3.6.50 \(\int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx\) [550]

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

[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 = 0.04, 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 = {5480, 2716, 2721, 2720} \begin {gather*} -\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)} \end {gather*}

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 2716

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 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 2721

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

Rule 5480

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 = 0.19, size = 67, normalized size = 0.68 \begin {gather*} -\frac {2 \left (3 b x-2 i F\left (\left .\frac {1}{4} (-2 i a+\pi -2 i b x)\right |2\right ) (i \sinh (a+b x))^{5/2}+\sinh (2 (a+b x))\right )}{15 b^2 \sinh ^{\frac {5}{2}}(a+b x)} \end {gather*}

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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Maple [F]
time = 0.40, 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 \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)/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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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)/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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3064 deep

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

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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