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

Optimal. Leaf size=128 \[ -\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}+\frac{315}{128} b^4 \sqrt{a+b x^2}-\frac{315}{128} \sqrt{a} b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6} \]

[Out]

(315*b^4*Sqrt[a + b*x^2])/128 - (105*b^3*(a + b*x^2)^(3/2))/(128*x^2) - (21*b^2*(a + b*x^2)^(5/2))/(64*x^4) -
(3*b*(a + b*x^2)^(7/2))/(16*x^6) - (a + b*x^2)^(9/2)/(8*x^8) - (315*Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a
]])/128

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Rubi [A]  time = 0.0793392, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ -\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}+\frac{315}{128} b^4 \sqrt{a+b x^2}-\frac{315}{128} \sqrt{a} b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^9,x]

[Out]

(315*b^4*Sqrt[a + b*x^2])/128 - (105*b^3*(a + b*x^2)^(3/2))/(128*x^2) - (21*b^2*(a + b*x^2)^(5/2))/(64*x^4) -
(3*b*(a + b*x^2)^(7/2))/(16*x^6) - (a + b*x^2)^(9/2)/(8*x^8) - (315*Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a
]])/128

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]]

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 50

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{9/2}}{x^9} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{9/2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{16} (9 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{7/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{32} \left (21 b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{128} \left (105 b^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{256} \left (315 b^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{315}{128} b^4 \sqrt{a+b x^2}-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{256} \left (315 a b^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{315}{128} b^4 \sqrt{a+b x^2}-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}+\frac{1}{128} \left (315 a b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{315}{128} b^4 \sqrt{a+b x^2}-\frac{105 b^3 \left (a+b x^2\right )^{3/2}}{128 x^2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{64 x^4}-\frac{3 b \left (a+b x^2\right )^{7/2}}{16 x^6}-\frac{\left (a+b x^2\right )^{9/2}}{8 x^8}-\frac{315}{128} \sqrt{a} b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0123607, size = 39, normalized size = 0.3 \[ -\frac{b^4 \left (a+b x^2\right )^{11/2} \, _2F_1\left (5,\frac{11}{2};\frac{13}{2};\frac{b x^2}{a}+1\right )}{11 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^9,x]

[Out]

-(b^4*(a + b*x^2)^(11/2)*Hypergeometric2F1[5, 11/2, 13/2, 1 + (b*x^2)/a])/(11*a^5)

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Maple [A]  time = 0.018, size = 190, normalized size = 1.5 \begin{align*} -{\frac{1}{8\,a{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{b}{16\,{a}^{2}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{5\,{b}^{2}}{64\,{a}^{3}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{35\,{b}^{3}}{128\,{a}^{4}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}+{\frac{35\,{b}^{4}}{128\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{9}{2}}}}+{\frac{45\,{b}^{4}}{128\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{63\,{b}^{4}}{128\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{105\,{b}^{4}}{128\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{315\,{b}^{4}}{128}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{315\,{b}^{4}}{128}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a/x^8*(b*x^2+a)^(11/2)-1/16*b/a^2/x^6*(b*x^2+a)^(11/2)-5/64*b^2/a^3/x^4*(b*x^2+a)^(11/2)-35/128*b^3/a^4/x
^2*(b*x^2+a)^(11/2)+35/128*b^4/a^4*(b*x^2+a)^(9/2)+45/128*b^4/a^3*(b*x^2+a)^(7/2)+63/128*b^4/a^2*(b*x^2+a)^(5/
2)+105/128*b^4/a*(b*x^2+a)^(3/2)-315/128*b^4*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+315/128*b^4*(b*x^2+
a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66289, size = 456, normalized size = 3.56 \begin{align*} \left [\frac{315 \, \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 (128 \, b^{4} x^{8} - 325 \, a b^{3} x^{6} - 210 \, a^{2} b^{2} x^{4} - 88 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt{b x^{2} + a}}{256 \, x^{8}}, \frac{315 \, \sqrt{-a} b^{4} x^{8} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (128 \, b^{4} x^{8} - 325 \, a b^{3} x^{6} - 210 \, a^{2} b^{2} x^{4} - 88 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt{b x^{2} + a}}{128 \, x^{8}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/256*(315*sqrt(a)*b^4*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(128*b^4*x^8 - 325*a*b^3*x
^6 - 210*a^2*b^2*x^4 - 88*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/x^8, 1/128*(315*sqrt(-a)*b^4*x^8*arctan(sqrt(-a
)/sqrt(b*x^2 + a)) + (128*b^4*x^8 - 325*a*b^3*x^6 - 210*a^2*b^2*x^4 - 88*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/
x^8]

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Sympy [A]  time = 8.46165, size = 173, normalized size = 1.35 \begin{align*} - \frac{315 \sqrt{a} b^{4} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{128} - \frac{a^{5}}{8 \sqrt{b} x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{13 a^{4} \sqrt{b}}{16 x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{149 a^{3} b^{\frac{3}{2}}}{64 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{535 a^{2} b^{\frac{5}{2}}}{128 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{197 a b^{\frac{7}{2}}}{128 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b^{\frac{9}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-315*sqrt(a)*b**4*asinh(sqrt(a)/(sqrt(b)*x))/128 - a**5/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 13*a**4*sqrt(b
)/(16*x**7*sqrt(a/(b*x**2) + 1)) - 149*a**3*b**(3/2)/(64*x**5*sqrt(a/(b*x**2) + 1)) - 535*a**2*b**(5/2)/(128*x
**3*sqrt(a/(b*x**2) + 1)) - 197*a*b**(7/2)/(128*x*sqrt(a/(b*x**2) + 1)) + b**(9/2)*x/sqrt(a/(b*x**2) + 1)

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Giac [A]  time = 2.65703, size = 140, normalized size = 1.09 \begin{align*} \frac{1}{128} \,{\left (\frac{315 \, a \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 128 \, \sqrt{b x^{2} + a} - \frac{325 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a - 765 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} + 643 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} - 187 \, \sqrt{b x^{2} + a} a^{4}}{b^{4} x^{8}}\right )} b^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(315*a*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 128*sqrt(b*x^2 + a) - (325*(b*x^2 + a)^(7/2)*a - 765*
(b*x^2 + a)^(5/2)*a^2 + 643*(b*x^2 + a)^(3/2)*a^3 - 187*sqrt(b*x^2 + a)*a^4)/(b^4*x^8))*b^4