( dx+c \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( dx+c \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) -3\,\sqrt{1-i\sinh \left ( dx+c \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( dx+c \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) -2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{b\sinh \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sinh \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sinh \left (d x + c\right )} b^{2} \sinh \left (d x + c\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(d*x + c))*b^2*sinh(d*x + c)^2, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sinh \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(5/2), x)