3.395 \(\int \frac {(a+b x^2)^{5/2}}{x^9} \, dx\)

Optimal. Leaf size=113 \[ \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{3/2}}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6} \]

[Out]

-5/48*b*(b*x^2+a)^(3/2)/x^6-1/8*(b*x^2+a)^(5/2)/x^8+5/128*b^4*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(3/2)-5/64*b^
2*(b*x^2+a)^(1/2)/x^4-5/128*b^3*(b*x^2+a)^(1/2)/a/x^2

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Rubi [A]  time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \[ \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{3/2}}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^9,x]

[Out]

(-5*b^2*Sqrt[a + b*x^2])/(64*x^4) - (5*b^3*Sqrt[a + b*x^2])/(128*a*x^2) - (5*b*(a + b*x^2)^(3/2))/(48*x^6) - (
a + b*x^2)^(5/2)/(8*x^8) + (5*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(3/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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+b x^2\right )^{5/2}}{x^9} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}+\frac {1}{16} (5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}+\frac {1}{32} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}+\frac {1}{128} \left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}-\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{256 a}\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{128 a}\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{64 x^4}-\frac {5 b^3 \sqrt {a+b x^2}}{128 a x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{48 x^6}-\frac {\left (a+b x^2\right )^{5/2}}{8 x^8}+\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.35 \[ -\frac {b^4 \left (a+b x^2\right )^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {b x^2}{a}+1\right )}{7 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^9,x]

[Out]

-1/7*(b^4*(a + b*x^2)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, 1 + (b*x^2)/a])/a^5

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fricas [A]  time = 0.86, size = 179, normalized size = 1.58 \[ \left [\frac {15 \, \sqrt {a} b^{4} x^{8} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{3} x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a^{3} b x^{2} + 48 \, a^{4}\right )} \sqrt {b x^{2} + a}}{768 \, a^{2} x^{8}}, -\frac {15 \, \sqrt {-a} b^{4} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, a b^{3} x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a^{3} b x^{2} + 48 \, a^{4}\right )} \sqrt {b x^{2} + a}}{384 \, a^{2} x^{8}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="fricas")

[Out]

[1/768*(15*sqrt(a)*b^4*x^8*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(15*a*b^3*x^6 + 118*a^2*b^2
*x^4 + 136*a^3*b*x^2 + 48*a^4)*sqrt(b*x^2 + a))/(a^2*x^8), -1/384*(15*sqrt(-a)*b^4*x^8*arctan(sqrt(-a)/sqrt(b*
x^2 + a)) + (15*a*b^3*x^6 + 118*a^2*b^2*x^4 + 136*a^3*b*x^2 + 48*a^4)*sqrt(b*x^2 + a))/(a^2*x^8)]

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giac [A]  time = 1.11, size = 109, normalized size = 0.96 \[ -\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x^{2} + a} a^{3} b^{5}}{a b^{4} x^{8}}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="giac")

[Out]

-1/384*(15*b^5*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) + (15*(b*x^2 + a)^(7/2)*b^5 + 73*(b*x^2 + a)^(5/2
)*a*b^5 - 55*(b*x^2 + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x^2 + a)*a^3*b^5)/(a*b^4*x^8))/b

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maple [A]  time = 0.02, size = 159, normalized size = 1.41 \[ \frac {5 b^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {3}{2}}}-\frac {5 \sqrt {b \,x^{2}+a}\, b^{4}}{128 a^{2}}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{4}}{384 a^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{4}}{128 a^{4}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{3}}{128 a^{4} x^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}{192 a^{3} x^{4}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b}{48 a^{2} x^{6}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^9,x)

[Out]

-1/8/a/x^8*(b*x^2+a)^(7/2)+1/48/a^2*b/x^6*(b*x^2+a)^(7/2)+1/192/a^3*b^2/x^4*(b*x^2+a)^(7/2)+1/128/a^4*b^3/x^2*
(b*x^2+a)^(7/2)-1/128/a^4*b^4*(b*x^2+a)^(5/2)-5/384/a^3*b^4*(b*x^2+a)^(3/2)+5/128/a^(3/2)*b^4*ln((2*a+2*(b*x^2
+a)^(1/2)*a^(1/2))/x)-5/128/a^2*b^4*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.38, size = 147, normalized size = 1.30 \[ \frac {5 \, b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4}}{128 \, a^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{384 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} b^{4}}{128 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{192 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{8 \, a x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^9,x, algorithm="maxima")

[Out]

5/128*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 1/128*(b*x^2 + a)^(5/2)*b^4/a^4 - 5/384*(b*x^2 + a)^(3/2)*b^
4/a^3 - 5/128*sqrt(b*x^2 + a)*b^4/a^2 + 1/128*(b*x^2 + a)^(7/2)*b^3/(a^4*x^2) + 1/192*(b*x^2 + a)^(7/2)*b^2/(a
^3*x^4) + 1/48*(b*x^2 + a)^(7/2)*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(7/2)/(a*x^8)

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mupad [B]  time = 5.43, size = 89, normalized size = 0.79 \[ \frac {55\,a\,{\left (b\,x^2+a\right )}^{3/2}}{384\,x^8}-\frac {73\,{\left (b\,x^2+a\right )}^{5/2}}{384\,x^8}-\frac {5\,a^2\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^8}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^9,x)

[Out]

(55*a*(a + b*x^2)^(3/2))/(384*x^8) - (b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(3/2)) - (73*(a + b*
x^2)^(5/2))/(384*x^8) - (5*a^2*(a + b*x^2)^(1/2))/(128*x^8) - (5*(a + b*x^2)^(7/2))/(128*a*x^8)

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sympy [A]  time = 7.58, size = 150, normalized size = 1.33 \[ - \frac {a^{3}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {23 a^{2} \sqrt {b}}{48 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {127 a b^{\frac {3}{2}}}{192 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {133 b^{\frac {5}{2}}}{384 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 b^{\frac {7}{2}}}{128 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**9,x)

[Out]

-a**3/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 23*a**2*sqrt(b)/(48*x**7*sqrt(a/(b*x**2) + 1)) - 127*a*b**(3/2)/
(192*x**5*sqrt(a/(b*x**2) + 1)) - 133*b**(5/2)/(384*x**3*sqrt(a/(b*x**2) + 1)) - 5*b**(7/2)/(128*a*x*sqrt(a/(b
*x**2) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(3/2))

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