3.572 \(\int x^8 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=167 \[ \frac{b^3 x^{15} \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}+\frac{3 a b^2 x^{13} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac{3 a^2 b x^{11} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac{a^3 x^9 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

[Out]

(a^3*x^9*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(9*(a + b*x^2)) + (3*a^2*b*x^11*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11
*(a + b*x^2)) + (3*a*b^2*x^13*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*(a + b*x^2)) + (b^3*x^15*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])/(15*(a + b*x^2))

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Rubi [A]  time = 0.0420411, antiderivative size = 167, 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} \[ \frac{b^3 x^{15} \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )}+\frac{3 a b^2 x^{13} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac{3 a^2 b x^{11} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac{a^3 x^9 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0150427, size = 61, normalized size = 0.37 \[ \frac{x^9 \sqrt{\left (a+b x^2\right )^2} \left (1755 a^2 b x^2+715 a^3+1485 a b^2 x^4+429 b^3 x^6\right )}{6435 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^9*Sqrt[(a + b*x^2)^2]*(715*a^3 + 1755*a^2*b*x^2 + 1485*a*b^2*x^4 + 429*b^3*x^6))/(6435*(a + b*x^2))

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Maple [A]  time = 0.164, size = 58, normalized size = 0.4 \begin{align*}{\frac{{x}^{9} \left ( 429\,{b}^{3}{x}^{6}+1485\,a{x}^{4}{b}^{2}+1755\,{a}^{2}b{x}^{2}+715\,{a}^{3} \right ) }{6435\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/6435*x^9*(429*b^3*x^6+1485*a*b^2*x^4+1755*a^2*b*x^2+715*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [A]  time = 1.00826, size = 47, normalized size = 0.28 \begin{align*} \frac{1}{15} \, b^{3} x^{15} + \frac{3}{13} \, a b^{2} x^{13} + \frac{3}{11} \, a^{2} b x^{11} + \frac{1}{9} \, a^{3} x^{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/15*b^3*x^15 + 3/13*a*b^2*x^13 + 3/11*a^2*b*x^11 + 1/9*a^3*x^9

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Fricas [A]  time = 1.43779, size = 88, normalized size = 0.53 \begin{align*} \frac{1}{15} \, b^{3} x^{15} + \frac{3}{13} \, a b^{2} x^{13} + \frac{3}{11} \, a^{2} b x^{11} + \frac{1}{9} \, a^{3} x^{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/15*b^3*x^15 + 3/13*a*b^2*x^13 + 3/11*a^2*b*x^11 + 1/9*a^3*x^9

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{8} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**8*((a + b*x**2)**2)**(3/2), x)

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Giac [A]  time = 1.12603, size = 90, normalized size = 0.54 \begin{align*} \frac{1}{15} \, b^{3} x^{15} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{3}{13} \, a b^{2} x^{13} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{3}{11} \, a^{2} b x^{11} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{9} \, a^{3} x^{9} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/15*b^3*x^15*sgn(b*x^2 + a) + 3/13*a*b^2*x^13*sgn(b*x^2 + a) + 3/11*a^2*b*x^11*sgn(b*x^2 + a) + 1/9*a^3*x^9*s
gn(b*x^2 + a)