3.550 \(\int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx\)

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

[Out]

-2/5*x/b/sinh(b*x+a)^(5/2)-4/15*cosh(b*x+a)/b^2/sinh(b*x+a)^(3/2)-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))*(I*sinh(b*x+a))^(1/2)/b^2/si
nh(b*x+a)^(1/2)

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Rubi [A]  time = 0.05, 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 = {5372, 2636, 2642, 2641} \[ -\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i \sqrt {i \sinh (a+b x)} F\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{15 b^2 \sqrt {\sinh (a+b x)}}-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/(5*b*Sinh[a + b*x]^(5/2)) - (4*Cosh[a + b*x])/(15*b^2*Sinh[a + b*x]^(3/2)) + (((4*I)/15)*EllipticF[(I*a
- Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b^2*Sqrt[Sinh[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 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 5372

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*} \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx &=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx}{5 b}\\ &=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}-\frac {2 \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx}{15 b}\\ &=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\left (2 \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{15 b \sqrt {\sinh (a+b x)}}\\ &=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i 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 \sqrt {\sinh (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 67, normalized size = 0.68 \[ -\frac {2 \left (\sinh (2 (a+b x))-2 i (i \sinh (a+b x))^{5/2} F\left (\left .\frac {1}{4} (-2 i a-2 i b x+\pi )\right |2\right )+3 b x\right )}{15 b^2 \sinh ^{\frac {5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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)/sinh(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {x \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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