### 3.549 $$\int \frac{x \cosh (a+b x)}{\sinh ^{\frac{5}{2}}(a+b x)} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0516689, antiderivative size = 98, normalized size of antiderivative = 1., 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, 2640, 2639} $-\frac{4 \cosh (a+b x)}{3 b^2 \sqrt{\sinh (a+b x)}}-\frac{4 i \sqrt{\sinh (a+b x)} E\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{3 b^2 \sqrt{i \sinh (a+b x)}}-\frac{2 x}{3 b \sinh ^{\frac{3}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 2640

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

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]

Rubi steps

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

Mathematica [A]  time = 0.203667, size = 66, normalized size = 0.67 $-\frac{2 \left (\sinh (2 (a+b x))+2 i (i \sinh (a+b x))^{3/2} E\left (\left .\frac{1}{4} (-2 i a-2 i b x+\pi )\right |2\right )+b x\right )}{3 b^2 \sinh ^{\frac{3}{2}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{x\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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