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

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

[Out]

a^3*x*(b^2*x^6+2*a*b*x^3+a^2)^(3/2)/(b*x^3+a)^3+3/4*a^2*b*x^4*(b^2*x^6+2*a*b*x^3+a^2)^(3/2)/(b*x^3+a)^3+3/7*a*
b^2*x^7*(b^2*x^6+2*a*b*x^3+a^2)^(3/2)/(b*x^3+a)^3+1/10*b^3*x^10*(b^2*x^6+2*a*b*x^3+a^2)^(3/2)/(b*x^3+a)^3

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Rubi [A]  time = 0.03, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 59, normalized size = 0.36 \[ \frac {x \sqrt {\left (a+b x^3\right )^2} \left (140 a^3+105 a^2 b x^3+60 a b^2 x^6+14 b^3 x^9\right )}{140 \left (a+b x^3\right )} \]

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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fricas [A]  time = 0.72, size = 32, normalized size = 0.20 \[ \frac {1}{10} \, b^{3} x^{10} + \frac {3}{7} \, a b^{2} x^{7} + \frac {3}{4} \, a^{2} b x^{4} + a^{3} x \]

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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giac [A]  time = 0.36, size = 64, normalized size = 0.40 \[ \frac {1}{10} \, b^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{7} \, a b^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{4} \, a^{2} b x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{3} + a\right ) \]

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)

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maple [A]  time = 0.00, size = 56, normalized size = 0.35 \[ \frac {\left (14 b^{3} x^{9}+60 a \,b^{2} x^{6}+105 a^{2} b \,x^{3}+140 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}} x}{140 \left (b \,x^{3}+a \right )^{3}} \]

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 = 0.71, size = 32, normalized size = 0.20 \[ \frac {1}{10} \, b^{3} x^{10} + \frac {3}{7} \, a b^{2} x^{7} + \frac {3}{4} \, a^{2} b x^{4} + a^{3} x \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{\frac {3}{2}}\, dx \]

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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