\(\int \frac {x^7}{2+3 x^4} \, dx\) [688]

Optimal result

Integrand size = 13, antiderivative size = 20 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {x^4}{12}-\frac {1}{18} \log \left (2+3 x^4\right ) \]

[Out]

1/12*x^4-1/18*ln(3*x^4+2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {x^4}{12}-\frac {1}{18} \log \left (3 x^4+2\right ) \]

[In]

Int[x^7/(2 + 3*x^4),x]

[Out]

x^4/12 - Log[2 + 3*x^4]/18

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {x}{2+3 x} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{3}-\frac {2}{3 (2+3 x)}\right ) \, dx,x,x^4\right ) \\ & = \frac {x^4}{12}-\frac {1}{18} \log \left (2+3 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {1}{36} \left (2+3 x^4-2 \log \left (2+3 x^4\right )\right ) \]

[In]

Integrate[x^7/(2 + 3*x^4),x]

[Out]

(2 + 3*x^4 - 2*Log[2 + 3*x^4])/36

Maple [A] (verified)

Time = 3.92 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {x^{4}}{12}-\frac {\ln \left (x^{4}+\frac {2}{3}\right )}{18}\)	\(15\)
default	\(\frac {x^{4}}{12}-\frac {\ln \left (3 x^{4}+2\right )}{18}\)	\(17\)
norman	\(\frac {x^{4}}{12}-\frac {\ln \left (3 x^{4}+2\right )}{18}\)	\(17\)
meijerg	\(\frac {x^{4}}{12}-\frac {\ln \left (\frac {3 x^{4}}{2}+1\right )}{18}\)	\(17\)
risch	\(\frac {x^{4}}{12}-\frac {\ln \left (3 x^{4}+2\right )}{18}\)	\(17\)

[In]

int(x^7/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/12*x^4-1/18*ln(x^4+2/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {1}{12} \, x^{4} - \frac {1}{18} \, \log \left (3 \, x^{4} + 2\right ) \]

[In]

integrate(x^7/(3*x^4+2),x, algorithm="fricas")

[Out]

1/12*x^4 - 1/18*log(3*x^4 + 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {x^{4}}{12} - \frac {\log {\left (3 x^{4} + 2 \right )}}{18} \]

[In]

integrate(x**7/(3*x**4+2),x)

[Out]

x**4/12 - log(3*x**4 + 2)/18

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {1}{12} \, x^{4} - \frac {1}{18} \, \log \left (3 \, x^{4} + 2\right ) \]

[In]

integrate(x^7/(3*x^4+2),x, algorithm="maxima")

[Out]

1/12*x^4 - 1/18*log(3*x^4 + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {1}{12} \, x^{4} - \frac {1}{18} \, \log \left (3 \, x^{4} + 2\right ) \]

[In]

integrate(x^7/(3*x^4+2),x, algorithm="giac")

[Out]

1/12*x^4 - 1/18*log(3*x^4 + 2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {x^7}{2+3 x^4} \, dx=\frac {x^4}{12}-\frac {\ln \left (x^4+\frac {2}{3}\right )}{18} \]

[In]

int(x^7/(3*x^4 + 2),x)

[Out]

x^4/12 - log(x^4 + 2/3)/18