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

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

[Out]

-1/16*b*(b*x^2+a)^(3/2)/x^8-1/10*(b*x^2+a)^(5/2)/x^10-3/256*b^5*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)-1/32*
b^2*(b*x^2+a)^(1/2)/x^6-1/128*b^3*(b*x^2+a)^(1/2)/a/x^4+3/256*b^4*(b*x^2+a)^(1/2)/a^2/x^2

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Rubi [A]  time = 0.08, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.70, size = 201, normalized size = 1.47 \[ \left [\frac {15 \, \sqrt {a} b^{5} x^{10} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, a b^{4} x^{8} - 10 \, a^{2} b^{3} x^{6} - 248 \, a^{3} b^{2} x^{4} - 336 \, a^{4} b x^{2} - 128 \, a^{5}\right )} \sqrt {b x^{2} + a}}{2560 \, a^{3} x^{10}}, \frac {15 \, \sqrt {-a} b^{5} x^{10} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, a b^{4} x^{8} - 10 \, a^{2} b^{3} x^{6} - 248 \, a^{3} b^{2} x^{4} - 336 \, a^{4} b x^{2} - 128 \, a^{5}\right )} \sqrt {b x^{2} + a}}{1280 \, a^{3} x^{10}}\right ] \]

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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giac [A]  time = 1.16, size = 126, normalized size = 0.92 \[ \frac {\frac {15 \, b^{6} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} b^{6} - 70 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a b^{6} - 128 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} b^{6} + 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} b^{6} - 15 \, \sqrt {b x^{2} + a} a^{4} b^{6}}{a^{2} b^{5} x^{10}}}{1280 \, b} \]

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*(15*b^6*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (15*(b*x^2 + a)^(9/2)*b^6 - 70*(b*x^2 + a)^(7
/2)*a*b^6 - 128*(b*x^2 + a)^(5/2)*a^2*b^6 + 70*(b*x^2 + a)^(3/2)*a^3*b^6 - 15*sqrt(b*x^2 + a)*a^4*b^6)/(a^2*b^
5*x^10))/b

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maple [A]  time = 0.04, size = 179, normalized size = 1.31 \[ -\frac {3 b^{5} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{256 a^{\frac {5}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, b^{5}}{256 a^{3}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{5}}{256 a^{4}}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{5}}{1280 a^{5}}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{4}}{1280 a^{5} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{3}}{640 a^{4} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}{160 a^{3} x^{6}}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}{80 a^{2} x^{8}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 a \,x^{10}} \]

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/a^2*b/x^8*(b*x^2+a)^(7/2)-1/160/a^3*b^2/x^6*(b*x^2+a)^(7/2)-1/640/a^4*b^3/x^
4*(b*x^2+a)^(7/2)-3/1280/a^5*b^4/x^2*(b*x^2+a)^(7/2)+3/1280/a^5*b^5*(b*x^2+a)^(5/2)+1/256/a^4*b^5*(b*x^2+a)^(3
/2)-3/256/a^(5/2)*b^5*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+3/256/a^3*b^5*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.44, size = 167, normalized size = 1.22 \[ -\frac {3 \, b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{5}}{1280 \, a^{5}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{5}}{256 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} b^{5}}{256 \, a^{3}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}}{1280 \, a^{5} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}}{640 \, a^{4} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{160 \, a^{3} x^{6}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{80 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{10 \, a x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/256*b^5*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 3/1280*(b*x^2 + a)^(5/2)*b^5/a^5 + 1/256*(b*x^2 + a)^(3/2)*
b^5/a^4 + 3/256*sqrt(b*x^2 + a)*b^5/a^3 - 3/1280*(b*x^2 + a)^(7/2)*b^4/(a^5*x^2) - 1/640*(b*x^2 + a)^(7/2)*b^3
/(a^4*x^4) - 1/160*(b*x^2 + a)^(7/2)*b^2/(a^3*x^6) + 3/80*(b*x^2 + a)^(7/2)*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(7/
2)/(a*x^10)

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mupad [B]  time = 5.68, size = 106, normalized size = 0.77 \[ \frac {7\,a\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^{10}}-\frac {{\left (b\,x^2+a\right )}^{5/2}}{10\,x^{10}}-\frac {3\,a^2\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {7\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^{10}}+\frac {3\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^2\,x^{10}}+\frac {b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{256\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^11,x)

[Out]

(b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(256*a^(5/2)) - (a + b*x^2)^(5/2)/(10*x^10) + (7*a*(a + b*x^2)^(
3/2))/(128*x^10) - (3*a^2*(a + b*x^2)^(1/2))/(256*x^10) - (7*(a + b*x^2)^(7/2))/(128*a*x^10) + (3*(a + b*x^2)^
(9/2))/(256*a^2*x^10)

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sympy [A]  time = 11.54, size = 175, normalized size = 1.28 \[ - \frac {a^{3}}{10 \sqrt {b} x^{11} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {29 a^{2} \sqrt {b}}{80 x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {73 a b^{\frac {3}{2}}}{160 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {129 b^{\frac {5}{2}}}{640 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {7}{2}}}{256 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b^{\frac {9}{2}}}{256 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 b^{5} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{256 a^{\frac {5}{2}}} \]

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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