### 3.544 $$\int x \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x) \, dx$$

Optimal. Leaf size=121 $\frac{20 i \sqrt{i \sinh (a+b x)} \text{EllipticF}\left (\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right ),2\right )}{147 b^2 \sqrt{\sinh (a+b x)}}-\frac{4 \sinh ^{\frac{5}{2}}(a+b x) \cosh (a+b x)}{49 b^2}+\frac{20 \sqrt{\sinh (a+b x)} \cosh (a+b x)}{147 b^2}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}$

[Out]

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

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Rubi [A]  time = 0.0702647, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {5372, 2635, 2642, 2641} $-\frac{4 \sinh ^{\frac{5}{2}}(a+b x) \cosh (a+b x)}{49 b^2}+\frac{20 \sqrt{\sinh (a+b x)} \cosh (a+b x)}{147 b^2}+\frac{20 i \sqrt{i \sinh (a+b x)} F\left (\left .\frac{1}{2} \left (i a+i b x-\frac{\pi }{2}\right )\right |2\right )}{147 b^2 \sqrt{\sinh (a+b x)}}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(((20*I)/147)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b^2*Sqrt[Sinh[a + b*x]]) + (20*Cosh
[a + b*x]*Sqrt[Sinh[a + b*x]])/(147*b^2) - (4*Cosh[a + b*x]*Sinh[a + b*x]^(5/2))/(49*b^2) + (2*x*Sinh[a + b*x]
^(7/2))/(7*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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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]

Rubi steps

\begin{align*} \int x \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x) \, dx &=\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{2 \int \sinh ^{\frac{7}{2}}(a+b x) \, dx}{7 b}\\ &=-\frac{4 \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x)}{49 b^2}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}+\frac{10 \int \sinh ^{\frac{3}{2}}(a+b x) \, dx}{49 b}\\ &=\frac{20 \cosh (a+b x) \sqrt{\sinh (a+b x)}}{147 b^2}-\frac{4 \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x)}{49 b^2}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{10 \int \frac{1}{\sqrt{\sinh (a+b x)}} \, dx}{147 b}\\ &=\frac{20 \cosh (a+b x) \sqrt{\sinh (a+b x)}}{147 b^2}-\frac{4 \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x)}{49 b^2}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}-\frac{\left (10 \sqrt{i \sinh (a+b x)}\right ) \int \frac{1}{\sqrt{i \sinh (a+b x)}} \, dx}{147 b \sqrt{\sinh (a+b x)}}\\ &=\frac{20 i F\left (\left .\frac{1}{2} \left (i a-\frac{\pi }{2}+i b x\right )\right |2\right ) \sqrt{i \sinh (a+b x)}}{147 b^2 \sqrt{\sinh (a+b x)}}+\frac{20 \cosh (a+b x) \sqrt{\sinh (a+b x)}}{147 b^2}-\frac{4 \cosh (a+b x) \sinh ^{\frac{5}{2}}(a+b x)}{49 b^2}+\frac{2 x \sinh ^{\frac{7}{2}}(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.310207, size = 103, normalized size = 0.85 $\frac{-80 i \sqrt{i \sinh (a+b x)} \text{EllipticF}\left (\frac{1}{4} (-2 i a-2 i b x+\pi ),2\right )+52 \sinh (2 (a+b x))-6 \sinh (4 (a+b x))-84 b x \cosh (2 (a+b x))+21 b x \cosh (4 (a+b x))+63 b x}{588 b^2 \sqrt{\sinh (a+b x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(63*b*x - 84*b*x*Cosh[2*(a + b*x)] + 21*b*x*Cosh[4*(a + b*x)] - (80*I)*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)/4
, 2]*Sqrt[I*Sinh[a + b*x]] + 52*Sinh[2*(a + b*x)] - 6*Sinh[4*(a + b*x)])/(588*b^2*Sqrt[Sinh[a + b*x]])

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Maple [F]  time = 0.027, 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 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 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)