\(\int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx\) [287]

Optimal result

Integrand size = 19, antiderivative size = 34 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=-\frac {1}{8} \sqrt {-3 x^2-4 x^4}-\frac {3}{32} \arcsin \left (1+\frac {8 x^2}{3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2043, 654, 633, 222} \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=-\frac {3}{32} \arcsin \left (\frac {8 x^2}{3}+1\right )-\frac {1}{8} \sqrt {-4 x^4-3 x^2} \]

[In]

Int[x^3/Sqrt[-3*x^2 - 4*x^4],x]

[Out]

-1/8*Sqrt[-3*x^2 - 4*x^4] - (3*ArcSin[1 + (8*x^2)/3])/32

Rule 222

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

Rule 633

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

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*} \text {integral}& = \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 {1}{32} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{9}}} \, dx,x,-3-8 x^2\right ) \\ & = -\frac {1}{8} \sqrt {-3 x^2-4 x^4}-\frac {3}{32} \sin ^{-1}\left (1+\frac {8 x^2}{3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=\frac {x \left (6 x+8 x^3+3 \sqrt {3+4 x^2} \log \left (-2 x+\sqrt {3+4 x^2}\right )\right )}{16 \sqrt {-x^2 \left (3+4 x^2\right )}} \]

[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))])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {3 i \operatorname {arcsinh}\left (\frac {8}{3} i x^{2}+i\right )}{32}-\frac {\sqrt {-4 x^{4}-3 x^{2}}}{8}\)	\(30\)
meijerg	\(-\frac {3 i \left (\frac {2 \sqrt {\pi }\, x \sqrt {3}\, \sqrt {\frac {4 x^{2}}{3}+1}}{3}-\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {2 x \sqrt {3}}{3}\right )\right )}{16 \sqrt {\pi }}\)	\(38\)
default	\(-\frac {x \sqrt {-4 x^{2}-3}\, \left (2 x \sqrt {-4 x^{2}-3}+3 \arctan \left (\frac {2 x}{\sqrt {-4 x^{2}-3}}\right )\right )}{16 \sqrt {-4 x^{4}-3 x^{2}}}\)	\(54\)
trager	\(-\frac {\sqrt {-4 x^{4}-3 x^{2}}}{8}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\sqrt {-4 x^{4}-3 x^{2}}}{x}\right )}{16}\)	\(55\)
risch	\(\frac {x^{2} \left (4 x^{2}+3\right )}{8 \sqrt {-x^{2} \left (4 x^{2}+3\right )}}-\frac {3 \arctan \left (\frac {2 x}{\sqrt {-4 x^{2}-3}}\right ) x \sqrt {-4 x^{2}-3}}{16 \sqrt {-x^{2} \left (4 x^{2}+3\right )}}\)	\(67\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=-\frac {1}{8} \, \sqrt {-4 \, x^{2} - 3} x - \frac {3}{32} i \, \log \left (-\frac {4 \, {\left (2 \, x + i \, \sqrt {-4 \, x^{2} - 3}\right )}}{x}\right ) + \frac {3}{32} i \, \log \left (-\frac {4 \, {\left (2 \, x - i \, \sqrt {-4 \, x^{2} - 3}\right )}}{x}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=\int \frac {x^{3}}{\sqrt {- x^{2} \cdot \left (4 x^{2} + 3\right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=-\frac {1}{8} \, \sqrt {-4 \, x^{4} - 3 \, x^{2}} + \frac {3}{32} \, \arcsin \left (-\frac {8}{3} \, x^{2} - 1\right ) \]

[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*arcsin(-8/3*x^2 - 1)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=\frac {3}{32} i \, \log \left (3\right ) \mathrm {sgn}\left (x\right ) - \frac {i \, \sqrt {4 \, x^{2} + 3} x}{8 \, \mathrm {sgn}\left (x\right )} - \frac {3 i \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 3}\right )}{16 \, \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.85 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{\sqrt {-3 x^2-4 x^4}} \, dx=-\frac {\sqrt {-4\,x^4-3\,x^2}}{8}+\frac {\ln \left (\frac {\sqrt {4\,x^2+3}\,\sqrt {x^2}}{2}+x^2+\frac {3}{8}\right )\,3{}\mathrm {i}}{32} \]

[In]

int(x^3/(- 3*x^2 - 4*x^4)^(1/2),x)

[Out]

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