3.404 \(\int \frac {(a+b x^2)^{5/2}}{x^{10}} \, dx\)

Optimal. Leaf size=44 \[ \frac {2 b \left (a+b x^2\right )^{7/2}}{63 a^2 x^7}-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9} \]

[Out]

-1/9*(b*x^2+a)^(7/2)/a/x^9+2/63*b*(b*x^2+a)^(7/2)/a^2/x^7

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {271, 264} \[ \frac {2 b \left (a+b x^2\right )^{7/2}}{63 a^2 x^7}-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^10,x]

[Out]

-(a + b*x^2)^(7/2)/(9*a*x^9) + (2*b*(a + b*x^2)^(7/2))/(63*a^2*x^7)

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]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{5/2}}{x^{10}} \, dx &=-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9}-\frac {(2 b) \int \frac {\left (a+b x^2\right )^{5/2}}{x^8} \, dx}{9 a}\\ &=-\frac {\left (a+b x^2\right )^{7/2}}{9 a x^9}+\frac {2 b \left (a+b x^2\right )^{7/2}}{63 a^2 x^7}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.70 \[ \frac {\left (a+b x^2\right )^{7/2} \left (2 b x^2-7 a\right )}{63 a^2 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^10,x]

[Out]

((a + b*x^2)^(7/2)*(-7*a + 2*b*x^2))/(63*a^2*x^9)

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fricas [A]  time = 0.68, size = 60, normalized size = 1.36 \[ \frac {{\left (2 \, b^{4} x^{8} - a b^{3} x^{6} - 15 \, a^{2} b^{2} x^{4} - 19 \, a^{3} b x^{2} - 7 \, a^{4}\right )} \sqrt {b x^{2} + a}}{63 \, a^{2} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^10,x, algorithm="fricas")

[Out]

1/63*(2*b^4*x^8 - a*b^3*x^6 - 15*a^2*b^2*x^4 - 19*a^3*b*x^2 - 7*a^4)*sqrt(b*x^2 + a)/(a^2*x^9)

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giac [B]  time = 1.14, size = 220, normalized size = 5.00 \[ \frac {4 \, {\left (63 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{14} b^{\frac {9}{2}} + 105 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} a b^{\frac {9}{2}} + 315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} a^{2} b^{\frac {9}{2}} + 189 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{3} b^{\frac {9}{2}} + 189 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{4} b^{\frac {9}{2}} + 27 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{5} b^{\frac {9}{2}} + 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{6} b^{\frac {9}{2}} - a^{7} b^{\frac {9}{2}}\right )}}{63 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^10,x, algorithm="giac")

[Out]

4/63*(63*(sqrt(b)*x - sqrt(b*x^2 + a))^14*b^(9/2) + 105*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a*b^(9/2) + 315*(sqrt
(b)*x - sqrt(b*x^2 + a))^10*a^2*b^(9/2) + 189*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(9/2) + 189*(sqrt(b)*x - s
qrt(b*x^2 + a))^6*a^4*b^(9/2) + 27*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^5*b^(9/2) + 9*(sqrt(b)*x - sqrt(b*x^2 + a
))^2*a^6*b^(9/2) - a^7*b^(9/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^9

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maple [A]  time = 0.01, size = 28, normalized size = 0.64 \[ -\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-2 b \,x^{2}+7 a \right )}{63 a^{2} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^10,x)

[Out]

-1/63*(b*x^2+a)^(7/2)*(-2*b*x^2+7*a)/x^9/a^2

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maxima [A]  time = 1.53, size = 36, normalized size = 0.82 \[ \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{63 \, a^{2} x^{7}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{9 \, a x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^10,x, algorithm="maxima")

[Out]

2/63*(b*x^2 + a)^(7/2)*b/(a^2*x^7) - 1/9*(b*x^2 + a)^(7/2)/(a*x^9)

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mupad [B]  time = 5.37, size = 91, normalized size = 2.07 \[ \frac {2\,b^4\,\sqrt {b\,x^2+a}}{63\,a^2\,x}-\frac {5\,b^2\,\sqrt {b\,x^2+a}}{21\,x^5}-\frac {b^3\,\sqrt {b\,x^2+a}}{63\,a\,x^3}-\frac {a^2\,\sqrt {b\,x^2+a}}{9\,x^9}-\frac {19\,a\,b\,\sqrt {b\,x^2+a}}{63\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^10,x)

[Out]

(2*b^4*(a + b*x^2)^(1/2))/(63*a^2*x) - (5*b^2*(a + b*x^2)^(1/2))/(21*x^5) - (b^3*(a + b*x^2)^(1/2))/(63*a*x^3)
- (a^2*(a + b*x^2)^(1/2))/(9*x^9) - (19*a*b*(a + b*x^2)^(1/2))/(63*x^7)

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sympy [B]  time = 1.59, size = 121, normalized size = 2.75 \[ - \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{9 x^{8}} - \frac {19 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{63 x^{6}} - \frac {5 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{21 x^{4}} - \frac {b^{\frac {7}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{63 a x^{2}} + \frac {2 b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{63 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**10,x)

[Out]

-a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(9*x**8) - 19*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(63*x**6) - 5*b**(5/2)*sqrt(a
/(b*x**2) + 1)/(21*x**4) - b**(7/2)*sqrt(a/(b*x**2) + 1)/(63*a*x**2) + 2*b**(9/2)*sqrt(a/(b*x**2) + 1)/(63*a**
2)

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