3.619 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{16}} \, dx\)

Optimal. Leaf size=255 \[ -\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 x^{15} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )} \]

[Out]

-1/15*a^5*((b*x^2+a)^2)^(1/2)/x^15/(b*x^2+a)-5/13*a^4*b*((b*x^2+a)^2)^(1/2)/x^13/(b*x^2+a)-10/11*a^3*b^2*((b*x
^2+a)^2)^(1/2)/x^11/(b*x^2+a)-10/9*a^2*b^3*((b*x^2+a)^2)^(1/2)/x^9/(b*x^2+a)-5/7*a*b^4*((b*x^2+a)^2)^(1/2)/x^7
/(b*x^2+a)-1/5*b^5*((b*x^2+a)^2)^(1/2)/x^5/(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} \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 x^{15} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^16,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*x^15*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*x
^13*(a + b*x^2)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*x^11*(a + b*x^2)) - (10*a^2*b^3*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(9*x^9*(a + b*x^2)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^7*(a + b*x^2)) - (
b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*x^5*(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^{16}} \, 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^{16}} \, 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^{16}}+\frac {5 a^4 b^6}{x^{14}}+\frac {10 a^3 b^7}{x^{12}}+\frac {10 a^2 b^8}{x^{10}}+\frac {5 a b^9}{x^8}+\frac {b^{10}}{x^6}\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}}{15 x^{15} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {10 a^3 b^2 \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}}{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}}{7 x^7 \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (3003 a^5+17325 a^4 b x^2+40950 a^3 b^2 x^4+50050 a^2 b^3 x^6+32175 a b^4 x^8+9009 b^5 x^{10}\right )}{45045 x^{15} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^16,x]

[Out]

-1/45045*(Sqrt[(a + b*x^2)^2]*(3003*a^5 + 17325*a^4*b*x^2 + 40950*a^3*b^2*x^4 + 50050*a^2*b^3*x^6 + 32175*a*b^
4*x^8 + 9009*b^5*x^10))/(x^15*(a + b*x^2))

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fricas [A]  time = 0.97, size = 59, normalized size = 0.23 \[ -\frac {9009 \, b^{5} x^{10} + 32175 \, a b^{4} x^{8} + 50050 \, a^{2} b^{3} x^{6} + 40950 \, a^{3} b^{2} x^{4} + 17325 \, a^{4} b x^{2} + 3003 \, a^{5}}{45045 \, x^{15}} \]

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^16,x, algorithm="fricas")

[Out]

-1/45045*(9009*b^5*x^10 + 32175*a*b^4*x^8 + 50050*a^2*b^3*x^6 + 40950*a^3*b^2*x^4 + 17325*a^4*b*x^2 + 3003*a^5
)/x^15

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giac [A]  time = 0.16, size = 107, normalized size = 0.42 \[ -\frac {9009 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 32175 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 50050 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 40950 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 17325 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3003 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{45045 \, x^{15}} \]

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^16,x, algorithm="giac")

[Out]

-1/45045*(9009*b^5*x^10*sgn(b*x^2 + a) + 32175*a*b^4*x^8*sgn(b*x^2 + a) + 50050*a^2*b^3*x^6*sgn(b*x^2 + a) + 4
0950*a^3*b^2*x^4*sgn(b*x^2 + a) + 17325*a^4*b*x^2*sgn(b*x^2 + a) + 3003*a^5*sgn(b*x^2 + a))/x^15

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \[ -\frac {\left (9009 b^{5} x^{10}+32175 a \,b^{4} x^{8}+50050 a^{2} b^{3} x^{6}+40950 a^{3} b^{2} x^{4}+17325 a^{4} b \,x^{2}+3003 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{45045 \left (b \,x^{2}+a \right )^{5} x^{15}} \]

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^16,x)

[Out]

-1/45045*(9009*b^5*x^10+32175*a*b^4*x^8+50050*a^2*b^3*x^6+40950*a^3*b^2*x^4+17325*a^4*b*x^2+3003*a^5)*((b*x^2+
a)^2)^(5/2)/x^15/(b*x^2+a)^5

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maxima [A]  time = 1.35, size = 57, normalized size = 0.22 \[ -\frac {b^{5}}{5 \, x^{5}} - \frac {5 \, a b^{4}}{7 \, x^{7}} - \frac {10 \, a^{2} b^{3}}{9 \, x^{9}} - \frac {10 \, a^{3} b^{2}}{11 \, x^{11}} - \frac {5 \, a^{4} b}{13 \, x^{13}} - \frac {a^{5}}{15 \, x^{15}} \]

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^16,x, algorithm="maxima")

[Out]

-1/5*b^5/x^5 - 5/7*a*b^4/x^7 - 10/9*a^2*b^3/x^9 - 10/11*a^3*b^2/x^11 - 5/13*a^4*b/x^13 - 1/15*a^5/x^15

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mupad [B]  time = 4.21, size = 231, normalized size = 0.91 \[ -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{15\,x^{15}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{5\,x^5\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {10\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )} \]

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^16,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(15*x^15*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(5*x
^5*(a + b*x^2)) - (5*a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 +
2*a*b*x^2)^(1/2))/(13*x^13*(a + b*x^2)) - (10*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(9*x^9*(a + b*x^2)) -
(10*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(11*x^11*(a + b*x^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{16}}\, dx \]

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**16,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**16, x)

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