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

Optimal. Leaf size=162 \[ \frac{b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{10 \left (a+b x^3\right )^3}+\frac{3 a b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{7 \left (a+b x^3\right )^3}+\frac{3 a^2 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{4 \left (a+b x^3\right )^3}+\frac{a^3 x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{\left (a+b x^3\right )^3} \]

[Out]

(a^3*x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(a + b*x^3)^3 + (3*a^2*b*x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(4*(
a + b*x^3)^3) + (3*a*b^2*x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(7*(a + b*x^3)^3) + (b^3*x^10*(a^2 + 2*a*b*x^3
+ b^2*x^6)^(3/2))/(10*(a + b*x^3)^3)

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Rubi [A]  time = 0.0327202, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1343, 194} \[ \frac{b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{10 \left (a+b x^3\right )^3}+\frac{3 a b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{7 \left (a+b x^3\right )^3}+\frac{3 a^2 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{4 \left (a+b x^3\right )^3}+\frac{a^3 x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{\left (a+b x^3\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(a + b*x^3)^3 + (3*a^2*b*x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(4*(
a + b*x^3)^3) + (3*a*b^2*x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2))/(7*(a + b*x^3)^3) + (b^3*x^10*(a^2 + 2*a*b*x^3
+ b^2*x^6)^(3/2))/(10*(a + b*x^3)^3)

Rule 1343

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx &=\frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \int \left (2 a b+2 b^2 x^3\right )^3 \, dx}{\left (2 a b+2 b^2 x^3\right )^3}\\ &=\frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \int \left (8 a^3 b^3+24 a^2 b^4 x^3+24 a b^5 x^6+8 b^6 x^9\right ) \, dx}{\left (2 a b+2 b^2 x^3\right )^3}\\ &=\frac{a^3 x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{\left (a+b x^3\right )^3}+\frac{3 a^2 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{4 \left (a+b x^3\right )^3}+\frac{3 a b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{7 \left (a+b x^3\right )^3}+\frac{b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{10 \left (a+b x^3\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.0153953, size = 59, normalized size = 0.36 \[ \frac{x \sqrt{\left (a+b x^3\right )^2} \left (105 a^2 b x^3+140 a^3+60 a b^2 x^6+14 b^3 x^9\right )}{140 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x^3)^2]*(140*a^3 + 105*a^2*b*x^3 + 60*a*b^2*x^6 + 14*b^3*x^9))/(140*(a + b*x^3))

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Maple [A]  time = 0.003, size = 56, normalized size = 0.4 \begin{align*}{\frac{x \left ( 14\,{b}^{3}{x}^{9}+60\,a{b}^{2}{x}^{6}+105\,{a}^{2}b{x}^{3}+140\,{a}^{3} \right ) }{140\, \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((b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)

[Out]

1/140*x*(14*b^3*x^9+60*a*b^2*x^6+105*a^2*b*x^3+140*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

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Maxima [A]  time = 1.05412, size = 43, normalized size = 0.27 \begin{align*} \frac{1}{10} \, b^{3} x^{10} + \frac{3}{7} \, a b^{2} x^{7} + \frac{3}{4} \, a^{2} b x^{4} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*x^10 + 3/7*a*b^2*x^7 + 3/4*a^2*b*x^4 + a^3*x

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Fricas [A]  time = 1.81017, size = 74, normalized size = 0.46 \begin{align*} \frac{1}{10} \, b^{3} x^{10} + \frac{3}{7} \, a b^{2} x^{7} + \frac{3}{4} \, a^{2} b x^{4} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*x^10 + 3/7*a*b^2*x^7 + 3/4*a^2*b*x^4 + a^3*x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12193, size = 86, normalized size = 0.53 \begin{align*} \frac{1}{10} \, b^{3} x^{10} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{3}{7} \, a b^{2} x^{7} \mathrm{sgn}\left (b x^{3} + a\right ) + \frac{3}{4} \, a^{2} b x^{4} \mathrm{sgn}\left (b x^{3} + a\right ) + a^{3} x \mathrm{sgn}\left (b x^{3} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*x^10*sgn(b*x^3 + a) + 3/7*a*b^2*x^7*sgn(b*x^3 + a) + 3/4*a^2*b*x^4*sgn(b*x^3 + a) + a^3*x*sgn(b*x^3 +
a)