\(\int \frac {c x^2}{2+3 x^4} \, dx\) [155]

Optimal result

Integrand size = 14, antiderivative size = 101 \[ \int \frac {c x^2}{2+3 x^4} \, dx=-\frac {c \arctan \left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \arctan \left (1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}} \]

[Out]

1/12*c*arctan(-1+6^(1/4)*x)*6^(1/4)+1/12*c*arctan(1+6^(1/4)*x)*6^(1/4)+1/24*c*ln(-6^(3/4)*x+3*x^2+6^(1/2))*6^(
1/4)-1/24*c*ln(6^(3/4)*x+3*x^2+6^(1/2))*6^(1/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {12, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {c x^2}{2+3 x^4} \, dx=-\frac {c \arctan \left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \arctan \left (\sqrt [4]{6} x+1\right )}{2\ 6^{3/4}}+\frac {c \log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{4\ 6^{3/4}}-\frac {c \log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{4\ 6^{3/4}} \]

[In]

Int[(c*x^2)/(2 + 3*x^4),x]

[Out]

-1/2*(c*ArcTan[1 - 6^(1/4)*x])/6^(3/4) + (c*ArcTan[1 + 6^(1/4)*x])/(2*6^(3/4)) + (c*Log[Sqrt[6] - 6^(3/4)*x +
3*x^2])/(4*6^(3/4)) - (c*Log[Sqrt[6] + 6^(3/4)*x + 3*x^2])/(4*6^(3/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = c \int \frac {x^2}{2+3 x^4} \, dx \\ & = -\frac {c \int \frac {\sqrt {2}-\sqrt {3} x^2}{2+3 x^4} \, dx}{2 \sqrt {3}}+\frac {c \int \frac {\sqrt {2}+\sqrt {3} x^2}{2+3 x^4} \, dx}{2 \sqrt {3}} \\ & = \frac {1}{12} c \int \frac {1}{\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx+\frac {1}{12} c \int \frac {1}{\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx+\frac {c \int \frac {\frac {2^{3/4}}{\sqrt [4]{3}}+2 x}{-\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}-x^2} \, dx}{4\ 6^{3/4}}+\frac {c \int \frac {\frac {2^{3/4}}{\sqrt [4]{3}}-2 x}{-\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}-x^2} \, dx}{4\ 6^{3/4}} \\ & = \frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}+\frac {c \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}-\frac {c \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}} \\ & = -\frac {c \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \tan ^{-1}\left (1+\sqrt [4]{6} x\right )}{2\ 6^{3/4}}+\frac {c \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}}-\frac {c \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{4\ 6^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.77 \[ \int \frac {c x^2}{2+3 x^4} \, dx=\frac {c \left (-2 \arctan \left (1-\sqrt [4]{6} x\right )+2 \arctan \left (1+\sqrt [4]{6} x\right )+\log \left (2-2 \sqrt [4]{6} x+\sqrt {6} x^2\right )-\log \left (2+2 \sqrt [4]{6} x+\sqrt {6} x^2\right )\right )}{4\ 6^{3/4}} \]

[In]

Integrate[(c*x^2)/(2 + 3*x^4),x]

[Out]

(c*(-2*ArcTan[1 - 6^(1/4)*x] + 2*ArcTan[1 + 6^(1/4)*x] + Log[2 - 2*6^(1/4)*x + Sqrt[6]*x^2] - Log[2 + 2*6^(1/4
)*x + Sqrt[6]*x^2]))/(4*6^(3/4))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.25

method	result	size
risch	\(\frac {c \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{12}\)	\(25\)
default	\(\frac {c \sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )\right )}{144}\)	\(94\)
meijerg	\(\frac {54^{\frac {3}{4}} c \left (\frac {x^{3} \sqrt {2}\, \ln \left (1-6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8-3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8+3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}\right )}{216}\)	\(171\)

[In]

int(c*x^2/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/12*c*sum(1/_R*ln(x-_R),_R=RootOf(3*_Z^4+2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int \frac {c x^2}{2+3 x^4} \, dx=\frac {1}{216} \cdot 54^{\frac {3}{4}} \left (-c^{4}\right )^{\frac {1}{4}} \log \left (3 \, c^{3} x + 54^{\frac {1}{4}} \left (-c^{4}\right )^{\frac {3}{4}}\right ) - \frac {1}{216} i \cdot 54^{\frac {3}{4}} \left (-c^{4}\right )^{\frac {1}{4}} \log \left (3 \, c^{3} x + i \cdot 54^{\frac {1}{4}} \left (-c^{4}\right )^{\frac {3}{4}}\right ) + \frac {1}{216} i \cdot 54^{\frac {3}{4}} \left (-c^{4}\right )^{\frac {1}{4}} \log \left (3 \, c^{3} x - i \cdot 54^{\frac {1}{4}} \left (-c^{4}\right )^{\frac {3}{4}}\right ) - \frac {1}{216} \cdot 54^{\frac {3}{4}} \left (-c^{4}\right )^{\frac {1}{4}} \log \left (3 \, c^{3} x - 54^{\frac {1}{4}} \left (-c^{4}\right )^{\frac {3}{4}}\right ) \]

[In]

integrate(c*x^2/(3*x^4+2),x, algorithm="fricas")

[Out]

1/216*54^(3/4)*(-c^4)^(1/4)*log(3*c^3*x + 54^(1/4)*(-c^4)^(3/4)) - 1/216*I*54^(3/4)*(-c^4)^(1/4)*log(3*c^3*x +
I*54^(1/4)*(-c^4)^(3/4)) + 1/216*I*54^(3/4)*(-c^4)^(1/4)*log(3*c^3*x - I*54^(1/4)*(-c^4)^(3/4)) - 1/216*54^(3
/4)*(-c^4)^(1/4)*log(3*c^3*x - 54^(1/4)*(-c^4)^(3/4))

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {c x^2}{2+3 x^4} \, dx=c \left (\frac {\sqrt [4]{6} \log {\left (x^{2} - \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{24} - \frac {\sqrt [4]{6} \log {\left (x^{2} + \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{24} + \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x - 1 \right )}}{12} + \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x + 1 \right )}}{12}\right ) \]

[In]

integrate(c*x**2/(3*x**4+2),x)

[Out]

c*(6**(1/4)*log(x**2 - 6**(3/4)*x/3 + sqrt(6)/3)/24 - 6**(1/4)*log(x**2 + 6**(3/4)*x/3 + sqrt(6)/3)/24 + 6**(1
/4)*atan(6**(1/4)*x - 1)/12 + 6**(1/4)*atan(6**(1/4)*x + 1)/12)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.22 \[ \int \frac {c x^2}{2+3 x^4} \, dx=\frac {1}{24} \, {\left (2 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + 2 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) - 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) + 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right )\right )} c \]

[In]

integrate(c*x^2/(3*x^4+2),x, algorithm="maxima")

[Out]

1/24*(2*3^(1/4)*2^(1/4)*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x + 3^(1/4)*2^(3/4))) + 2*3^(1/4)*2^(1/4)*arctan
(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x - 3^(1/4)*2^(3/4))) - 3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/4)*x +
sqrt(2)) + 3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2)))*c

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {c x^2}{2+3 x^4} \, dx=\frac {1}{24} \, {\left (2 \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + 2 \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - 6^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) + 6^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right )\right )} c \]

[In]

integrate(c*x^2/(3*x^4+2),x, algorithm="giac")

[Out]

1/24*(2*6^(1/4)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) + 2*6^(1/4)*arctan(3/4*sqrt(2)*(2/
3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) - 6^(1/4)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) + 6^(1/4)*log(x^2
- sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)))*c

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.32 \[ \int \frac {c x^2}{2+3 x^4} \, dx=\frac {{\left (-1\right )}^{1/4}\,{24}^{1/4}\,c\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,{24}^{1/4}\,x}{2}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,{24}^{1/4}\,x}{2}\right )\right )}{12} \]

[In]

int((c*x^2)/(3*x^4 + 2),x)

[Out]

((-1)^(1/4)*24^(1/4)*c*(atan(((-1)^(1/4)*24^(1/4)*x)/2) - atanh(((-1)^(1/4)*24^(1/4)*x)/2)))/12