3.533 \(\int \frac {x \sinh (a+b x)}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=64 \[ \frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{3 b^2}+\frac {4 \sinh (a+b x)}{3 b^2 \sqrt {\cosh (a+b x)}}-\frac {2 x}{3 b \cosh ^{\frac {3}{2}}(a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5373, 2636, 2639} \[ \frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{3 b^2}+\frac {4 \sinh (a+b x)}{3 b^2 \sqrt {\cosh (a+b x)}}-\frac {2 x}{3 b \cosh ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2636

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

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]

Rule 5373

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(x^(m -
n + 1)*Cosh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[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*} \int \frac {x \sinh (a+b x)}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx &=-\frac {2 x}{3 b \cosh ^{\frac {3}{2}}(a+b x)}+\frac {2 \int \frac {1}{\cosh ^{\frac {3}{2}}(a+b x)} \, dx}{3 b}\\ &=-\frac {2 x}{3 b \cosh ^{\frac {3}{2}}(a+b x)}+\frac {4 \sinh (a+b x)}{3 b^2 \sqrt {\cosh (a+b x)}}-\frac {2 \int \sqrt {\cosh (a+b x)} \, dx}{3 b}\\ &=-\frac {2 x}{3 b \cosh ^{\frac {3}{2}}(a+b x)}+\frac {4 i E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{3 b^2}+\frac {4 \sinh (a+b x)}{3 b^2 \sqrt {\cosh (a+b x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 57, normalized size = 0.89 \[ \frac {2 \left (\sinh (2 (a+b x))+2 i \cosh ^{\frac {3}{2}}(a+b x) E\left (\left .\frac {1}{2} i (a+b x)\right |2\right )-b x\right )}{3 b^2 \cosh ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*Cosh[a + b*x
]^(3/2))

________________________________________________________________________________________

fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,\mathrm {sinh}\left (a+b\,x\right )}{{\mathrm {cosh}\left (a+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________