3.2014 \(\int \frac{(a+\frac{b}{x^3})^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=38 \[ \frac{2 a \left (a+\frac{b}{x^3}\right )^{5/2}}{15 b^2}-\frac{2 \left (a+\frac{b}{x^3}\right )^{7/2}}{21 b^2} \]

[Out]

(2*a*(a + b/x^3)^(5/2))/(15*b^2) - (2*(a + b/x^3)^(7/2))/(21*b^2)

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Rubi [A]  time = 0.0226198, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{2 a \left (a+\frac{b}{x^3}\right )^{5/2}}{15 b^2}-\frac{2 \left (a+\frac{b}{x^3}\right )^{7/2}}{21 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^3)^(3/2)/x^7,x]

[Out]

(2*a*(a + b/x^3)^(5/2))/(15*b^2) - (2*(a + b/x^3)^(7/2))/(21*b^2)

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x^3}\right )^{3/2}}{x^7} \, dx &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int x (a+b x)^{3/2} \, dx,x,\frac{1}{x^3}\right )\right )\\ &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{a (a+b x)^{3/2}}{b}+\frac{(a+b x)^{5/2}}{b}\right ) \, dx,x,\frac{1}{x^3}\right )\right )\\ &=\frac{2 a \left (a+\frac{b}{x^3}\right )^{5/2}}{15 b^2}-\frac{2 \left (a+\frac{b}{x^3}\right )^{7/2}}{21 b^2}\\ \end{align*}

Mathematica [A]  time = 0.011973, size = 40, normalized size = 1.05 \[ \frac{2 \sqrt{a+\frac{b}{x^3}} \left (a x^3+b\right )^2 \left (2 a x^3-5 b\right )}{105 b^2 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^3)^(3/2)/x^7,x]

[Out]

(2*Sqrt[a + b/x^3]*(b + a*x^3)^2*(-5*b + 2*a*x^3))/(105*b^2*x^9)

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Maple [A]  time = 0.004, size = 39, normalized size = 1. \begin{align*}{\frac{ \left ( 2\,a{x}^{3}+2\,b \right ) \left ( 2\,a{x}^{3}-5\,b \right ) }{105\,{b}^{2}{x}^{6}} \left ({\frac{a{x}^{3}+b}{{x}^{3}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(a*x^3+b)*(2*a*x^3-5*b)*((a*x^3+b)/x^3)^(3/2)/b^2/x^6

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Maxima [A]  time = 0.955871, size = 41, normalized size = 1.08 \begin{align*} -\frac{2 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{7}{2}}}{21 \, b^{2}} + \frac{2 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{5}{2}} a}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(a + b/x^3)^(7/2)/b^2 + 2/15*(a + b/x^3)^(5/2)*a/b^2

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Fricas [A]  time = 1.48881, size = 115, normalized size = 3.03 \begin{align*} \frac{2 \,{\left (2 \, a^{3} x^{9} - a^{2} b x^{6} - 8 \, a b^{2} x^{3} - 5 \, b^{3}\right )} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{105 \, b^{2} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(2*a^3*x^9 - a^2*b*x^6 - 8*a*b^2*x^3 - 5*b^3)*sqrt((a*x^3 + b)/x^3)/(b^2*x^9)

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Sympy [B]  time = 2.33394, size = 371, normalized size = 9.76 \begin{align*} \frac{4 a^{\frac{15}{2}} b^{\frac{3}{2}} x^{12} \sqrt{\frac{a x^{3}}{b} + 1}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} + \frac{2 a^{\frac{13}{2}} b^{\frac{5}{2}} x^{9} \sqrt{\frac{a x^{3}}{b} + 1}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} - \frac{18 a^{\frac{11}{2}} b^{\frac{7}{2}} x^{6} \sqrt{\frac{a x^{3}}{b} + 1}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} - \frac{26 a^{\frac{9}{2}} b^{\frac{9}{2}} x^{3} \sqrt{\frac{a x^{3}}{b} + 1}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} - \frac{10 a^{\frac{7}{2}} b^{\frac{11}{2}} \sqrt{\frac{a x^{3}}{b} + 1}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} - \frac{4 a^{8} b x^{\frac{27}{2}}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} - \frac{4 a^{7} b^{2} x^{\frac{21}{2}}}{105 a^{\frac{9}{2}} b^{3} x^{\frac{27}{2}} + 105 a^{\frac{7}{2}} b^{4} x^{\frac{21}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**(15/2)*b**(3/2)*x**12*sqrt(a*x**3/b + 1)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) + 2*
a**(13/2)*b**(5/2)*x**9*sqrt(a*x**3/b + 1)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) - 18*a*
*(11/2)*b**(7/2)*x**6*sqrt(a*x**3/b + 1)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) - 26*a**(
9/2)*b**(9/2)*x**3*sqrt(a*x**3/b + 1)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) - 10*a**(7/2
)*b**(11/2)*sqrt(a*x**3/b + 1)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) - 4*a**8*b*x**(27/2
)/(105*a**(9/2)*b**3*x**(27/2) + 105*a**(7/2)*b**4*x**(21/2)) - 4*a**7*b**2*x**(21/2)/(105*a**(9/2)*b**3*x**(2
7/2) + 105*a**(7/2)*b**4*x**(21/2))

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Giac [B]  time = 1.21181, size = 105, normalized size = 2.76 \begin{align*} -\frac{2 \,{\left (\frac{7 \,{\left (3 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{5}{2}} - 5 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{3}{2}} a\right )} a}{b} + \frac{15 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{7}{2}} - 42 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{5}{2}} a + 35 \,{\left (a + \frac{b}{x^{3}}\right )}^{\frac{3}{2}} a^{2}}{b}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(7*(3*(a + b/x^3)^(5/2) - 5*(a + b/x^3)^(3/2)*a)*a/b + (15*(a + b/x^3)^(7/2) - 42*(a + b/x^3)^(5/2)*a +
35*(a + b/x^3)^(3/2)*a^2)/b)/b