3.189 \(\int x^2 \sqrt{a+b x^3+c x^6} \, dx\)

Optimal. Leaf size=83 \[ \frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{24 c^{3/2}} \]

[Out]

((b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(12*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^
3 + c*x^6])])/(24*c^(3/2))

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Rubi [A]  time = 0.0601921, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1352, 612, 621, 206} \[ \frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{24 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

((b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(12*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^
3 + c*x^6])])/(24*c^(3/2))

Rule 1352

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]

Rule 612

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 621

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 206

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

Rubi steps

\begin{align*} \int x^2 \sqrt{a+b x^3+c x^6} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^3\right )\\ &=\frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^3\right )}{24 c}\\ &=\frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (b^2-4 a c\right ) \operatorname{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 )}{12 c}\\ &=\frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{12 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{24 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.010546, size = 87, normalized size = 1.05 \[ \frac{1}{3} \left (\frac{\left (b+2 c x^3\right ) \sqrt{a+b x^3+c x^6}}{4 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^3}{2 \sqrt{c} \sqrt{a+b x^3+c x^6}}\right )}{8 c^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\sqrt{c{x}^{6}+b{x}^{3}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(x^2*(c*x^6+b*x^3+a)^(1/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54722, size = 463, normalized size = 5.58 \begin{align*} \left [-\frac{{\left (b^{2} - 4 \, a c\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 \, \sqrt{c x^{6} + b x^{3} + a}{\left (2 \, c^{2} x^{3} + b c\right )}}{48 \, c^{2}}, \frac{{\left (b^{2} - 4 \, a c\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 \, \sqrt{c x^{6} + b x^{3} + a}{\left (2 \, c^{2} x^{3} + b c\right )}}{24 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*((b^2 - 4*a*c)*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*sqrt(c*x^6 + b*x^3 + a)*(2*c^2*x^3 + b*c))/c^2, 1/24*((b^2 - 4*a*c)*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*sqrt(c*x^6 + b*x^3 + a)*(2*c^2*x^3 + b
*c))/c^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c*x**6+b*x**3+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.15953, size = 103, normalized size = 1.24 \begin{align*} \frac{1}{12} \, \sqrt{c x^{6} + b x^{3} + a}{\left (2 \, x^{3} + \frac{b}{c}\right )} + \frac{{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{3} - \sqrt{c x^{6} + b x^{3} + a}\right )} \sqrt{c} - b \right |}\right )}{24 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(c*x^6 + b*x^3 + a)*(2*x^3 + b/c) + 1/24*(b^2 - 4*a*c)*log(abs(-2*(sqrt(c)*x^3 - sqrt(c*x^6 + b*x^3 +
a))*sqrt(c) - b))/c^(3/2)