3.384 \(\int \frac {(a+b x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=44 \[ \frac {2 b \left (a+b x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7} \]

[Out]

-1/7*(b*x^2+a)^(5/2)/a/x^7+2/35*b*(b*x^2+a)^(5/2)/a^2/x^5

________________________________________________________________________________________

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} \[ \frac {2 b \left (a+b x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^8,x]

[Out]

-(a + b*x^2)^(5/2)/(7*a*x^7) + (2*b*(a + b*x^2)^(5/2))/(35*a^2*x^5)

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 )^{3/2}}{x^8} \, dx &=-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}-\frac {(2 b) \int \frac {\left (a+b x^2\right )^{3/2}}{x^6} \, dx}{7 a}\\ &=-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}+\frac {2 b \left (a+b x^2\right )^{5/2}}{35 a^2 x^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 31, normalized size = 0.70 \[ \frac {\left (a+b x^2\right )^{5/2} \left (2 b x^2-5 a\right )}{35 a^2 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^8,x]

[Out]

((a + b*x^2)^(5/2)*(-5*a + 2*b*x^2))/(35*a^2*x^7)

________________________________________________________________________________________

fricas [A]  time = 1.32, size = 49, normalized size = 1.11 \[ \frac {{\left (2 \, b^{3} x^{6} - a b^{2} x^{4} - 8 \, a^{2} b x^{2} - 5 \, a^{3}\right )} \sqrt {b x^{2} + a}}{35 \, a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^8,x, algorithm="fricas")

[Out]

1/35*(2*b^3*x^6 - a*b^2*x^4 - 8*a^2*b*x^2 - 5*a^3)*sqrt(b*x^2 + a)/(a^2*x^7)

________________________________________________________________________________________

giac [B]  time = 0.69, size = 166, normalized size = 3.77 \[ \frac {4 \, {\left (35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} b^{\frac {7}{2}} + 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a b^{\frac {7}{2}} + 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} b^{\frac {7}{2}} + 14 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} b^{\frac {7}{2}} + 7 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} b^{\frac {7}{2}} - a^{5} b^{\frac {7}{2}}\right )}}{35 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^8,x, algorithm="giac")

[Out]

4/35*(35*(sqrt(b)*x - sqrt(b*x^2 + a))^10*b^(7/2) + 35*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(7/2) + 70*(sqrt(b)
*x - sqrt(b*x^2 + a))^6*a^2*b^(7/2) + 14*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*b^(7/2) + 7*(sqrt(b)*x - sqrt(b*x
^2 + a))^2*a^4*b^(7/2) - a^5*b^(7/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^7

________________________________________________________________________________________

maple [A]  time = 0.00, size = 28, normalized size = 0.64 \[ -\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (-2 b \,x^{2}+5 a \right )}{35 a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/x^8,x)

[Out]

-1/35*(b*x^2+a)^(5/2)*(-2*b*x^2+5*a)/x^7/a^2

________________________________________________________________________________________

maxima [A]  time = 1.39, size = 36, normalized size = 0.82 \[ \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{35 \, a^{2} x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{7 \, a x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^8,x, algorithm="maxima")

[Out]

2/35*(b*x^2 + a)^(5/2)*b/(a^2*x^5) - 1/7*(b*x^2 + a)^(5/2)/(a*x^7)

________________________________________________________________________________________

mupad [B]  time = 5.65, size = 71, normalized size = 1.61 \[ \frac {2\,b^3\,\sqrt {b\,x^2+a}}{35\,a^2\,x}-\frac {8\,b\,\sqrt {b\,x^2+a}}{35\,x^5}-\frac {b^2\,\sqrt {b\,x^2+a}}{35\,a\,x^3}-\frac {a\,\sqrt {b\,x^2+a}}{7\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x^8,x)

[Out]

(2*b^3*(a + b*x^2)^(1/2))/(35*a^2*x) - (8*b*(a + b*x^2)^(1/2))/(35*x^5) - (b^2*(a + b*x^2)^(1/2))/(35*a*x^3) -
(a*(a + b*x^2)^(1/2))/(7*x^7)

________________________________________________________________________________________

sympy [B]  time = 1.18, size = 94, normalized size = 2.14 \[ - \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{7 x^{6}} - \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{35 x^{4}} - \frac {b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a x^{2}} + \frac {2 b^{\frac {7}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**8,x)

[Out]

-a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(7*x**6) - 8*b**(3/2)*sqrt(a/(b*x**2) + 1)/(35*x**4) - b**(5/2)*sqrt(a/(b*x**2
) + 1)/(35*a*x**2) + 2*b**(7/2)*sqrt(a/(b*x**2) + 1)/(35*a**2)

________________________________________________________________________________________