3.559 \(\int \frac {x \cosh (a+b x)}{\text {csch}^{\frac {5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=121 \[ -\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {20 \cosh (a+b x)}{147 b^2 \sqrt {\text {csch}(a+b x)}}+\frac {20 i \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{147 b^2}+\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)} \]

[Out]

2/7*x/b/csch(b*x+a)^(7/2)-4/49*cosh(b*x+a)/b^2/csch(b*x+a)^(5/2)+20/147*cosh(b*x+a)/b^2/csch(b*x+a)^(1/2)-20/1
47*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 = 0.07, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5445, 3769, 3771, 2641} \[ -\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {20 \cosh (a+b x)}{147 b^2 \sqrt {\text {csch}(a+b x)}}+\frac {20 i \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{147 b^2}+\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/(7*b*Csch[a + b*x]^(7/2)) - (4*Cosh[a + b*x])/(49*b^2*Csch[a + b*x]^(5/2)) + (20*Cosh[a + b*x])/(147*b^2
*Sqrt[Csch[a + b*x]]) + (((20*I)/147)*Sqrt[Csch[a + b*x]]*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a +
b*x]])/b^2

Rule 2641

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

Rule 3769

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

Rule 3771

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 5445

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 \frac {x \cosh (a+b x)}{\text {csch}^{\frac {5}{2}}(a+b x)} \, dx &=\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)}-\frac {2 \int \frac {1}{\text {csch}^{\frac {7}{2}}(a+b x)} \, dx}{7 b}\\ &=\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {10 \int \frac {1}{\text {csch}^{\frac {3}{2}}(a+b x)} \, dx}{49 b}\\ &=\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {20 \cosh (a+b x)}{147 b^2 \sqrt {\text {csch}(a+b x)}}-\frac {10 \int \sqrt {\text {csch}(a+b x)} \, dx}{147 b}\\ &=\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {20 \cosh (a+b x)}{147 b^2 \sqrt {\text {csch}(a+b x)}}-\frac {\left (10 \sqrt {\text {csch}(a+b x)} \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{147 b}\\ &=\frac {2 x}{7 b \text {csch}^{\frac {7}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{49 b^2 \text {csch}^{\frac {5}{2}}(a+b x)}+\frac {20 \cosh (a+b x)}{147 b^2 \sqrt {\text {csch}(a+b x)}}+\frac {20 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)}}{147 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 103, normalized size = 0.85 \[ \frac {\sqrt {\text {csch}(a+b x)} \left (52 \sinh (2 (a+b x))-6 \sinh (4 (a+b x))-84 b x \cosh (2 (a+b x))+21 b x \cosh (4 (a+b x))-80 i \sqrt {i \sinh (a+b x)} F\left (\left .\frac {1}{4} (-2 i a-2 i b x+\pi )\right |2\right )+63 b x\right )}{588 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Csch[a + b*x]]*(63*b*x - 84*b*x*Cosh[2*(a + b*x)] + 21*b*x*Cosh[4*(a + b*x)] - (80*I)*EllipticF[((-2*I)*
a + Pi - (2*I)*b*x)/4, 2]*Sqrt[I*Sinh[a + b*x]] + 52*Sinh[2*(a + b*x)] - 6*Sinh[4*(a + b*x)]))/(588*b^2)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {x \cosh \left (b x +a \right )}{\mathrm {csch}\left (b x +a \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\mathrm {cosh}\left (a+b\,x\right )}{{\left (\frac {1}{\mathrm {sinh}\left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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