\(\int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx\) [538]

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

Optimal result

Integrand size = 18, antiderivative size = 84 \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\frac {4 i \sqrt {\cosh (a+b x)} E\left (\left .\frac {1}{2} i (a+b x)\right |2\right ) \sqrt {\text {sech}(a+b x)}}{3 b^2}-\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {4 \sqrt {\text {sech}(a+b x)} \sinh (a+b x)}{3 b^2} \]

[Out]

-2/3*x*sech(b*x+a)^(3/2)/b+4/3*sinh(b*x+a)*sech(b*x+a)^(1/2)/b^2+4/3*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*
a+1/2*b*x)*EllipticE(I*sinh(1/2*a+1/2*b*x),2^(1/2))*cosh(b*x+a)^(1/2)*sech(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, 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 = {5552, 3853, 3856, 2719} \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\frac {4 \sinh (a+b x) \sqrt {\text {sech}(a+b x)}}{3 b^2}+\frac {4 i \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)} E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{3 b^2}-\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b} \]

[In]

Int[x*Sech[a + b*x]^(5/2)*Sinh[a + b*x],x]

[Out]

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

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
 1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 5552

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 \int \text {sech}^{\frac {3}{2}}(a+b x) \, dx}{3 b} \\ & = -\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {4 \sqrt {\text {sech}(a+b x)} \sinh (a+b x)}{3 b^2}-\frac {2 \int \frac {1}{\sqrt {\text {sech}(a+b x)}} \, dx}{3 b} \\ & = -\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {4 \sqrt {\text {sech}(a+b x)} \sinh (a+b x)}{3 b^2}-\frac {\left (2 \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)}\right ) \int \sqrt {\cosh (a+b x)} \, dx}{3 b} \\ & = \frac {4 i \sqrt {\cosh (a+b x)} E\left (\left .\frac {1}{2} i (a+b x)\right |2\right ) \sqrt {\text {sech}(a+b x)}}{3 b^2}-\frac {2 x \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {4 \sqrt {\text {sech}(a+b x)} \sinh (a+b x)}{3 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68 \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \left (-b x+2 i \cosh ^{\frac {3}{2}}(a+b x) E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )+\sinh (2 (a+b x))\right )}{3 b^2} \]

[In]

Integrate[x*Sech[a + b*x]^(5/2)*Sinh[a + b*x],x]

[Out]

(2*Sech[a + b*x]^(3/2)*(-(b*x) + (2*I)*Cosh[a + b*x]^(3/2)*EllipticE[(I/2)*(a + b*x), 2] + Sinh[2*(a + b*x)]))
/(3*b^2)

Maple [F]

\[\int x \operatorname {sech}\left (b x +a \right )^{\frac {5}{2}} \sinh \left (b x +a \right )d x\]

[In]

int(x*sech(b*x+a)^(5/2)*sinh(b*x+a),x)

[Out]

int(x*sech(b*x+a)^(5/2)*sinh(b*x+a),x)

Fricas [F(-2)]

Exception generated. \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F(-1)]

Timed out. \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\text {Timed out} \]

[In]

integrate(x*sech(b*x+a)**(5/2)*sinh(b*x+a),x)

[Out]

Timed out

Maxima [F]

\[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\int { x \operatorname {sech}\left (b x + a\right )^{\frac {5}{2}} \sinh \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

integrate(x*sech(b*x + a)^(5/2)*sinh(b*x + a), x)

Giac [F]

\[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\int { x \operatorname {sech}\left (b x + a\right )^{\frac {5}{2}} \sinh \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

integrate(x*sech(b*x + a)^(5/2)*sinh(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int x \text {sech}^{\frac {5}{2}}(a+b x) \sinh (a+b x) \, dx=\int x\,\mathrm {sinh}\left (a+b\,x\right )\,{\left (\frac {1}{\mathrm {cosh}\left (a+b\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(x*sinh(a + b*x)*(1/cosh(a + b*x))^(5/2),x)

[Out]

int(x*sinh(a + b*x)*(1/cosh(a + b*x))^(5/2), x)