\(\int \frac {1}{x^7 (3+4 x^3+x^6)} \, dx\) [157]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=-\frac {1}{18 x^6}+\frac {4}{27 x^3}+\frac {13 \log (x)}{27}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{162} \log \left (3+x^3\right ) \]

[Out]

-1/18/x^6+4/27/x^3+13/27*ln(x)-1/6*ln(x^3+1)+1/162*ln(x^3+3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1371, 723, 814} \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=-\frac {1}{18 x^6}+\frac {4}{27 x^3}-\frac {1}{6} \log \left (x^3+1\right )+\frac {1}{162} \log \left (x^3+3\right )+\frac {13 \log (x)}{27} \]

[In]

Int[1/(x^7*(3 + 4*x^3 + x^6)),x]

[Out]

-1/18*1/x^6 + 4/(27*x^3) + (13*Log[x])/27 - Log[1 + x^3]/6 + Log[3 + x^3]/162

Rule 723

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m + 1)/((m
+ 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d + e*x)^(m + 1)*(Simp[c*d - b*e - c
*e*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x^3 \left (3+4 x+x^2\right )} \, dx,x,x^3\right ) \\ & = -\frac {1}{18 x^6}+\frac {1}{9} \text {Subst}\left (\int \frac {-4-x}{x^2 \left (3+4 x+x^2\right )} \, dx,x,x^3\right ) \\ & = -\frac {1}{18 x^6}+\frac {1}{9} \text {Subst}\left (\int \left (-\frac {4}{3 x^2}+\frac {13}{9 x}-\frac {3}{2 (1+x)}+\frac {1}{18 (3+x)}\right ) \, dx,x,x^3\right ) \\ & = -\frac {1}{18 x^6}+\frac {4}{27 x^3}+\frac {13 \log (x)}{27}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{162} \log \left (3+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=-\frac {1}{18 x^6}+\frac {4}{27 x^3}+\frac {13 \log (x)}{27}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{162} \log \left (3+x^3\right ) \]

[In]

Integrate[1/(x^7*(3 + 4*x^3 + x^6)),x]

[Out]

-1/18*1/x^6 + 4/(27*x^3) + (13*Log[x])/27 - Log[1 + x^3]/6 + Log[3 + x^3]/162

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80

method	result	size
risch	\(\frac {-\frac {1}{18}+\frac {4 x^{3}}{27}}{x^{6}}+\frac {13 \ln \left (x \right )}{27}-\frac {\ln \left (x^{3}+1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{162}\)	\(33\)
default	\(-\frac {1}{18 x^{6}}+\frac {4}{27 x^{3}}+\frac {13 \ln \left (x \right )}{27}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{162}-\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(41\)
norman	\(\frac {-\frac {1}{18}+\frac {4 x^{3}}{27}}{x^{6}}+\frac {13 \ln \left (x \right )}{27}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{162}-\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(42\)
parallelrisch	\(\frac {78 \ln \left (x \right ) x^{6}-27 \ln \left (x +1\right ) x^{6}+\ln \left (x^{3}+3\right ) x^{6}-27 \ln \left (x^{2}-x +1\right ) x^{6}-9+24 x^{3}}{162 x^{6}}\)	\(53\)

[In]

int(1/x^7/(x^6+4*x^3+3),x,method=_RETURNVERBOSE)

[Out]

(-1/18+4/27*x^3)/x^6+13/27*ln(x)-1/6*ln(x^3+1)+1/162*ln(x^3+3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=\frac {x^{6} \log \left (x^{3} + 3\right ) - 27 \, x^{6} \log \left (x^{3} + 1\right ) + 78 \, x^{6} \log \left (x\right ) + 24 \, x^{3} - 9}{162 \, x^{6}} \]

[In]

integrate(1/x^7/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

1/162*(x^6*log(x^3 + 3) - 27*x^6*log(x^3 + 1) + 78*x^6*log(x) + 24*x^3 - 9)/x^6

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=\frac {13 \log {\left (x \right )}}{27} - \frac {\log {\left (x^{3} + 1 \right )}}{6} + \frac {\log {\left (x^{3} + 3 \right )}}{162} + \frac {8 x^{3} - 3}{54 x^{6}} \]

[In]

integrate(1/x**7/(x**6+4*x**3+3),x)

[Out]

13*log(x)/27 - log(x**3 + 1)/6 + log(x**3 + 3)/162 + (8*x**3 - 3)/(54*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=\frac {8 \, x^{3} - 3}{54 \, x^{6}} + \frac {1}{162} \, \log \left (x^{3} + 3\right ) - \frac {1}{6} \, \log \left (x^{3} + 1\right ) + \frac {13}{81} \, \log \left (x^{3}\right ) \]

[In]

integrate(1/x^7/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

1/54*(8*x^3 - 3)/x^6 + 1/162*log(x^3 + 3) - 1/6*log(x^3 + 1) + 13/81*log(x^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=-\frac {13 \, x^{6} - 8 \, x^{3} + 3}{54 \, x^{6}} + \frac {1}{162} \, \log \left ({\left | x^{3} + 3 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) + \frac {13}{27} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x^7/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

-1/54*(13*x^6 - 8*x^3 + 3)/x^6 + 1/162*log(abs(x^3 + 3)) - 1/6*log(abs(x^3 + 1)) + 13/27*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^7 \left (3+4 x^3+x^6\right )} \, dx=\frac {\ln \left (x^3+3\right )}{162}-\frac {\ln \left (x^3+1\right )}{6}+\frac {13\,\ln \left (x\right )}{27}+\frac {\frac {4\,x^3}{27}-\frac {1}{18}}{x^6} \]

[In]

int(1/(x^7*(4*x^3 + x^6 + 3)),x)

[Out]

log(x^3 + 3)/162 - log(x^3 + 1)/6 + (13*log(x))/27 + ((4*x^3)/27 - 1/18)/x^6