\(\int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx\) [6]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {a x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {b x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )} \]

[Out]

1/6*a*x^6*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/9*b*x^9*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 14} \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {a x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {b x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )} \]

[In]

Int[x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1369

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^5 \left (a b+b^2 x^3\right ) \, dx}{a b+b^2 x^3} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (a b x^5+b^2 x^8\right ) \, dx}{a b+b^2 x^3} \\ & = \frac {a x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {b x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (3 a x^6+2 b x^9\right )}{18 \left (a+b x^3\right )} \]

[In]

Integrate[x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

(Sqrt[(a + b*x^3)^2]*(3*a*x^6 + 2*b*x^9))/(18*(a + b*x^3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.39

method result size
pseudoelliptic \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{2} \left (-2 b \,x^{3}+a \right )}{18 b^{2}}\) \(31\)
gosper \(\frac {x^{6} \left (2 b \,x^{3}+3 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{18 b \,x^{3}+18 a}\) \(36\)
default \(\frac {x^{6} \left (2 b \,x^{3}+3 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{18 b \,x^{3}+18 a}\) \(36\)
risch \(\frac {a \,x^{6} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{6 b \,x^{3}+6 a}+\frac {b \,x^{9} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{9 b \,x^{3}+9 a}\) \(54\)

[In]

int(x^5*((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/18*csgn(b*x^3+a)*(b*x^3+a)^2*(-2*b*x^3+a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.16 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {1}{9} \, b x^{9} + \frac {1}{6} \, a x^{6} \]

[In]

integrate(x^5*((b*x^3+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/9*b*x^9 + 1/6*a*x^6

Sympy [F(-1)]

Timed out. \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\text {Timed out} \]

[In]

integrate(x**5*((b*x**3+a)**2)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=-\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a x^{3}}{6 \, b} - \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2}}{6 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}}}{9 \, b^{2}} \]

[In]

integrate(x^5*((b*x^3+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a*x^3/b - 1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2/b^2 + 1/9*(b^2*x^6 + 2*
a*b*x^3 + a^2)^(3/2)/b^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.29 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {1}{18} \, {\left (2 \, b x^{9} + 3 \, a x^{6}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \]

[In]

integrate(x^5*((b*x^3+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/18*(2*b*x^9 + 3*a*x^6)*sgn(b*x^3 + a)

Mupad [B] (verification not implemented)

Time = 8.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.75 \[ \int x^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}\,\left (8\,b^2\,\left (a^2+b^2\,x^6\right )-12\,a^2\,b^2+4\,a\,b^3\,x^3\right )}{72\,b^4} \]

[In]

int(x^5*((a + b*x^3)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2)*(8*b^2*(a^2 + b^2*x^6) - 12*a^2*b^2 + 4*a*b^3*x^3))/(72*b^4)