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

Optimal. Leaf size=87 \[ \frac {20 i F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{147 b^2}-\frac {4 \sinh (a+b x) \cosh ^{\frac {5}{2}}(a+b x)}{49 b^2}-\frac {20 \sinh (a+b x) \sqrt {\cosh (a+b x)}}{147 b^2}+\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b} \]

[Out]

2/7*x*cosh(b*x+a)^(7/2)/b+20/147*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/cosh(1/2*a+1/2*b*x)*EllipticF(I*sinh(1/2*a+1/
2*b*x),2^(1/2))/b^2-4/49*cosh(b*x+a)^(5/2)*sinh(b*x+a)/b^2-20/147*sinh(b*x+a)*cosh(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5373, 2635, 2641} \[ \frac {20 i F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{147 b^2}-\frac {4 \sinh (a+b x) \cosh ^{\frac {5}{2}}(a+b x)}{49 b^2}-\frac {20 \sinh (a+b x) \sqrt {\cosh (a+b x)}}{147 b^2}+\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Cosh[a + b*x]^(7/2))/(7*b) + (((20*I)/147)*EllipticF[(I/2)*(a + b*x), 2])/b^2 - (20*Sqrt[Cosh[a + b*x]]*S
inh[a + b*x])/(147*b^2) - (4*Cosh[a + b*x]^(5/2)*Sinh[a + b*x])/(49*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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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 {5}{2}}(a+b x) \sinh (a+b x) \, dx &=\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \int \cosh ^{\frac {7}{2}}(a+b x) \, dx}{7 b}\\ &=\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {4 \cosh ^{\frac {5}{2}}(a+b x) \sinh (a+b x)}{49 b^2}-\frac {10 \int \cosh ^{\frac {3}{2}}(a+b x) \, dx}{49 b}\\ &=\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {20 \sqrt {\cosh (a+b x)} \sinh (a+b x)}{147 b^2}-\frac {4 \cosh ^{\frac {5}{2}}(a+b x) \sinh (a+b x)}{49 b^2}-\frac {10 \int \frac {1}{\sqrt {\cosh (a+b x)}} \, dx}{147 b}\\ &=\frac {2 x \cosh ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {20 i F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{147 b^2}-\frac {20 \sqrt {\cosh (a+b x)} \sinh (a+b x)}{147 b^2}-\frac {4 \cosh ^{\frac {5}{2}}(a+b x) \sinh (a+b x)}{49 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 77, normalized size = 0.89 \[ \frac {\sqrt {\cosh (a+b x)} (-46 \sinh (a+b x)-6 \sinh (3 (a+b x))+63 b x \cosh (a+b x)+21 b x \cosh (3 (a+b x)))+40 i F\left (\left .\frac {1}{2} i (a+b x)\right |2\right )}{294 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((40*I)*EllipticF[(I/2)*(a + b*x), 2] + Sqrt[Cosh[a + b*x]]*(63*b*x*Cosh[a + b*x] + 21*b*x*Cosh[3*(a + b*x)] -
46*Sinh[a + b*x] - 6*Sinh[3*(a + b*x)]))/(294*b^2)

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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)^(5/2)*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 \[ \int x \cosh \left (b x + a\right )^{\frac {5}{2}} \sinh \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int x \left (\cosh ^{\frac {5}{2}}\left (b x +a \right )\right ) \sinh \left (b x +a \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cosh \left (b x + a\right )^{\frac {5}{2}} \sinh \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {cosh}\left (a+b\,x\right )}^{5/2}\,\mathrm {sinh}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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