\(\int \frac {-12-36 x^4+e^2 (-18 x-6 x^2+18 x^5-18 x^6)}{e^2 (x^2+3 x^6)} \, dx\) [8886]

Optimal result

Integrand size = 45, antiderivative size = 24 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=6 \left (\frac {2}{e^2 x}-x+\log \left (\frac {1}{3 x^3}+x\right )\right ) \]

[Out]

12/exp(2)/x-6*x+6*ln(x+1/3/x^3)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 1607, 1847, 457, 78, 1600, 14} \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=6 \log \left (3 x^4+1\right )-6 x+\frac {12}{e^2 x}-18 \log (x) \]

[In]

Int[(-12 - 36*x^4 + E^2*(-18*x - 6*x^2 + 18*x^5 - 18*x^6))/(E^2*(x^2 + 3*x^6)),x]

[Out]

12/(E^2*x) - 6*x - 18*Log[x] + 6*Log[1 + 3*x^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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1847

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{x^2+3 x^6} \, dx}{e^2} \\ & = \frac {\int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{x^2 \left (1+3 x^4\right )} \, dx}{e^2} \\ & = \frac {\int \left (\frac {-18 e^2+18 e^2 x^4}{x \left (1+3 x^4\right )}+\frac {-12-6 e^2 x^2-36 x^4-18 e^2 x^6}{x^2 \left (1+3 x^4\right )}\right ) \, dx}{e^2} \\ & = \frac {\int \frac {-18 e^2+18 e^2 x^4}{x \left (1+3 x^4\right )} \, dx}{e^2}+\frac {\int \frac {-12-6 e^2 x^2-36 x^4-18 e^2 x^6}{x^2 \left (1+3 x^4\right )} \, dx}{e^2} \\ & = \frac {\text {Subst}\left (\int \frac {-18 e^2+18 e^2 x}{x (1+3 x)} \, dx,x,x^4\right )}{4 e^2}+\frac {\int \frac {-12-6 e^2 x^2}{x^2} \, dx}{e^2} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {18 e^2}{x}+\frac {72 e^2}{1+3 x}\right ) \, dx,x,x^4\right )}{4 e^2}+\frac {\int \left (-6 e^2-\frac {12}{x^2}\right ) \, dx}{e^2} \\ & = \frac {12}{e^2 x}-6 x-18 \log (x)+6 \log \left (1+3 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=-\frac {6 \left (-\frac {2}{x}+e^2 x+3 e^2 \log (x)-e^2 \log \left (1+3 x^4\right )\right )}{e^2} \]

[In]

Integrate[(-12 - 36*x^4 + E^2*(-18*x - 6*x^2 + 18*x^5 - 18*x^6))/(E^2*(x^2 + 3*x^6)),x]

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08

method	result	size
risch	\(-6 x +\frac {12 \,{\mathrm e}^{-2}}{x}-18 \ln \left (x \right )+6 \ln \left (-3 x^{4}-1\right )\)	\(26\)
norman	\(\frac {-6 x^{2}+12 \,{\mathrm e}^{-2}}{x}-18 \ln \left (x \right )+6 \ln \left (3 x^{4}+1\right )\)	\(32\)
default	\({\mathrm e}^{-2} \left (-6 \,{\mathrm e}^{2} x +\frac {12}{x}-18 \,{\mathrm e}^{2} \ln \left (x \right )+6 \,{\mathrm e}^{2} \ln \left (3 x^{4}+1\right )\right )\)	\(35\)
parallelrisch	\(-\frac {{\mathrm e}^{-2} \left (18 x \,{\mathrm e}^{2} \ln \left (x \right )-6 \,{\mathrm e}^{2} \ln \left (x^{4}+\frac {1}{3}\right ) x +6 x^{2} {\mathrm e}^{2}-12\right )}{x}\)	\(37\)
meijerg	\(\text {Expression too large to display}\)	\(718\)

[In]

int(((-18*x^6+18*x^5-6*x^2-18*x)*exp(2)-36*x^4-12)/(3*x^6+x^2)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-6*x+12/x*exp(-2)-18*ln(x)+6*ln(-3*x^4-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=-\frac {6 \, {\left (x^{2} e^{2} - x e^{2} \log \left (3 \, x^{4} + 1\right ) + 3 \, x e^{2} \log \left (x\right ) - 2\right )} e^{\left (-2\right )}}{x} \]

[In]

integrate(((-18*x^6+18*x^5-6*x^2-18*x)*exp(2)-36*x^4-12)/(3*x^6+x^2)/exp(2),x, algorithm="fricas")

[Out]

-6*(x^2*e^2 - x*e^2*log(3*x^4 + 1) + 3*x*e^2*log(x) - 2)*e^(-2)/x

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=- 6 x - 18 \log {\left (x \right )} + 6 \log {\left (3 x^{4} + 1 \right )} + \frac {12}{x e^{2}} \]

[In]

integrate(((-18*x**6+18*x**5-6*x**2-18*x)*exp(2)-36*x**4-12)/(3*x**6+x**2)/exp(2),x)

[Out]

-6*x - 18*log(x) + 6*log(3*x**4 + 1) + 12*exp(-2)/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=-6 \, {\left (x e^{2} - e^{2} \log \left (3 \, x^{4} + 1\right ) + 3 \, e^{2} \log \left (x\right ) - \frac {2}{x}\right )} e^{\left (-2\right )} \]

[In]

integrate(((-18*x^6+18*x^5-6*x^2-18*x)*exp(2)-36*x^4-12)/(3*x^6+x^2)/exp(2),x, algorithm="maxima")

[Out]

-6*(x*e^2 - e^2*log(3*x^4 + 1) + 3*e^2*log(x) - 2/x)*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=-6 \, {\left (x e^{2} - e^{2} \log \left (3 \, x^{4} + 1\right ) + 3 \, e^{2} \log \left ({\left | x \right |}\right ) - \frac {2}{x}\right )} e^{\left (-2\right )} \]

[In]

integrate(((-18*x^6+18*x^5-6*x^2-18*x)*exp(2)-36*x^4-12)/(3*x^6+x^2)/exp(2),x, algorithm="giac")

[Out]

-6*(x*e^2 - e^2*log(3*x^4 + 1) + 3*e^2*log(abs(x)) - 2/x)*e^(-2)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-12-36 x^4+e^2 \left (-18 x-6 x^2+18 x^5-18 x^6\right )}{e^2 \left (x^2+3 x^6\right )} \, dx=6\,\ln \left (x^4+\frac {1}{3}\right )-6\,x-18\,\ln \left (x\right )+\frac {12\,{\mathrm {e}}^{-2}}{x} \]

[In]

int(-(exp(-2)*(exp(2)*(18*x + 6*x^2 - 18*x^5 + 18*x^6) + 36*x^4 + 12))/(x^2 + 3*x^6),x)

[Out]

6*log(x^4 + 1/3) - 6*x - 18*log(x) + (12*exp(-2))/x