\(\int x^2 (a+b x^3+c x^6)^{3/2} \, dx\) [205]

Optimal result

Integrand size = 20, antiderivative size = 124 \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{64 c^2}+\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c}+\frac {\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{128 c^{5/2}} \]

[Out]

1/24*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(3/2)/c+1/128*(-4*a*c+b^2)^2*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^
(1/2))/c^(5/2)-1/64*(-4*a*c+b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1366, 626, 635, 212} \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {\left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{128 c^{5/2}}-\frac {\left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{64 c^2}+\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c} \]

[In]

Int[x^2*(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

-1/64*((b^2 - 4*a*c)*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/c^2 + ((b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(3/2))/(2
4*c) + ((b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(128*c^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^3\right ) \\ & = \frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c}-\frac {\left (b^2-4 a c\right ) \text {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{16 c} \\ & = -\frac {\left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{64 c^2}+\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{128 c^2} \\ & = -\frac {\left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{64 c^2}+\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c}+\frac {\left (b^2-4 a c\right )^2 \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^3}{\sqrt {a+b x^3+c x^6}}\right )}{64 c^2} \\ & = -\frac {\left (b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{64 c^2}+\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{24 c}+\frac {\left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{128 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6} \left (-3 b^2+20 a c+8 b c x^3+8 c^2 x^6\right )}{192 c^2}+\frac {\left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^3}{-\sqrt {a}+\sqrt {a+b x^3+c x^6}}\right )}{64 c^{5/2}} \]

[In]

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

[Out]

((b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6]*(-3*b^2 + 20*a*c + 8*b*c*x^3 + 8*c^2*x^6))/(192*c^2) + ((-b^2 + 4*a*c)^
2*ArcTanh[(Sqrt[c]*x^3)/(-Sqrt[a] + Sqrt[a + b*x^3 + c*x^6])])/(64*c^(5/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.40 \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{9} + 24 \, b c^{3} x^{6} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{768 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{9} + 24 \, b c^{3} x^{6} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{384 \, c^{3}}\right ] \]

[In]

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

[Out]

[1/768*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 - 4*sqrt(c*x^6 + b*x^3 + a)*
(2*c*x^3 + b)*sqrt(c) - 4*a*c) + 4*(16*c^4*x^9 + 24*b*c^3*x^6 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*
x^3)*sqrt(c*x^6 + b*x^3 + a))/c^3, -1/384*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^6 + b
*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 2*(16*c^4*x^9 + 24*b*c^3*x^6 - 3*b^3*c + 20*a*b*
c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^3]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {1}{192} \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, {\left (4 \, {\left (2 \, c x^{3} + 3 \, b\right )} x^{3} + \frac {b^{2} c^{2} + 20 \, a c^{3}}{c^{3}}\right )} x^{3} - \frac {3 \, b^{3} c - 20 \, a b c^{2}}{c^{3}}\right )} - \frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{3} - \sqrt {c x^{6} + b x^{3} + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/192*sqrt(c*x^6 + b*x^3 + a)*(2*(4*(2*c*x^3 + 3*b)*x^3 + (b^2*c^2 + 20*a*c^3)/c^3)*x^3 - (3*b^3*c - 20*a*b*c^
2)/c^3) - 1/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x^3 - sqrt(c*x^6 + b*x^3 + a))*sqrt(c) + b))
/c^(5/2)

Mupad [B] (verification not implemented)

Time = 8.53 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x^2 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {\left (c\,x^3+\frac {b}{2}\right )\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{12\,c}+\frac {\left (3\,a\,c-\frac {3\,b^2}{4}\right )\,\left (\left (\frac {b}{4\,c}+\frac {x^3}{2}\right )\,\sqrt {c\,x^6+b\,x^3+a}+\frac {\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{12\,c} \]

[In]

int(x^2*(a + b*x^3 + c*x^6)^(3/2),x)

[Out]

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