3.31 \(\int x (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\)

Optimal. Leaf size=167 \[ \frac{b^3 x^{11} \sqrt{a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac{3 a b^2 x^8 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac{3 a^2 b x^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac{a^3 x^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )} \]

[Out]

(a^3*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a + b*x^3)) + (3*a^2*b*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(
a + b*x^3)) + (3*a*b^2*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*(a + b*x^3)) + (b^3*x^11*Sqrt[a^2 + 2*a*b*x^3 +
b^2*x^6])/(11*(a + b*x^3))

________________________________________________________________________________________

Rubi [A]  time = 0.0378921, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1355, 270} \[ \frac{b^3 x^{11} \sqrt{a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac{3 a b^2 x^8 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac{3 a^2 b x^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac{a^3 x^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(a^3*x^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*(a + b*x^3)) + (3*a^2*b*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(
a + b*x^3)) + (3*a*b^2*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*(a + b*x^3)) + (b^3*x^11*Sqrt[a^2 + 2*a*b*x^3 +
b^2*x^6])/(11*(a + b*x^3))

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && 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 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int x \left (a b+b^2 x^3\right )^3 \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int \left (a^3 b^3 x+3 a^2 b^4 x^4+3 a b^5 x^7+b^6 x^{10}\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{a^3 x^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac{3 a^2 b x^5 \sqrt{a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac{3 a b^2 x^8 \sqrt{a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac{b^3 x^{11} \sqrt{a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}\\ \end{align*}

Mathematica [A]  time = 0.015788, size = 61, normalized size = 0.37 \[ \frac{x^2 \sqrt{\left (a+b x^3\right )^2} \left (264 a^2 b x^3+220 a^3+165 a b^2 x^6+40 b^3 x^9\right )}{440 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x^2*Sqrt[(a + b*x^3)^2]*(220*a^3 + 264*a^2*b*x^3 + 165*a*b^2*x^6 + 40*b^3*x^9))/(440*(a + b*x^3))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 58, normalized size = 0.4 \begin{align*}{\frac{{x}^{2} \left ( 40\,{b}^{3}{x}^{9}+165\,a{b}^{2}{x}^{6}+264\,{a}^{2}b{x}^{3}+220\,{a}^{3} \right ) }{440\, \left ( b{x}^{3}+a \right ) ^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)

[Out]

1/440*x^2*(40*b^3*x^9+165*a*b^2*x^6+264*a^2*b*x^3+220*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

________________________________________________________________________________________

Maxima [A]  time = 1.02098, size = 47, normalized size = 0.28 \begin{align*} \frac{1}{11} \, b^{3} x^{11} + \frac{3}{8} \, a b^{2} x^{8} + \frac{3}{5} \, a^{2} b x^{5} + \frac{1}{2} \, a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/11*b^3*x^11 + 3/8*a*b^2*x^8 + 3/5*a^2*b*x^5 + 1/2*a^3*x^2

________________________________________________________________________________________

Fricas [A]  time = 1.78368, size = 82, normalized size = 0.49 \begin{align*} \frac{1}{11} \, b^{3} x^{11} + \frac{3}{8} \, a b^{2} x^{8} + \frac{3}{5} \, a^{2} b x^{5} + \frac{1}{2} \, a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/11*b^3*x^11 + 3/8*a*b^2*x^8 + 3/5*a^2*b*x^5 + 1/2*a^3*x^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.1024, size = 90, normalized size = 0.54 \begin{align*} \frac{1}{11} \, b^{3} x^{11} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{3}{8} \, a b^{2} x^{8} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{3}{5} \, a^{2} b x^{5} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{1}{2} \, a^{3} x^{2} \mathrm{sgn}\left (b x^{3} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

1/11*b^3*x^11*sgn(b*x^3 + a) + 3/8*a*b^2*x^8*sgn(b*x^3 + a) + 3/5*a^2*b*x^5*sgn(b*x^3 + a) + 1/2*a^3*x^2*sgn(b
*x^3 + a)