3.616 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{10}} \, dx\)

Optimal. Leaf size=246 \[ \frac {b^5 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 246, 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} \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5 \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {b^5 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^10,x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.34 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (7 a^5+45 a^4 b x^2+126 a^3 b^2 x^4+210 a^2 b^3 x^6+315 a b^4 x^8-63 b^5 x^{10}\right )}{63 x^9 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^10,x]

[Out]

-1/63*(Sqrt[(a + b*x^2)^2]*(7*a^5 + 45*a^4*b*x^2 + 126*a^3*b^2*x^4 + 210*a^2*b^3*x^6 + 315*a*b^4*x^8 - 63*b^5*
x^10))/(x^9*(a + b*x^2))

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fricas [A]  time = 1.07, size = 59, normalized size = 0.24 \[ \frac {63 \, b^{5} x^{10} - 315 \, a b^{4} x^{8} - 210 \, a^{2} b^{3} x^{6} - 126 \, a^{3} b^{2} x^{4} - 45 \, a^{4} b x^{2} - 7 \, a^{5}}{63 \, x^{9}} \]

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^10,x, algorithm="fricas")

[Out]

1/63*(63*b^5*x^10 - 315*a*b^4*x^8 - 210*a^2*b^3*x^6 - 126*a^3*b^2*x^4 - 45*a^4*b*x^2 - 7*a^5)/x^9

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giac [A]  time = 0.17, size = 105, normalized size = 0.43 \[ b^{5} x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {315 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 210 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 126 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 45 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 7 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{63 \, x^{9}} \]

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^10,x, algorithm="giac")

[Out]

b^5*x*sgn(b*x^2 + a) - 1/63*(315*a*b^4*x^8*sgn(b*x^2 + a) + 210*a^2*b^3*x^6*sgn(b*x^2 + a) + 126*a^3*b^2*x^4*s
gn(b*x^2 + a) + 45*a^4*b*x^2*sgn(b*x^2 + a) + 7*a^5*sgn(b*x^2 + a))/x^9

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maple [A]  time = 0.01, size = 80, normalized size = 0.33 \[ -\frac {\left (-63 b^{5} x^{10}+315 a \,b^{4} x^{8}+210 a^{2} b^{3} x^{6}+126 a^{3} b^{2} x^{4}+45 a^{4} b \,x^{2}+7 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{63 \left (b \,x^{2}+a \right )^{5} x^{9}} \]

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^10,x)

[Out]

-1/63*(-63*b^5*x^10+315*a*b^4*x^8+210*a^2*b^3*x^6+126*a^3*b^2*x^4+45*a^4*b*x^2+7*a^5)*((b*x^2+a)^2)^(5/2)/x^9/
(b*x^2+a)^5

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maxima [A]  time = 1.35, size = 54, normalized size = 0.22 \[ b^{5} x - \frac {5 \, a b^{4}}{x} - \frac {10 \, a^{2} b^{3}}{3 \, x^{3}} - \frac {2 \, a^{3} b^{2}}{x^{5}} - \frac {5 \, a^{4} b}{7 \, x^{7}} - \frac {a^{5}}{9 \, x^{9}} \]

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^10,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^{10}} \,d x \]

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^10,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^10, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{10}}\, dx \]

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**10,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**10, x)

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