3.6.53 \(\int x \cosh (a+b x) \text {csch}^{\frac {7}{2}}(a+b x) \, dx\) [553]

Optimal. Leaf size=98 \[ -\frac {4 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b}+\frac {4 i \sqrt {\text {csch}(a+b x)} F\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {i \sinh (a+b x)}}{15 b^2} \]

[Out]

-4/15*cosh(b*x+a)*csch(b*x+a)^(3/2)/b^2-2/5*x*csch(b*x+a)^(5/2)/b-4/15*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)*EllipticF(cos(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*csch(b*x+a)^(1/2)*(I*sinh(b*
x+a))^(1/2)/b^2

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Rubi [A]
time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5553, 3853, 3856, 2720} \begin {gather*} -\frac {4 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{15 b^2}+\frac {4 i \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{15 b^2}-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x]*Csch[a + b*x]^(7/2),x]

[Out]

(-4*Cosh[a + b*x]*Csch[a + b*x]^(3/2))/(15*b^2) - (2*x*Csch[a + b*x]^(5/2))/(5*b) + (((4*I)/15)*Sqrt[Csch[a +
b*x]]*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/b^2

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 5553

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_)*(x_)^(m_.), x_Symbol] :> Simp[(-x^(m -
n + 1))*(Csch[a + b*x^n]^(p - 1)/(b*n*(p - 1))), x] + Dist[(m - n + 1)/(b*n*(p - 1)), Int[x^(m - n)*Csch[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*} \int x \cosh (a+b x) \text {csch}^{\frac {7}{2}}(a+b x) \, dx &=-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b}+\frac {2 \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx}{5 b}\\ &=-\frac {4 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b}-\frac {2 \int \sqrt {\text {csch}(a+b x)} \, dx}{15 b}\\ &=-\frac {4 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b}-\frac {\left (2 \sqrt {\text {csch}(a+b x)} \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{15 b}\\ &=-\frac {4 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \text {csch}^{\frac {5}{2}}(a+b x)}{5 b}+\frac {4 i \sqrt {\text {csch}(a+b x)} F\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {i \sinh (a+b x)}}{15 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 75, normalized size = 0.77 \begin {gather*} -\frac {2 \sqrt {\text {csch}(a+b x)} \left (2 \coth (a+b x)+3 b x \text {csch}^2(a+b x)+2 i F\left (\left .\frac {1}{4} (-2 i a+\pi -2 i b x)\right |2\right ) \sqrt {i \sinh (a+b x)}\right )}{15 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x]*Csch[a + b*x]^(7/2),x]

[Out]

(-2*Sqrt[Csch[a + b*x]]*(2*Coth[a + b*x] + 3*b*x*Csch[a + b*x]^2 + (2*I)*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)
/4, 2]*Sqrt[I*Sinh[a + b*x]]))/(15*b^2)

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Maple [F]
time = 0.95, size = 0, normalized size = 0.00 \[\int x \cosh \left (b x +a \right ) \mathrm {csch}\left (b x +a \right )^{\frac {7}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x+a)*csch(b*x+a)^(7/2),x)

[Out]

int(x*cosh(b*x+a)*csch(b*x+a)^(7/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(x*cosh(b*x + a)*csch(b*x + a)^(7/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)*csch(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)*csch(b*x + a)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {cosh}\left (a+b\,x\right )\,{\left (\frac {1}{\mathrm {sinh}\left (a+b\,x\right )}\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(a + b*x)*(1/sinh(a + b*x))^(7/2),x)

[Out]

int(x*cosh(a + b*x)*(1/sinh(a + b*x))^(7/2), x)

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