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

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

[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)

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Rubi [A]  time = 0.0580791, antiderivative size = 246, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0181717, size = 83, normalized size = 0.34 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (210 a^2 b^3 x^6+126 a^3 b^2 x^4+45 a^4 b x^2+7 a^5+315 a b^4 x^8-63 b^5 x^{10}\right )}{63 x^9 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)
)/(63*x^9*(a + b*x^2))

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Maple [A]  time = 0.171, size = 80, normalized size = 0.3 \begin{align*} -{\frac{-63\,{b}^{5}{x}^{10}+315\,a{b}^{4}{x}^{8}+210\,{a}^{2}{b}^{3}{x}^{6}+126\,{b}^{2}{a}^{3}{x}^{4}+45\,{a}^{4}b{x}^{2}+7\,{a}^{5}}{63\,{x}^{9} \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

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 = 1.00842, size = 80, normalized size = 0.33 \begin{align*} \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}} \end{align*}

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]

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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Fricas [A]  time = 1.26493, size = 134, normalized size = 0.54 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}{x^{10}}\, dx \end{align*}

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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Giac [A]  time = 1.1324, size = 142, normalized size = 0.58 \begin{align*} 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}} \end{align*}

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