3.621 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{20}} \, dx\)

Optimal. Leaf size=255 \[ -\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )} \]

[Out]

-1/19*a^5*((b*x^2+a)^2)^(1/2)/x^19/(b*x^2+a)-5/17*a^4*b*((b*x^2+a)^2)^(1/2)/x^17/(b*x^2+a)-2/3*a^3*b^2*((b*x^2
+a)^2)^(1/2)/x^15/(b*x^2+a)-10/13*a^2*b^3*((b*x^2+a)^2)^(1/2)/x^13/(b*x^2+a)-5/11*a*b^4*((b*x^2+a)^2)^(1/2)/x^
11/(b*x^2+a)-1/9*b^5*((b*x^2+a)^2)^(1/2)/x^9/(b*x^2+a)

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Rubi [A]  time = 0.06, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 270} \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^20,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(19*x^19*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*x
^17*(a + b*x^2)) - (2*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*x^15*(a + b*x^2)) - (10*a^2*b^3*Sqrt[a^2 + 2
*a*b*x^2 + b^2*x^4])/(13*x^13*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*x^11*(a + b*x^2)) -
(b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*x^9*(a + b*x^2))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{x^{20}} \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5 b^5}{x^{20}}+\frac {5 a^4 b^6}{x^{18}}+\frac {10 a^3 b^7}{x^{16}}+\frac {10 a^2 b^8}{x^{14}}+\frac {5 a b^9}{x^{12}}+\frac {b^{10}}{x^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (21879 a^5+122265 a^4 b x^2+277134 a^3 b^2 x^4+319770 a^2 b^3 x^6+188955 a b^4 x^8+46189 b^5 x^{10}\right )}{415701 x^{19} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^20,x]

[Out]

-1/415701*(Sqrt[(a + b*x^2)^2]*(21879*a^5 + 122265*a^4*b*x^2 + 277134*a^3*b^2*x^4 + 319770*a^2*b^3*x^6 + 18895
5*a*b^4*x^8 + 46189*b^5*x^10))/(x^19*(a + b*x^2))

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fricas [A]  time = 0.58, size = 59, normalized size = 0.23 \[ -\frac {46189 \, b^{5} x^{10} + 188955 \, a b^{4} x^{8} + 319770 \, a^{2} b^{3} x^{6} + 277134 \, a^{3} b^{2} x^{4} + 122265 \, a^{4} b x^{2} + 21879 \, a^{5}}{415701 \, x^{19}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^20,x, algorithm="fricas")

[Out]

-1/415701*(46189*b^5*x^10 + 188955*a*b^4*x^8 + 319770*a^2*b^3*x^6 + 277134*a^3*b^2*x^4 + 122265*a^4*b*x^2 + 21
879*a^5)/x^19

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giac [A]  time = 0.20, size = 107, normalized size = 0.42 \[ -\frac {46189 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 188955 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 319770 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 277134 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 122265 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 21879 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{415701 \, x^{19}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^20,x, algorithm="giac")

[Out]

-1/415701*(46189*b^5*x^10*sgn(b*x^2 + a) + 188955*a*b^4*x^8*sgn(b*x^2 + a) + 319770*a^2*b^3*x^6*sgn(b*x^2 + a)
+ 277134*a^3*b^2*x^4*sgn(b*x^2 + a) + 122265*a^4*b*x^2*sgn(b*x^2 + a) + 21879*a^5*sgn(b*x^2 + a))/x^19

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ -\frac {\left (46189 b^{5} x^{10}+188955 a \,b^{4} x^{8}+319770 a^{2} b^{3} x^{6}+277134 a^{3} b^{2} x^{4}+122265 a^{4} b \,x^{2}+21879 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{415701 \left (b \,x^{2}+a \right )^{5} x^{19}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^20,x)

[Out]

-1/415701*(46189*b^5*x^10+188955*a*b^4*x^8+319770*a^2*b^3*x^6+277134*a^3*b^2*x^4+122265*a^4*b*x^2+21879*a^5)*(
(b*x^2+a)^2)^(5/2)/x^19/(b*x^2+a)^5

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maxima [A]  time = 1.34, size = 57, normalized size = 0.22 \[ -\frac {b^{5}}{9 \, x^{9}} - \frac {5 \, a b^{4}}{11 \, x^{11}} - \frac {10 \, a^{2} b^{3}}{13 \, x^{13}} - \frac {2 \, a^{3} b^{2}}{3 \, x^{15}} - \frac {5 \, a^{4} b}{17 \, x^{17}} - \frac {a^{5}}{19 \, x^{19}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^20,x, algorithm="maxima")

[Out]

-1/9*b^5/x^9 - 5/11*a*b^4/x^11 - 10/13*a^2*b^3/x^13 - 2/3*a^3*b^2/x^15 - 5/17*a^4*b/x^17 - 1/19*a^5/x^19

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mupad [B]  time = 4.27, size = 231, normalized size = 0.91 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{19\,x^{19}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{17\,x^{17}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {2\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^{15}\,\left (b\,x^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^20,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(19*x^19*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(9*x
^9*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(11*x^11*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4
+ 2*a*b*x^2)^(1/2))/(17*x^17*(a + b*x^2)) - (10*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(13*x^13*(a + b*x^2
)) - (2*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(3*x^15*(a + b*x^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{20}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**20,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**20, x)

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