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]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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])

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]]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.84, size = 53, normalized size = 1.39 \[ \frac {2 \, {\left (2 \, a^{3} x^{9} - a^{2} b x^{6} - 8 \, a b^{2} x^{3} - 5 \, b^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{105 \, b^{2} x^{9}} \]

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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giac [A]  time = 0.20, size = 29, normalized size = 0.76 \[ -\frac {2 \, {\left (5 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {7}{2}} - 7 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {5}{2}} a\right )}}{105 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(5*(a + b/x^3)^(7/2) - 7*(a + b/x^3)^(5/2)*a)/b^2

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maple [A]  time = 0.01, size = 39, normalized size = 1.03 \[ \frac {2 \left (a \,x^{3}+b \right ) \left (2 a \,x^{3}-5 b \right ) \left (\frac {a \,x^{3}+b}{x^{3}}\right )^{\frac {3}{2}}}{105 b^{2} x^{6}} \]

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 = 0.87, size = 30, normalized size = 0.79 \[ -\frac {2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {7}{2}}}{21 \, b^{2}} + \frac {2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {5}{2}} a}{15 \, b^{2}} \]

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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mupad [B]  time = 2.36, size = 68, normalized size = 1.79 \[ \frac {4\,a^3\,\sqrt {a+\frac {b}{x^3}}}{105\,b^2}-\frac {2\,b\,\sqrt {a+\frac {b}{x^3}}}{21\,x^9}-\frac {16\,a\,\sqrt {a+\frac {b}{x^3}}}{105\,x^6}-\frac {2\,a^2\,\sqrt {a+\frac {b}{x^3}}}{105\,b\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^3)^(3/2)/x^7,x)

[Out]

(4*a^3*(a + b/x^3)^(1/2))/(105*b^2) - (2*b*(a + b/x^3)^(1/2))/(21*x^9) - (16*a*(a + b/x^3)^(1/2))/(105*x^6) -
(2*a^2*(a + b/x^3)^(1/2))/(105*b*x^3)

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sympy [B]  time = 2.05, size = 371, normalized size = 9.76 \[ \frac {4 a^{\frac {15}{2}} b^{\frac {3}{2}} x^{12} \sqrt {\frac {a x^{3}}{b} + 1}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} + \frac {2 a^{\frac {13}{2}} b^{\frac {5}{2}} x^{9} \sqrt {\frac {a x^{3}}{b} + 1}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} - \frac {18 a^{\frac {11}{2}} b^{\frac {7}{2}} x^{6} \sqrt {\frac {a x^{3}}{b} + 1}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} - \frac {26 a^{\frac {9}{2}} b^{\frac {9}{2}} x^{3} \sqrt {\frac {a x^{3}}{b} + 1}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} - \frac {10 a^{\frac {7}{2}} b^{\frac {11}{2}} \sqrt {\frac {a x^{3}}{b} + 1}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} - \frac {4 a^{8} b x^{\frac {27}{2}}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} - \frac {4 a^{7} b^{2} x^{\frac {21}{2}}}{105 a^{\frac {9}{2}} b^{3} x^{\frac {27}{2}} + 105 a^{\frac {7}{2}} b^{4} x^{\frac {21}{2}}} \]

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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