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]

2/5*x*cosh(b*x+a)^(5/2)/b+12/25*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*a+1/2*b*x)*EllipticE(I*sinh(1/2*a+1/2
*b*x),2^(1/2))/b^2-4/25*cosh(b*x+a)^(3/2)*sinh(b*x+a)/b^2

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Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 2.01, 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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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)^(3/2)*sinh(b*x+a),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh \left (b x + a\right )^{\frac {3}{2}} \sinh \left (b x + a\right )\,{d x} \]

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)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x \left (\cosh ^{\frac {3}{2}}\left (b x +a \right )\right ) \sinh \left (b x +a \right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh \left (b x + a\right )^{\frac {3}{2}} \sinh \left (b x + a\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,{\mathrm {cosh}\left (a+b\,x\right )}^{3/2}\,\mathrm {sinh}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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