3.3.89 \(\int \frac {x^3}{\sqrt {-3 x^2+4 x^4}} \, dx\) [289]

Optimal. Leaf size=45 \[ \frac {1}{8} \sqrt {-3 x^2+4 x^4}+\frac {3}{16} \tanh ^{-1}\left (\frac {2 x^2}{\sqrt {-3 x^2+4 x^4}}\right ) \]

[Out]

3/16*arctanh(2*x^2/(4*x^4-3*x^2)^(1/2))+1/8*(4*x^4-3*x^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2043, 654, 634, 212} \begin {gather*} \frac {1}{8} \sqrt {4 x^4-3 x^2}+\frac {3}{16} \tanh ^{-1}\left (\frac {2 x^2}{\sqrt {4 x^4-3 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[-3*x^2 + 4*x^4],x]

[Out]

Sqrt[-3*x^2 + 4*x^4]/8 + (3*ArcTanh[(2*x^2)/Sqrt[-3*x^2 + 4*x^4]])/16

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 634

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

Rule 654

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

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {-3 x^2+4 x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x}{\sqrt {-3 x+4 x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{8} \sqrt {-3 x^2+4 x^4}+\frac {3}{16} \text {Subst}\left (\int \frac {1}{\sqrt {-3 x+4 x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{8} \sqrt {-3 x^2+4 x^4}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-4 x^2} \, dx,x,\frac {x^2}{\sqrt {-3 x^2+4 x^4}}\right )\\ &=\frac {1}{8} \sqrt {-3 x^2+4 x^4}+\frac {3}{16} \tanh ^{-1}\left (\frac {2 x^2}{\sqrt {-3 x^2+4 x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 58, normalized size = 1.29 \begin {gather*} \frac {x \left (-6 x+8 x^3-3 \sqrt {-3+4 x^2} \log \left (-2 x+\sqrt {-3+4 x^2}\right )\right )}{16 \sqrt {x^2 \left (-3+4 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[-3*x^2 + 4*x^4],x]

[Out]

(x*(-6*x + 8*x^3 - 3*Sqrt[-3 + 4*x^2]*Log[-2*x + Sqrt[-3 + 4*x^2]]))/(16*Sqrt[x^2*(-3 + 4*x^2)])

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Maple [A]
time = 0.12, size = 60, normalized size = 1.33

method	result	size
trager	\(\frac {\sqrt {4 x^{4}-3 x^{2}}}{8}+\frac {3 \ln \left (\frac {2 x^{2}+\sqrt {4 x^{4}-3 x^{2}}}{x}\right )}{16}\)	\(43\)
default	\(\frac {x \sqrt {4 x^{2}-3}\, \left (3 \ln \left (\sqrt {4}\, x +\sqrt {4 x^{2}-3}\right ) \sqrt {4}+4 x \sqrt {4 x^{2}-3}\right )}{32 \sqrt {4 x^{4}-3 x^{2}}}\)	\(60\)
meijerg	\(\frac {3 i \sqrt {-\mathrm {signum}\left (-1+\frac {4 x^{2}}{3}\right )}\, \left (\frac {2 i \sqrt {\pi }\, x \sqrt {3}\, \sqrt {-\frac {4 x^{2}}{3}+1}}{3}-i \sqrt {\pi }\, \arcsin \left (\frac {2 x \sqrt {3}}{3}\right )\right )}{16 \sqrt {\pi }\, \sqrt {\mathrm {signum}\left (-1+\frac {4 x^{2}}{3}\right )}}\)	\(62\)
risch	\(\frac {x^{2} \left (4 x^{2}-3\right )}{8 \sqrt {x^{2} \left (4 x^{2}-3\right )}}+\frac {3 \ln \left (\sqrt {4}\, x +\sqrt {4 x^{2}-3}\right ) \sqrt {4}\, x \sqrt {4 x^{2}-3}}{32 \sqrt {x^{2} \left (4 x^{2}-3\right )}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(4*x^4-3*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/32*x*(4*x^2-3)^(1/2)*(3*ln(4^(1/2)*x+(4*x^2-3)^(1/2))*4^(1/2)+4*x*(4*x^2-3)^(1/2))/(4*x^4-3*x^2)^(1/2)

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Maxima [A]
time = 0.50, size = 41, normalized size = 0.91 \begin {gather*} \frac {1}{8} \, \sqrt {4 \, x^{4} - 3 \, x^{2}} + \frac {3}{32} \, \log \left (8 \, x^{2} + 4 \, \sqrt {4 \, x^{4} - 3 \, x^{2}} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(4*x^4-3*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/8*sqrt(4*x^4 - 3*x^2) + 3/32*log(8*x^2 + 4*sqrt(4*x^4 - 3*x^2) - 3)

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Fricas [A]
time = 0.34, size = 45, normalized size = 1.00 \begin {gather*} \frac {1}{8} \, \sqrt {4 \, x^{4} - 3 \, x^{2}} - \frac {3}{16} \, \log \left (-\frac {2 \, x^{2} - \sqrt {4 \, x^{4} - 3 \, x^{2}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(4*x^4-3*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(4*x^4 - 3*x^2) - 3/16*log(-(2*x^2 - sqrt(4*x^4 - 3*x^2))/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {x^{2} \cdot \left (4 x^{2} - 3\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(4*x**4-3*x**2)**(1/2),x)

[Out]

Integral(x**3/sqrt(x**2*(4*x**2 - 3)), x)

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Giac [A]
time = 3.76, size = 44, normalized size = 0.98 \begin {gather*} \frac {3}{32} \, \log \left (3\right ) \mathrm {sgn}\left (x\right ) + \frac {\sqrt {4 \, x^{2} - 3} x}{8 \, \mathrm {sgn}\left (x\right )} - \frac {3 \, \log \left ({\left | -2 \, x + \sqrt {4 \, x^{2} - 3} \right |}\right )}{16 \, \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(4*x^4-3*x^2)^(1/2),x, algorithm="giac")

[Out]

3/32*log(3)*sgn(x) + 1/8*sqrt(4*x^2 - 3)*x/sgn(x) - 3/16*log(abs(-2*x + sqrt(4*x^2 - 3)))/sgn(x)

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Mupad [B]
time = 4.46, size = 40, normalized size = 0.89 \begin {gather*} \frac {3\,\ln \left (\frac {\sqrt {4\,x^2-3}\,\sqrt {x^2}}{2}+x^2-\frac {3}{8}\right )}{32}+\frac {\sqrt {4\,x^4-3\,x^2}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(4*x^4 - 3*x^2)^(1/2),x)

[Out]

(3*log(((4*x^2 - 3)^(1/2)*(x^2)^(1/2))/2 + x^2 - 3/8))/32 + (4*x^4 - 3*x^2)^(1/2)/8

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