\(\int (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [32]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a^3 x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{\left (a+b x^3\right )^3}+\frac {3 a^2 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{4 \left (a+b x^3\right )^3}+\frac {3 a b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{7 \left (a+b x^3\right )^3}+\frac {b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{10 \left (a+b x^3\right )^3} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1357, 200} \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {3 a b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{7 \left (a+b x^3\right )^3}+\frac {3 a^2 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{4 \left (a+b x^3\right )^3}+\frac {b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{10 \left (a+b x^3\right )^3}+\frac {a^3 x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{\left (a+b x^3\right )^3} \]

[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 200

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 1357

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*} \text {integral}& = \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] (verified)

Time = 1.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x \sqrt {\left (a+b x^3\right )^2} \left (140 a^3+105 a^2 b x^3+60 a b^2 x^6+14 b^3 x^9\right )}{140 \left (a+b x^3\right )} \]

[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))

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.35

method result size
gosper \(\frac {x \left (14 b^{3} x^{9}+60 b^{2} x^{6} a +105 a^{2} b \,x^{3}+140 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{140 \left (b \,x^{3}+a \right )^{3}}\) \(56\)
default \(\frac {x \left (14 b^{3} x^{9}+60 b^{2} x^{6} a +105 a^{2} b \,x^{3}+140 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{140 \left (b \,x^{3}+a \right )^{3}}\) \(56\)
risch \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} x^{10}}{10 b \,x^{3}+10 a}+\frac {3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \,b^{2} x^{7}}{7 \left (b \,x^{3}+a \right )}+\frac {3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b \,x^{4}}{4 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} x}{b \,x^{3}+a}\) \(113\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.20 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{10} \, b^{3} x^{10} + \frac {3}{7} \, a b^{2} x^{7} + \frac {3}{4} \, a^{2} b x^{4} + a^{3} x \]

[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

Sympy [F]

\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{\frac {3}{2}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.20 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{10} \, b^{3} x^{10} + \frac {3}{7} \, a b^{2} x^{7} + \frac {3}{4} \, a^{2} b x^{4} + a^{3} x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.40 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {1}{10} \, b^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{7} \, a b^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{4} \, a^{2} b x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{3} + a\right ) \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \]

[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)