3.422 \(\int \frac{(a+b x^2)^{9/2}}{x^{11}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{63 b^4 \sqrt{a+b x^2}}{256 x^2}-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 \sqrt{a}}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}} \]

[Out]

(-63*b^4*Sqrt[a + b*x^2])/(256*x^2) - (21*b^3*(a + b*x^2)^(3/2))/(128*x^4) - (21*b^2*(a + b*x^2)^(5/2))/(160*x
^6) - (9*b*(a + b*x^2)^(7/2))/(80*x^8) - (a + b*x^2)^(9/2)/(10*x^10) - (63*b^5*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]
])/(256*Sqrt[a])

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Rubi [A]  time = 0.0826313, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 47, 63, 208} \[ -\frac{63 b^4 \sqrt{a+b x^2}}{256 x^2}-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 \sqrt{a}}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^11,x]

[Out]

(-63*b^4*Sqrt[a + b*x^2])/(256*x^2) - (21*b^3*(a + b*x^2)^(3/2))/(128*x^4) - (21*b^2*(a + b*x^2)^(5/2))/(160*x
^6) - (9*b*(a + b*x^2)^(7/2))/(80*x^8) - (a + b*x^2)^(9/2)/(10*x^10) - (63*b^5*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]
])/(256*Sqrt[a])

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 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^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{9/2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{20} (9 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{7/2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{160} \left (63 b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{64} \left (21 b^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{256} \left (63 b^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{63 b^4 \sqrt{a+b x^2}}{256 x^2}-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{512} \left (63 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{63 b^4 \sqrt{a+b x^2}}{256 x^2}-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}+\frac{1}{256} \left (63 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=-\frac{63 b^4 \sqrt{a+b x^2}}{256 x^2}-\frac{21 b^3 \left (a+b x^2\right )^{3/2}}{128 x^4}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{160 x^6}-\frac{9 b \left (a+b x^2\right )^{7/2}}{80 x^8}-\frac{\left (a+b x^2\right )^{9/2}}{10 x^{10}}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0479415, size = 109, normalized size = 0.83 \[ -\frac{2858 a^2 b^3 x^6+2024 a^3 b^2 x^4+784 a^4 b x^2+128 a^5+2455 a b^4 x^8+315 b^5 x^{10} \sqrt{\frac{b x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )+965 b^5 x^{10}}{1280 x^{10} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^11,x]

[Out]

-(128*a^5 + 784*a^4*b*x^2 + 2024*a^3*b^2*x^4 + 2858*a^2*b^3*x^6 + 2455*a*b^4*x^8 + 965*b^5*x^10 + 315*b^5*x^10
*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(1280*x^10*Sqrt[a + b*x^2])

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Maple [B]  time = 0.046, size = 213, normalized size = 1.6 \begin{align*} -{\frac{1}{10\,a{x}^{10}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{b}{80\,{a}^{2}{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{{b}^{2}}{160\,{a}^{3}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{{b}^{3}}{128\,{a}^{4}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{7\,{b}^{4}}{256\,{a}^{5}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}+{\frac{7\,{b}^{5}}{256\,{a}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{9}{2}}}}+{\frac{9\,{b}^{5}}{256\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{63\,{b}^{5}}{1280\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{21\,{b}^{5}}{256\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{63\,{b}^{5}}{256}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{63\,{b}^{5}}{256\,a}\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^11,x)

[Out]

-1/10/a/x^10*(b*x^2+a)^(11/2)-1/80*b/a^2/x^8*(b*x^2+a)^(11/2)-1/160*b^2/a^3/x^6*(b*x^2+a)^(11/2)-1/128*b^3/a^4
/x^4*(b*x^2+a)^(11/2)-7/256*b^4/a^5/x^2*(b*x^2+a)^(11/2)+7/256*b^5/a^5*(b*x^2+a)^(9/2)+9/256*b^5/a^4*(b*x^2+a)
^(7/2)+63/1280*b^5/a^3*(b*x^2+a)^(5/2)+21/256*b^5/a^2*(b*x^2+a)^(3/2)-63/256*b^5/a^(1/2)*ln((2*a+2*a^(1/2)*(b*
x^2+a)^(1/2))/x)+63/256*b^5/a*(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^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72751, size = 497, normalized size = 3.79 \begin{align*} \left [\frac{315 \, \sqrt{a} b^{5} x^{10} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (965 \, a b^{4} x^{8} + 1490 \, a^{2} b^{3} x^{6} + 1368 \, a^{3} b^{2} x^{4} + 656 \, a^{4} b x^{2} + 128 \, a^{5}\right )} \sqrt{b x^{2} + a}}{2560 \, a x^{10}}, \frac{315 \, \sqrt{-a} b^{5} x^{10} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (965 \, a b^{4} x^{8} + 1490 \, a^{2} b^{3} x^{6} + 1368 \, a^{3} b^{2} x^{4} + 656 \, a^{4} b x^{2} + 128 \, a^{5}\right )} \sqrt{b x^{2} + a}}{1280 \, a x^{10}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^11,x, algorithm="fricas")

[Out]

[1/2560*(315*sqrt(a)*b^5*x^10*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(965*a*b^4*x^8 + 1490*a^
2*b^3*x^6 + 1368*a^3*b^2*x^4 + 656*a^4*b*x^2 + 128*a^5)*sqrt(b*x^2 + a))/(a*x^10), 1/1280*(315*sqrt(-a)*b^5*x^
10*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (965*a*b^4*x^8 + 1490*a^2*b^3*x^6 + 1368*a^3*b^2*x^4 + 656*a^4*b*x^2 + 1
28*a^5)*sqrt(b*x^2 + a))/(a*x^10)]

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Sympy [A]  time = 10.3829, size = 153, normalized size = 1.17 \begin{align*} - \frac{a^{4} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{10 x^{9}} - \frac{41 a^{3} b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{80 x^{7}} - \frac{171 a^{2} b^{\frac{5}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{160 x^{5}} - \frac{149 a b^{\frac{7}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{128 x^{3}} - \frac{193 b^{\frac{9}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{256 x} - \frac{63 b^{5} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{256 \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(9/2)/x**11,x)

[Out]

-a**4*sqrt(b)*sqrt(a/(b*x**2) + 1)/(10*x**9) - 41*a**3*b**(3/2)*sqrt(a/(b*x**2) + 1)/(80*x**7) - 171*a**2*b**(
5/2)*sqrt(a/(b*x**2) + 1)/(160*x**5) - 149*a*b**(7/2)*sqrt(a/(b*x**2) + 1)/(128*x**3) - 193*b**(9/2)*sqrt(a/(b
*x**2) + 1)/(256*x) - 63*b**5*asinh(sqrt(a)/(sqrt(b)*x))/(256*sqrt(a))

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Giac [A]  time = 1.66556, size = 139, normalized size = 1.06 \begin{align*} \frac{1}{1280} \, b^{5}{\left (\frac{315 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{965 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 2370 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a + 2688 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 1470 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} + 315 \, \sqrt{b x^{2} + a} a^{4}}{b^{5} x^{10}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^11,x, algorithm="giac")

[Out]

1/1280*b^5*(315*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (965*(b*x^2 + a)^(9/2) - 2370*(b*x^2 + a)^(7/2)*a
+ 2688*(b*x^2 + a)^(5/2)*a^2 - 1470*(b*x^2 + a)^(3/2)*a^3 + 315*sqrt(b*x^2 + a)*a^4)/(b^5*x^10))