3.1761 \(\int (a+\frac{b}{x})^{3/2} x^{7/2} \, dx\)

Optimal. Leaf size=74 \[ \frac{16 b^2 x^{5/2} \left (a+\frac{b}{x}\right )^{5/2}}{315 a^3}-\frac{8 b x^{7/2} \left (a+\frac{b}{x}\right )^{5/2}}{63 a^2}+\frac{2 x^{9/2} \left (a+\frac{b}{x}\right )^{5/2}}{9 a} \]

[Out]

(16*b^2*(a + b/x)^(5/2)*x^(5/2))/(315*a^3) - (8*b*(a + b/x)^(5/2)*x^(7/2))/(63*a^2) + (2*(a + b/x)^(5/2)*x^(9/
2))/(9*a)

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Rubi [A]  time = 0.0226424, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {271, 264} \[ \frac{16 b^2 x^{5/2} \left (a+\frac{b}{x}\right )^{5/2}}{315 a^3}-\frac{8 b x^{7/2} \left (a+\frac{b}{x}\right )^{5/2}}{63 a^2}+\frac{2 x^{9/2} \left (a+\frac{b}{x}\right )^{5/2}}{9 a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*x^(7/2),x]

[Out]

(16*b^2*(a + b/x)^(5/2)*x^(5/2))/(315*a^3) - (8*b*(a + b/x)^(5/2)*x^(7/2))/(63*a^2) + (2*(a + b/x)^(5/2)*x^(9/
2))/(9*a)

Rule 271

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

Rule 264

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

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x}\right )^{3/2} x^{7/2} \, dx &=\frac{2 \left (a+\frac{b}{x}\right )^{5/2} x^{9/2}}{9 a}-\frac{(4 b) \int \left (a+\frac{b}{x}\right )^{3/2} x^{5/2} \, dx}{9 a}\\ &=-\frac{8 b \left (a+\frac{b}{x}\right )^{5/2} x^{7/2}}{63 a^2}+\frac{2 \left (a+\frac{b}{x}\right )^{5/2} x^{9/2}}{9 a}+\frac{\left (8 b^2\right ) \int \left (a+\frac{b}{x}\right )^{3/2} x^{3/2} \, dx}{63 a^2}\\ &=\frac{16 b^2 \left (a+\frac{b}{x}\right )^{5/2} x^{5/2}}{315 a^3}-\frac{8 b \left (a+\frac{b}{x}\right )^{5/2} x^{7/2}}{63 a^2}+\frac{2 \left (a+\frac{b}{x}\right )^{5/2} x^{9/2}}{9 a}\\ \end{align*}

Mathematica [A]  time = 0.015161, size = 49, normalized size = 0.66 \[ \frac{2 \sqrt{x} \sqrt{a+\frac{b}{x}} (a x+b)^2 \left (35 a^2 x^2-20 a b x+8 b^2\right )}{315 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(3/2)*x^(7/2),x]

[Out]

(2*Sqrt[a + b/x]*Sqrt[x]*(b + a*x)^2*(8*b^2 - 20*a*b*x + 35*a^2*x^2))/(315*a^3)

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Maple [A]  time = 0.004, size = 44, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,ax+2\,b \right ) \left ( 35\,{a}^{2}{x}^{2}-20\,xab+8\,{b}^{2} \right ) }{315\,{a}^{3}}{x}^{{\frac{3}{2}}} \left ({\frac{ax+b}{x}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(3/2)*x^(7/2),x)

[Out]

2/315*(a*x+b)*(35*a^2*x^2-20*a*b*x+8*b^2)*x^(3/2)*((a*x+b)/x)^(3/2)/a^3

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Maxima [A]  time = 0.976777, size = 70, normalized size = 0.95 \begin{align*} \frac{2 \,{\left (35 \,{\left (a + \frac{b}{x}\right )}^{\frac{9}{2}} x^{\frac{9}{2}} - 90 \,{\left (a + \frac{b}{x}\right )}^{\frac{7}{2}} b x^{\frac{7}{2}} + 63 \,{\left (a + \frac{b}{x}\right )}^{\frac{5}{2}} b^{2} x^{\frac{5}{2}}\right )}}{315 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(3/2)*x^(7/2),x, algorithm="maxima")

[Out]

2/315*(35*(a + b/x)^(9/2)*x^(9/2) - 90*(a + b/x)^(7/2)*b*x^(7/2) + 63*(a + b/x)^(5/2)*b^2*x^(5/2))/a^3

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Fricas [A]  time = 1.52703, size = 136, normalized size = 1.84 \begin{align*} \frac{2 \,{\left (35 \, a^{4} x^{4} + 50 \, a^{3} b x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{315 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(3/2)*x^(7/2),x, algorithm="fricas")

[Out]

2/315*(35*a^4*x^4 + 50*a^3*b*x^3 + 3*a^2*b^2*x^2 - 4*a*b^3*x + 8*b^4)*sqrt(x)*sqrt((a*x + b)/x)/a^3

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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((a+b/x)**(3/2)*x**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.22338, size = 154, normalized size = 2.08 \begin{align*} -\frac{2}{105} \, b{\left (\frac{8 \, b^{\frac{7}{2}}}{a^{3}} - \frac{15 \,{\left (a x + b\right )}^{\frac{7}{2}} - 42 \,{\left (a x + b\right )}^{\frac{5}{2}} b + 35 \,{\left (a x + b\right )}^{\frac{3}{2}} b^{2}}{a^{3}}\right )} \mathrm{sgn}\left (x\right ) + \frac{2}{315} \, a{\left (\frac{16 \, b^{\frac{9}{2}}}{a^{4}} + \frac{35 \,{\left (a x + b\right )}^{\frac{9}{2}} - 135 \,{\left (a x + b\right )}^{\frac{7}{2}} b + 189 \,{\left (a x + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (a x + b\right )}^{\frac{3}{2}} b^{3}}{a^{4}}\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(3/2)*x^(7/2),x, algorithm="giac")

[Out]

-2/105*b*(8*b^(7/2)/a^3 - (15*(a*x + b)^(7/2) - 42*(a*x + b)^(5/2)*b + 35*(a*x + b)^(3/2)*b^2)/a^3)*sgn(x) + 2
/315*a*(16*b^(9/2)/a^4 + (35*(a*x + b)^(9/2) - 135*(a*x + b)^(7/2)*b + 189*(a*x + b)^(5/2)*b^2 - 105*(a*x + b)
^(3/2)*b^3)/a^4)*sgn(x)