### 3.545 $$\int x \cosh (a+b x) \sinh ^{\frac{3}{2}}(a+b x) \, dx$$

Optimal. Leaf size=98 $-\frac{4 \sinh ^{\frac{3}{2}}(a+b x) \cosh (a+b x)}{25 b^2}-\frac{12 i \sqrt{\sinh (a+b x)} E\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{25 b^2 \sqrt{i \sinh (a+b x)}}+\frac{2 x \sinh ^{\frac{5}{2}}(a+b x)}{5 b}$

[Out]

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

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Rubi [A]  time = 0.0524617, 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, 2635, 2640, 2639} $-\frac{4 \sinh ^{\frac{3}{2}}(a+b x) \cosh (a+b x)}{25 b^2}-\frac{12 i \sqrt{\sinh (a+b x)} E\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{25 b^2 \sqrt{i \sinh (a+b x)}}+\frac{2 x \sinh ^{\frac{5}{2}}(a+b x)}{5 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.1644, size = 143, normalized size = 1.46 $\frac{e^{-3 (a+b x)} \left (48 e^{2 (a+b x)} \sqrt{1-e^{2 (a+b x)}} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};e^{2 (a+b x)}\right )+\left (e^{2 (a+b x)}-1\right ) \left ((24-10 b x) e^{2 (a+b x)}+(5 b x-2) e^{4 (a+b x)}+5 b x+2\right )\right )}{50 \sqrt{2} b^2 \sqrt{e^{a+b x}-e^{-a-b x}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-1 + E^(2*(a + b*x)))*(2 + 5*b*x + E^(2*(a + b*x))*(24 - 10*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 = 0.028, size = 0, normalized size = 0. \begin{align*} \int x\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{{\frac{3}{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)^(3/2),x)

[Out]

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

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

[Out]

integrate(x*cosh(b*x + a)*sinh(b*x + a)^(3/2), 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)^(3/2),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)**(3/2),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 ) \sinh \left (b x + a\right )^{\frac{3}{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)^(3/2),x, algorithm="giac")

[Out]

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