3.377 \(\int \frac {(a+b x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=116 \[ -\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}}+\frac {3 b^3 \sqrt {a+b x^2}}{128 a^2 x^2}-\frac {b^2 \sqrt {a+b x^2}}{64 a x^4}-\frac {\left (a+b x^2\right )^{3/2}}{8 x^8}-\frac {b \sqrt {a+b x^2}}{16 x^6} \]

[Out]

-1/8*(b*x^2+a)^(3/2)/x^8-3/128*b^4*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)-1/16*b*(b*x^2+a)^(1/2)/x^6-1/64*b^
2*(b*x^2+a)^(1/2)/a/x^4+3/128*b^3*(b*x^2+a)^(1/2)/a^2/x^2

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Rubi [A]  time = 0.07, antiderivative size = 116, 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} \[ \frac {3 b^3 \sqrt {a+b x^2}}{128 a^2 x^2}-\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}}-\frac {b^2 \sqrt {a+b x^2}}{64 a x^4}-\frac {b \sqrt {a+b x^2}}{16 x^6}-\frac {\left (a+b x^2\right )^{3/2}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^9,x]

[Out]

-(b*Sqrt[a + b*x^2])/(16*x^6) - (b^2*Sqrt[a + b*x^2])/(64*a*x^4) + (3*b^3*Sqrt[a + b*x^2])/(128*a^2*x^2) - (a
+ b*x^2)^(3/2)/(8*x^8) - (3*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(5/2))

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

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.34 \[ -\frac {b^4 \left (a+b x^2\right )^{5/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b x^2}{a}+1\right )}{5 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^9,x]

[Out]

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

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fricas [A]  time = 0.64, size = 179, normalized size = 1.54 \[ \left [\frac {3 \, \sqrt {a} b^{4} x^{8} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{3} x^{6} - 2 \, a^{2} b^{2} x^{4} - 24 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{256 \, a^{3} x^{8}}, \frac {3 \, \sqrt {-a} b^{4} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, a b^{3} x^{6} - 2 \, a^{2} b^{2} x^{4} - 24 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt {b x^{2} + a}}{128 \, a^{3} x^{8}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="fricas")

[Out]

[1/256*(3*sqrt(a)*b^4*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*a*b^3*x^6 - 2*a^2*b^2*x^4
- 24*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/(a^3*x^8), 1/128*(3*sqrt(-a)*b^4*x^8*arctan(sqrt(-a)/sqrt(b*x^2 + a
)) + (3*a*b^3*x^6 - 2*a^2*b^2*x^4 - 24*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/(a^3*x^8)]

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giac [A]  time = 0.65, size = 109, normalized size = 0.94 \[ \frac {\frac {3 \, b^{5} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5} - 11 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{5} - 11 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 3 \, \sqrt {b x^{2} + a} a^{3} b^{5}}{a^{2} b^{4} x^{8}}}{128 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="giac")

[Out]

1/128*(3*b^5*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x^2 + a)^(7/2)*b^5 - 11*(b*x^2 + a)^(5/2)
*a*b^5 - 11*(b*x^2 + a)^(3/2)*a^2*b^5 + 3*sqrt(b*x^2 + a)*a^3*b^5)/(a^2*b^4*x^8))/b

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maple [A]  time = 0.02, size = 142, normalized size = 1.22 \[ -\frac {3 b^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {5}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, b^{4}}{128 a^{3}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{4}}{128 a^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3}}{128 a^{4} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2}}{64 a^{3} x^{4}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b}{16 a^{2} x^{6}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/x^9,x)

[Out]

-1/8/a/x^8*(b*x^2+a)^(5/2)+1/16/a^2*b/x^6*(b*x^2+a)^(5/2)-1/64/a^3*b^2/x^4*(b*x^2+a)^(5/2)-1/128/a^4*b^3/x^2*(
b*x^2+a)^(5/2)+1/128/a^4*b^4*(b*x^2+a)^(3/2)-3/128/a^(5/2)*b^4*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+3/128/a^3
*b^4*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.38, size = 130, normalized size = 1.12 \[ -\frac {3 \, b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{128 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} b^{4}}{128 \, a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{128 \, a^{4} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{64 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{8 \, a x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^9,x, algorithm="maxima")

[Out]

-3/128*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 1/128*(b*x^2 + a)^(3/2)*b^4/a^4 + 3/128*sqrt(b*x^2 + a)*b^4
/a^3 - 1/128*(b*x^2 + a)^(5/2)*b^3/(a^4*x^2) - 1/64*(b*x^2 + a)^(5/2)*b^2/(a^3*x^4) + 1/16*(b*x^2 + a)^(5/2)*b
/(a^2*x^6) - 1/8*(b*x^2 + a)^(5/2)/(a*x^8)

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mupad [B]  time = 5.21, size = 89, normalized size = 0.77 \[ \frac {3\,a\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {11\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}-\frac {11\,{\left (b\,x^2+a\right )}^{5/2}}{128\,a\,x^8}+\frac {3\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^8}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{128\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x^9,x)

[Out]

(b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(128*a^(5/2)) - (11*(a + b*x^2)^(3/2))/(128*x^8) + (3*a*(a + b*x
^2)^(1/2))/(128*x^8) - (11*(a + b*x^2)^(5/2))/(128*a*x^8) + (3*(a + b*x^2)^(7/2))/(128*a^2*x^8)

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sympy [A]  time = 8.26, size = 148, normalized size = 1.28 \[ - \frac {a^{2}}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 a \sqrt {b}}{16 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {13 b^{\frac {3}{2}}}{64 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {5}{2}}}{128 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b^{\frac {7}{2}}}{128 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**9,x)

[Out]

-a**2/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 5*a*sqrt(b)/(16*x**7*sqrt(a/(b*x**2) + 1)) - 13*b**(3/2)/(64*x**
5*sqrt(a/(b*x**2) + 1)) + b**(5/2)/(128*a*x**3*sqrt(a/(b*x**2) + 1)) + 3*b**(7/2)/(128*a**2*x*sqrt(a/(b*x**2)
+ 1)) - 3*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(5/2))

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