3.396 \(\int \frac{(a+b x^2)^{5/2}}{x^{11}} \, dx\)

Optimal. Leaf size=137 \[ \frac{3 b^4 \sqrt{a+b x^2}}{256 a^2 x^2}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{5/2}}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}} \]

[Out]

-(b^2*Sqrt[a + b*x^2])/(32*x^6) - (b^3*Sqrt[a + b*x^2])/(128*a*x^4) + (3*b^4*Sqrt[a + b*x^2])/(256*a^2*x^2) -
(b*(a + b*x^2)^(3/2))/(16*x^8) - (a + b*x^2)^(5/2)/(10*x^10) - (3*b^5*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a
^(5/2))

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Rubi [A]  time = 0.0834827, antiderivative size = 137, 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, 51, 63, 208} \[ \frac{3 b^4 \sqrt{a+b x^2}}{256 a^2 x^2}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{5/2}}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^11,x]

[Out]

-(b^2*Sqrt[a + b*x^2])/(32*x^6) - (b^3*Sqrt[a + b*x^2])/(128*a*x^4) + (3*b^4*Sqrt[a + b*x^2])/(256*a^2*x^2) -
(b*(a + b*x^2)^(3/2))/(16*x^8) - (a + b*x^2)^(5/2)/(10*x^10) - (3*b^5*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a
^(5/2))

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2}}{x^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}+\frac{1}{32} \left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}+\frac{1}{64} b^3 \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{256 a}\\ &=-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}+\frac{3 b^4 \sqrt{a+b x^2}}{256 a^2 x^2}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}+\frac{\left (3 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{512 a^2}\\ &=-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}+\frac{3 b^4 \sqrt{a+b x^2}}{256 a^2 x^2}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}+\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{256 a^2}\\ &=-\frac{b^2 \sqrt{a+b x^2}}{32 x^6}-\frac{b^3 \sqrt{a+b x^2}}{128 a x^4}+\frac{3 b^4 \sqrt{a+b x^2}}{256 a^2 x^2}-\frac{b \left (a+b x^2\right )^{3/2}}{16 x^8}-\frac{\left (a+b x^2\right )^{5/2}}{10 x^{10}}-\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{256 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0102124, size = 39, normalized size = 0.28 \[ \frac{b^5 \left (a+b x^2\right )^{7/2} \, _2F_1\left (\frac{7}{2},6;\frac{9}{2};\frac{b x^2}{a}+1\right )}{7 a^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^11,x]

[Out]

(b^5*(a + b*x^2)^(7/2)*Hypergeometric2F1[7/2, 6, 9/2, 1 + (b*x^2)/a])/(7*a^6)

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Maple [A]  time = 0.029, size = 179, normalized size = 1.3 \begin{align*} -{\frac{1}{10\,a{x}^{10}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,b}{80\,{a}^{2}{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{{b}^{2}}{160\,{a}^{3}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{{b}^{3}}{640\,{a}^{4}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,{b}^{4}}{1280\,{a}^{5}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{b}^{5}}{1280\,{a}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{5}}{256\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{5}}{256}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{3\,{b}^{5}}{256\,{a}^{3}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^11,x)

[Out]

-1/10/a/x^10*(b*x^2+a)^(7/2)+3/80*b/a^2/x^8*(b*x^2+a)^(7/2)-1/160*b^2/a^3/x^6*(b*x^2+a)^(7/2)-1/640*b^3/a^4/x^
4*(b*x^2+a)^(7/2)-3/1280*b^4/a^5/x^2*(b*x^2+a)^(7/2)+3/1280*b^5/a^5*(b*x^2+a)^(5/2)+1/256*b^5/a^4*(b*x^2+a)^(3
/2)-3/256*b^5/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+3/256*b^5/a^3*(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)^(5/2)/x^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7218, size = 489, normalized size = 3.57 \begin{align*} \left [\frac{15 \, \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 (15 \, a b^{4} x^{8} - 10 \, a^{2} b^{3} x^{6} - 248 \, a^{3} b^{2} x^{4} - 336 \, a^{4} b x^{2} - 128 \, a^{5}\right )} \sqrt{b x^{2} + a}}{2560 \, a^{3} x^{10}}, \frac{15 \, \sqrt{-a} b^{5} x^{10} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (15 \, a b^{4} x^{8} - 10 \, a^{2} b^{3} x^{6} - 248 \, a^{3} b^{2} x^{4} - 336 \, a^{4} b x^{2} - 128 \, a^{5}\right )} \sqrt{b x^{2} + a}}{1280 \, a^{3} x^{10}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^11,x, algorithm="fricas")

[Out]

[1/2560*(15*sqrt(a)*b^5*x^10*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*a*b^4*x^8 - 10*a^2*b^
3*x^6 - 248*a^3*b^2*x^4 - 336*a^4*b*x^2 - 128*a^5)*sqrt(b*x^2 + a))/(a^3*x^10), 1/1280*(15*sqrt(-a)*b^5*x^10*a
rctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*a*b^4*x^8 - 10*a^2*b^3*x^6 - 248*a^3*b^2*x^4 - 336*a^4*b*x^2 - 128*a^5)*
sqrt(b*x^2 + a))/(a^3*x^10)]

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Sympy [A]  time = 11.7643, size = 175, normalized size = 1.28 \begin{align*} - \frac{a^{3}}{10 \sqrt{b} x^{11} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{29 a^{2} \sqrt{b}}{80 x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{73 a b^{\frac{3}{2}}}{160 x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{129 b^{\frac{5}{2}}}{640 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b^{\frac{7}{2}}}{256 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 b^{\frac{9}{2}}}{256 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 b^{5} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{256 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**11,x)

[Out]

-a**3/(10*sqrt(b)*x**11*sqrt(a/(b*x**2) + 1)) - 29*a**2*sqrt(b)/(80*x**9*sqrt(a/(b*x**2) + 1)) - 73*a*b**(3/2)
/(160*x**7*sqrt(a/(b*x**2) + 1)) - 129*b**(5/2)/(640*x**5*sqrt(a/(b*x**2) + 1)) + b**(7/2)/(256*a*x**3*sqrt(a/
(b*x**2) + 1)) + 3*b**(9/2)/(256*a**2*x*sqrt(a/(b*x**2) + 1)) - 3*b**5*asinh(sqrt(a)/(sqrt(b)*x))/(256*a**(5/2
))

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Giac [A]  time = 1.8112, size = 146, normalized size = 1.07 \begin{align*} \frac{1}{1280} \, b^{5}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{15 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 70 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a - 128 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} + 70 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} - 15 \, \sqrt{b x^{2} + a} a^{4}}{a^{2} b^{5} x^{10}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^11,x, algorithm="giac")

[Out]

1/1280*b^5*(15*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (15*(b*x^2 + a)^(9/2) - 70*(b*x^2 + a)^(7/2)*
a - 128*(b*x^2 + a)^(5/2)*a^2 + 70*(b*x^2 + a)^(3/2)*a^3 - 15*sqrt(b*x^2 + a)*a^4)/(a^2*b^5*x^10))