3.560 \(\int x^5 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=106 \[ \frac{\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^3}-\frac{a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{5 b^3}+\frac{a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^3} \]

[Out]

(a^2*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(8*b^3) - (a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(5*b^3) +
((a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(12*b^3)

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Rubi [A]  time = 0.0834046, antiderivative size = 119, normalized size of antiderivative = 1.12, 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 = {1111, 645} \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^3}-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^4}{5 b^3}+\frac{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^3}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(a^2*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*b^3) - (a*(a + b*x^2)^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
)/(5*b^3) + ((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^3)

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 645

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

Rubi steps

\begin{align*} \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \left (\frac{a^2 \left (a b+b^2 x\right )^3}{b^2}-\frac{2 a \left (a b+b^2 x\right )^4}{b^3}+\frac{\left (a b+b^2 x\right )^5}{b^4}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac{a^2 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 b^3}-\frac{a \left (a+b x^2\right )^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 b^3}+\frac{\left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0153002, size = 61, normalized size = 0.58 \[ \frac{x^6 \sqrt{\left (a+b x^2\right )^2} \left (45 a^2 b x^2+20 a^3+36 a b^2 x^4+10 b^3 x^6\right )}{120 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^6*Sqrt[(a + b*x^2)^2]*(20*a^3 + 45*a^2*b*x^2 + 36*a*b^2*x^4 + 10*b^3*x^6))/(120*(a + b*x^2))

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Maple [A]  time = 0.175, size = 58, normalized size = 0.6 \begin{align*}{\frac{{x}^{6} \left ( 10\,{b}^{3}{x}^{6}+36\,a{b}^{2}{x}^{4}+45\,{a}^{2}b{x}^{2}+20\,{a}^{3} \right ) }{120\, \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^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/120*x^6*(10*b^3*x^6+36*a*b^2*x^4+45*a^2*b*x^2+20*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49195, size = 85, normalized size = 0.8 \begin{align*} \frac{1}{12} \, b^{3} x^{12} + \frac{3}{10} \, a b^{2} x^{10} + \frac{3}{8} \, a^{2} b x^{8} + \frac{1}{6} \, a^{3} x^{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/12*b^3*x^12 + 3/10*a*b^2*x^10 + 3/8*a^2*b*x^8 + 1/6*a^3*x^6

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \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**5*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**5*((a + b*x**2)**2)**(3/2), x)

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Giac [A]  time = 1.1235, size = 90, normalized size = 0.85 \begin{align*} \frac{1}{12} \, b^{3} x^{12} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{3}{10} \, a b^{2} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{3}{8} \, a^{2} b x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{6} \, a^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/12*b^3*x^12*sgn(b*x^2 + a) + 3/10*a*b^2*x^10*sgn(b*x^2 + a) + 3/8*a^2*b*x^8*sgn(b*x^2 + a) + 1/6*a^3*x^6*sgn
(b*x^2 + a)