### 3.529 $$\int x \cosh ^{\frac{3}{2}}(a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=64 $\frac{12 i E\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{25 b^2}-\frac{4 \sinh (a+b x) \cosh ^{\frac{3}{2}}(a+b x)}{25 b^2}+\frac{2 x \cosh ^{\frac{5}{2}}(a+b x)}{5 b}$

[Out]

(2*x*Cosh[a + b*x]^(5/2))/(5*b) + (((12*I)/25)*EllipticE[(I/2)*(a + b*x), 2])/b^2 - (4*Cosh[a + b*x]^(3/2)*Sin
h[a + b*x])/(25*b^2)

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Rubi [A]  time = 0.0426493, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {5373, 2635, 2639} $\frac{12 i E\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{25 b^2}-\frac{4 \sinh (a+b x) \cosh ^{\frac{3}{2}}(a+b x)}{25 b^2}+\frac{2 x \cosh ^{\frac{5}{2}}(a+b x)}{5 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Cosh[a + b*x]^(3/2)*Sinh[a + b*x],x]

[Out]

(2*x*Cosh[a + b*x]^(5/2))/(5*b) + (((12*I)/25)*EllipticE[(I/2)*(a + b*x), 2])/b^2 - (4*Cosh[a + b*x]^(3/2)*Sin
h[a + b*x])/(25*b^2)

Rule 5373

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(x^(m -
n + 1)*Cosh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cosh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 2635

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

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 x \cosh ^{\frac{3}{2}}(a+b x) \sinh (a+b x) \, dx &=\frac{2 x \cosh ^{\frac{5}{2}}(a+b x)}{5 b}-\frac{2 \int \cosh ^{\frac{5}{2}}(a+b x) \, dx}{5 b}\\ &=\frac{2 x \cosh ^{\frac{5}{2}}(a+b x)}{5 b}-\frac{4 \cosh ^{\frac{3}{2}}(a+b x) \sinh (a+b x)}{25 b^2}-\frac{6 \int \sqrt{\cosh (a+b x)} \, dx}{25 b}\\ &=\frac{2 x \cosh ^{\frac{5}{2}}(a+b x)}{5 b}+\frac{12 i E\left (\left .\frac{1}{2} i (a+b x)\right |2\right )}{25 b^2}-\frac{4 \cosh ^{\frac{3}{2}}(a+b x) \sinh (a+b x)}{25 b^2}\\ \end{align*}

Mathematica [C]  time = 1.84358, size = 142, normalized size = 2.22 $\frac{e^{-3 (a+b x)} \left (48 e^{2 (a+b x)} \sqrt{e^{2 (a+b x)}+1} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 (a+b x)}\right )+\left (e^{2 (a+b x)}+1\right ) \left (2 (5 b x-12) e^{2 (a+b x)}+(5 b x-2) e^{4 (a+b x)}+5 b x+2\right )\right )}{50 \sqrt{2} b^2 \sqrt{e^{-a-b x}+e^{a+b x}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Cosh[a + b*x]^(3/2)*Sinh[a + b*x],x]

[Out]

((1 + E^(2*(a + b*x)))*(2 + 5*b*x + 2*E^(2*(a + b*x))*(-12 + 5*b*x) + E^(4*(a + b*x))*(-2 + 5*b*x)) + 48*E^(2*
(a + b*x))*Sqrt[1 + E^(2*(a + b*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^(2*(a + b*x))])/(50*Sqrt[2]*b^2*E^(3
*(a + b*x))*Sqrt[E^(-a - b*x) + E^(a + b*x)])

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int x \left ( \cosh \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}}\sinh \left ( bx+a \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x+a)^(3/2)*sinh(b*x+a),x)

[Out]

int(x*cosh(b*x+a)^(3/2)*sinh(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cosh \left (b x + a\right )^{\frac{3}{2}} \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^(3/2)*sinh(b*x+a),x, algorithm="maxima")

[Out]

integrate(x*cosh(b*x + a)^(3/2)*sinh(b*x + a), 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)^(3/2)*sinh(b*x+a),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)**(3/2)*sinh(b*x+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cosh \left (b x + a\right )^{\frac{3}{2}} \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x+a)^(3/2)*sinh(b*x+a),x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)^(3/2)*sinh(b*x + a), x)