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

Optimal. Leaf size=43 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1352, 621, 206} \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.07, size = 118, normalized size = 2.74 \[ \left [\frac {\log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right )}{6 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right )}{3 \, c}\right ] \]

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]

[1/6*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)/sqrt(c), -1/3
*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))/c]

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giac [A]  time = 0.68, size = 40, normalized size = 0.93 \[ -\frac {\log \left ({\left | -2 \, {\left (\sqrt {c} x^{3} - \sqrt {c x^{6} + b x^{3} + a}\right )} \sqrt {c} - b \right |}\right )}{3 \, \sqrt {c}} \]

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]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {c \,x^{6}+b \,x^{3}+a}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is 4*a*c-b^2 zero or nonzero?

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mupad [B]  time = 1.58, size = 34, normalized size = 0.79 \[ \frac {\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )}{3\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

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

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

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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