\(\int \frac {d x^3}{2+3 x^4} \, dx\) [159]

Optimal result

Integrand size = 14, antiderivative size = 13 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} d \log \left (2+3 x^4\right ) \]

[Out]

1/12*d*ln(3*x^4+2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 266} \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} d \log \left (3 x^4+2\right ) \]

[In]

Int[(d*x^3)/(2 + 3*x^4),x]

[Out]

(d*Log[2 + 3*x^4])/12

Rule 12

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps \begin{align*} \text {integral}& = d \int \frac {x^3}{2+3 x^4} \, dx \\ & = \frac {1}{12} d \log \left (2+3 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} d \log \left (2+3 x^4\right ) \]

[In]

Integrate[(d*x^3)/(2 + 3*x^4),x]

[Out]

(d*Log[2 + 3*x^4])/12

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
parallelrisch	\(\frac {d \ln \left (x^{4}+\frac {2}{3}\right )}{12}\)	\(10\)
derivativedivides	\(\frac {d \ln \left (3 x^{4}+2\right )}{12}\)	\(12\)
default	\(\frac {d \ln \left (3 x^{4}+2\right )}{12}\)	\(12\)
norman	\(\frac {d \ln \left (3 x^{4}+2\right )}{12}\)	\(12\)
meijerg	\(\frac {d \ln \left (\frac {3 x^{4}}{2}+1\right )}{12}\)	\(12\)
risch	\(\frac {d \ln \left (3 x^{4}+2\right )}{12}\)	\(12\)

[In]

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

[Out]

1/12*d*ln(x^4+2/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} \, d \log \left (3 \, x^{4} + 2\right ) \]

[In]

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

[Out]

1/12*d*log(3*x^4 + 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {d \log {\left (3 x^{4} + 2 \right )}}{12} \]

[In]

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

[Out]

d*log(3*x**4 + 2)/12

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} \, d \log \left (3 \, x^{4} + 2\right ) \]

[In]

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

[Out]

1/12*d*log(3*x^4 + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {1}{12} \, d \log \left (3 \, x^{4} + 2\right ) \]

[In]

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

[Out]

1/12*d*log(3*x^4 + 2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {d x^3}{2+3 x^4} \, dx=\frac {d\,\ln \left (x^4+\frac {2}{3}\right )}{12} \]

[In]

int((d*x^3)/(3*x^4 + 2),x)

[Out]

(d*log(x^4 + 2/3))/12