\(\int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx\) [545]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 98 \[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=-\frac {12 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{25 b^2 \sqrt {i \sinh (a+b x)}}-\frac {4 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b} \]

[Out]

-4/25*cosh(b*x+a)*sinh(b*x+a)^(3/2)/b^2+2/5*x*sinh(b*x+a)^(5/2)/b+12/25*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^2/(I*si
nh(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5480, 2715, 2721, 2719} \[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=-\frac {4 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{25 b^2}-\frac {12 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{25 b^2 \sqrt {i \sinh (a+b x)}}+\frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b} \]

[In]

Int[x*Cosh[a + b*x]*Sinh[a + b*x]^(3/2),x]

[Out]

(((-12*I)/25)*EllipticE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/(b^2*Sqrt[I*Sinh[a + b*x]]) - (4*Cosh[
a + b*x]*Sinh[a + b*x]^(3/2))/(25*b^2) + (2*x*Sinh[a + b*x]^(5/2))/(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]

Rule 5480

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[x^(m - n
 + 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {2 \int \sinh ^{\frac {5}{2}}(a+b x) \, dx}{5 b} \\ & = -\frac {4 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b}+\frac {6 \int \sqrt {\sinh (a+b x)} \, dx}{25 b} \\ & = -\frac {4 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b}+\frac {\left (6 \sqrt {\sinh (a+b x)}\right ) \int \sqrt {i \sinh (a+b x)} \, dx}{25 b \sqrt {i \sinh (a+b x)}} \\ & = -\frac {12 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)}}{25 b^2 \sqrt {i \sinh (a+b x)}}-\frac {4 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sinh ^{\frac {5}{2}}(a+b x)}{5 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.95 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.46 \[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\frac {e^{-3 (a+b x)} \left (\left (-1+e^{2 (a+b x)}\right ) \left (2+5 b x+e^{2 (a+b x)} (24-10 b x)+e^{4 (a+b x)} (-2+5 b x)\right )+48 e^{2 (a+b x)} \sqrt {1-e^{2 (a+b x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},e^{2 (a+b x)}\right )\right )}{50 \sqrt {2} b^2 \sqrt {-e^{-a-b x}+e^{a+b x}}} \]

[In]

Integrate[x*Cosh[a + b*x]*Sinh[a + b*x]^(3/2),x]

[Out]

((-1 + E^(2*(a + b*x)))*(2 + 5*b*x + E^(2*(a + b*x))*(24 - 10*b*x) + E^(4*(a + b*x))*(-2 + 5*b*x)) + 48*E^(2*(
a + b*x))*Sqrt[1 - E^(2*(a + b*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, E^(2*(a + b*x))])/(50*Sqrt[2]*b^2*E^(3*(
a + b*x))*Sqrt[-E^(-a - b*x) + E^(a + b*x)])

Maple [F]

\[\int x \cosh \left (b x +a \right ) \sinh \left (b x +a \right )^{\frac {3}{2}}d x\]

[In]

int(x*cosh(b*x+a)*sinh(b*x+a)^(3/2),x)

[Out]

int(x*cosh(b*x+a)*sinh(b*x+a)^(3/2),x)

Fricas [F(-2)]

Exception generated. \[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int x \sinh ^{\frac {3}{2}}{\left (a + b x \right )} \cosh {\left (a + b x \right )}\, dx \]

[In]

integrate(x*cosh(b*x+a)*sinh(b*x+a)**(3/2),x)

[Out]

Integral(x*sinh(a + b*x)**(3/2)*cosh(a + b*x), x)

Maxima [F]

\[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int { x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate(x*cosh(b*x + a)*sinh(b*x + a)^(3/2), x)

Giac [F]

\[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int { x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate(x*cosh(b*x + a)*sinh(b*x + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int x \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int x\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^{3/2} \,d x \]

[In]

int(x*cosh(a + b*x)*sinh(a + b*x)^(3/2),x)

[Out]

int(x*cosh(a + b*x)*sinh(a + b*x)^(3/2), x)