3.43.99 \(\int \frac {-x+x^3+e^{12} (-x+x^3)+e^6 (-2 x+2 x^3)+(-1+2 x^2+e^{12} (-1+2 x^2)+e^6 (-2+4 x^2)) \log (3)+(x+2 e^6 x+e^{12} x) \log (x)}{x^3+2 x^2 \log (3)+x \log ^2(3)} \, dx\)

Optimal. Leaf size=26 \[ \left (1+e^6\right )^2 \left (1-\frac {-x^2+\log (x)}{x+\log (3)}\right ) \]

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Rubi [B]  time = 0.48, antiderivative size = 83, normalized size of antiderivative = 3.19, number of steps used = 14, number of rules used = 9, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 27, 6688, 12, 6742, 43, 44, 2314, 31} \begin {gather*} \left (1+e^6\right )^2 x-\frac {\left (1+e^6\right )^2 \log ^2(3)}{x+\log (3)}+\frac {\left (1+e^6\right )^2 x \log (x)}{\log (3) (x+\log (3))}-\frac {\left (1+e^6\right )^2 \log (x)}{\log (3)}+\frac {\left (1+e^6\right )^2 \log (3) \log (9)}{x+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + x^3 + E^12*(-x + x^3) + E^6*(-2*x + 2*x^3) + (-1 + 2*x^2 + E^12*(-1 + 2*x^2) + E^6*(-2 + 4*x^2))*Log
[3] + (x + 2*E^6*x + E^12*x)*Log[x])/(x^3 + 2*x^2*Log[3] + x*Log[3]^2),x]

[Out]

(1 + E^6)^2*x - ((1 + E^6)^2*Log[3]^2)/(x + Log[3]) + ((1 + E^6)^2*Log[3]*Log[9])/(x + Log[3]) - ((1 + E^6)^2*
Log[x])/Log[3] + ((1 + E^6)^2*x*Log[x])/(Log[3]*(x + Log[3]))

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

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

Rule 43

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1594

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

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x+x^3+e^{12} \left (-x+x^3\right )+e^6 \left (-2 x+2 x^3\right )+\left (-1+2 x^2+e^{12} \left (-1+2 x^2\right )+e^6 \left (-2+4 x^2\right )\right ) \log (3)+\left (x+2 e^6 x+e^{12} x\right ) \log (x)}{x \left (x^2+2 x \log (3)+\log ^2(3)\right )} \, dx\\ &=\int \frac {-x+x^3+e^{12} \left (-x+x^3\right )+e^6 \left (-2 x+2 x^3\right )+\left (-1+2 x^2+e^{12} \left (-1+2 x^2\right )+e^6 \left (-2+4 x^2\right )\right ) \log (3)+\left (x+2 e^6 x+e^{12} x\right ) \log (x)}{x (x+\log (3))^2} \, dx\\ &=\int \frac {\left (1+e^6\right )^2 \left (-x+x^3-\log (3)+x^2 \log (9)+x \log (x)\right )}{x (x+\log (3))^2} \, dx\\ &=\left (1+e^6\right )^2 \int \frac {-x+x^3-\log (3)+x^2 \log (9)+x \log (x)}{x (x+\log (3))^2} \, dx\\ &=\left (1+e^6\right )^2 \int \left (-\frac {1}{(x+\log (3))^2}+\frac {x^2}{(x+\log (3))^2}-\frac {\log (3)}{x (x+\log (3))^2}+\frac {x \log (9)}{(x+\log (3))^2}+\frac {\log (x)}{(x+\log (3))^2}\right ) \, dx\\ &=\frac {\left (1+e^6\right )^2}{x+\log (3)}+\left (1+e^6\right )^2 \int \frac {x^2}{(x+\log (3))^2} \, dx+\left (1+e^6\right )^2 \int \frac {\log (x)}{(x+\log (3))^2} \, dx-\left (\left (1+e^6\right )^2 \log (3)\right ) \int \frac {1}{x (x+\log (3))^2} \, dx+\left (\left (1+e^6\right )^2 \log (9)\right ) \int \frac {x}{(x+\log (3))^2} \, dx\\ &=\frac {\left (1+e^6\right )^2}{x+\log (3)}+\frac {\left (1+e^6\right )^2 x \log (x)}{\log (3) (x+\log (3))}+\left (1+e^6\right )^2 \int \left (1+\frac {\log ^2(3)}{(x+\log (3))^2}-\frac {\log (9)}{x+\log (3)}\right ) \, dx-\frac {\left (1+e^6\right )^2 \int \frac {1}{x+\log (3)} \, dx}{\log (3)}-\left (\left (1+e^6\right )^2 \log (3)\right ) \int \left (\frac {1}{x \log ^2(3)}-\frac {1}{\log (3) (x+\log (3))^2}-\frac {1}{\log ^2(3) (x+\log (3))}\right ) \, dx+\left (\left (1+e^6\right )^2 \log (9)\right ) \int \left (-\frac {\log (3)}{(x+\log (3))^2}+\frac {1}{x+\log (3)}\right ) \, dx\\ &=\left (1+e^6\right )^2 x-\frac {\left (1+e^6\right )^2 \log ^2(3)}{x+\log (3)}+\frac {\left (1+e^6\right )^2 \log (3) \log (9)}{x+\log (3)}-\frac {\left (1+e^6\right )^2 \log (x)}{\log (3)}+\frac {\left (1+e^6\right )^2 x \log (x)}{\log (3) (x+\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 30, normalized size = 1.15 \begin {gather*} \frac {\left (1+e^6\right )^2 \left (x^2+x \log (3)+\log ^2(3)-\log (x)\right )}{x+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^3 + E^12*(-x + x^3) + E^6*(-2*x + 2*x^3) + (-1 + 2*x^2 + E^12*(-1 + 2*x^2) + E^6*(-2 + 4*x^2
))*Log[3] + (x + 2*E^6*x + E^12*x)*Log[x])/(x^3 + 2*x^2*Log[3] + x*Log[3]^2),x]

[Out]

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

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fricas [B]  time = 0.63, size = 63, normalized size = 2.42 \begin {gather*} \frac {x^{2} e^{12} + 2 \, x^{2} e^{6} + {\left (e^{12} + 2 \, e^{6} + 1\right )} \log \relax (3)^{2} + x^{2} + {\left (x e^{12} + 2 \, x e^{6} + x\right )} \log \relax (3) - {\left (e^{12} + 2 \, e^{6} + 1\right )} \log \relax (x)}{x + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)^4+2*x*exp(3)^2+x)*log(x)+((2*x^2-1)*exp(3)^4+(4*x^2-2)*exp(3)^2+2*x^2-1)*log(3)+(x^3-x)*e
xp(3)^4+(2*x^3-2*x)*exp(3)^2+x^3-x)/(x*log(3)^2+2*x^2*log(3)+x^3),x, algorithm="fricas")

[Out]

(x^2*e^12 + 2*x^2*e^6 + (e^12 + 2*e^6 + 1)*log(3)^2 + x^2 + (x*e^12 + 2*x*e^6 + x)*log(3) - (e^12 + 2*e^6 + 1)
*log(x))/(x + log(3))

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giac [B]  time = 0.22, size = 76, normalized size = 2.92 \begin {gather*} \frac {x^{2} e^{12} + 2 \, x^{2} e^{6} + x e^{12} \log \relax (3) + 2 \, x e^{6} \log \relax (3) + e^{12} \log \relax (3)^{2} + 2 \, e^{6} \log \relax (3)^{2} + x^{2} + x \log \relax (3) + \log \relax (3)^{2} - e^{12} \log \relax (x) - 2 \, e^{6} \log \relax (x) - \log \relax (x)}{x + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)^4+2*x*exp(3)^2+x)*log(x)+((2*x^2-1)*exp(3)^4+(4*x^2-2)*exp(3)^2+2*x^2-1)*log(3)+(x^3-x)*e
xp(3)^4+(2*x^3-2*x)*exp(3)^2+x^3-x)/(x*log(3)^2+2*x^2*log(3)+x^3),x, algorithm="giac")

[Out]

(x^2*e^12 + 2*x^2*e^6 + x*e^12*log(3) + 2*x*e^6*log(3) + e^12*log(3)^2 + 2*e^6*log(3)^2 + x^2 + x*log(3) + log
(3)^2 - e^12*log(x) - 2*e^6*log(x) - log(x))/(x + log(3))

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maple [A]  time = 0.08, size = 42, normalized size = 1.62

method	result	size
norman	\(\frac {\left ({\mathrm e}^{12}+2 \,{\mathrm e}^{6}+1\right ) x^{2}+\left (-{\mathrm e}^{12}-2 \,{\mathrm e}^{6}-1\right ) \ln \relax (x )}{\ln \relax (3)+x}\)	\(42\)
risch	\(-\frac {\left ({\mathrm e}^{12}+2 \,{\mathrm e}^{6}+1\right ) \ln \relax (x )}{\ln \relax (3)+x}+\frac {\left ({\mathrm e}^{12}+2 \,{\mathrm e}^{6}+1\right ) \left (\ln \relax (3)^{2}+x \ln \relax (3)+x^{2}\right )}{\ln \relax (3)+x}\)	\(47\)
default	\(\frac {\ln \relax (x ) x \,{\mathrm e}^{12}}{\ln \relax (3) \left (\ln \relax (3)+x \right )}+\frac {2 \ln \relax (x ) x \,{\mathrm e}^{6}}{\ln \relax (3) \left (\ln \relax (3)+x \right )}+\frac {\ln \relax (x ) x}{\ln \relax (3) \left (\ln \relax (3)+x \right )}+x \,{\mathrm e}^{12}+2 x \,{\mathrm e}^{6}+x +\frac {\ln \relax (3)^{2} {\mathrm e}^{12}}{\ln \relax (3)+x}+\frac {2 \ln \relax (3)^{2} {\mathrm e}^{6}}{\ln \relax (3)+x}+\frac {\ln \relax (3)^{2}}{\ln \relax (3)+x}-\frac {\ln \relax (x ) {\mathrm e}^{12}}{\ln \relax (3)}-\frac {2 \ln \relax (x ) {\mathrm e}^{6}}{\ln \relax (3)}-\frac {\ln \relax (x )}{\ln \relax (3)}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(3)^4+2*x*exp(3)^2+x)*ln(x)+((2*x^2-1)*exp(3)^4+(4*x^2-2)*exp(3)^2+2*x^2-1)*ln(3)+(x^3-x)*exp(3)^4+
(2*x^3-2*x)*exp(3)^2+x^3-x)/(x*ln(3)^2+2*x^2*ln(3)+x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.37, size = 369, normalized size = 14.19 \begin {gather*} 2 \, {\left (\frac {\log \relax (3)}{x + \log \relax (3)} + \log \left (x + \log \relax (3)\right )\right )} e^{12} \log \relax (3) - {\left (\frac {1}{x \log \relax (3) + \log \relax (3)^{2}} - \frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)^{2}} + \frac {\log \relax (x)}{\log \relax (3)^{2}}\right )} e^{12} \log \relax (3) + 4 \, {\left (\frac {\log \relax (3)}{x + \log \relax (3)} + \log \left (x + \log \relax (3)\right )\right )} e^{6} \log \relax (3) - 2 \, {\left (\frac {1}{x \log \relax (3) + \log \relax (3)^{2}} - \frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)^{2}} + \frac {\log \relax (x)}{\log \relax (3)^{2}}\right )} e^{6} \log \relax (3) - {\left (2 \, \log \relax (3) \log \left (x + \log \relax (3)\right ) - x + \frac {\log \relax (3)^{2}}{x + \log \relax (3)}\right )} e^{12} - {\left (\frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)} - \frac {\log \relax (x)}{\log \relax (3)}\right )} e^{12} - 2 \, {\left (2 \, \log \relax (3) \log \left (x + \log \relax (3)\right ) - x + \frac {\log \relax (3)^{2}}{x + \log \relax (3)}\right )} e^{6} - 2 \, {\left (\frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)} - \frac {\log \relax (x)}{\log \relax (3)}\right )} e^{6} + 2 \, {\left (\frac {\log \relax (3)}{x + \log \relax (3)} + \log \left (x + \log \relax (3)\right )\right )} \log \relax (3) - {\left (\frac {1}{x \log \relax (3) + \log \relax (3)^{2}} - \frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)^{2}} + \frac {\log \relax (x)}{\log \relax (3)^{2}}\right )} \log \relax (3) - 2 \, \log \relax (3) \log \left (x + \log \relax (3)\right ) + x - \frac {\log \relax (3)^{2}}{x + \log \relax (3)} - \frac {e^{12} \log \relax (x)}{x + \log \relax (3)} - \frac {2 \, e^{6} \log \relax (x)}{x + \log \relax (3)} + \frac {e^{12}}{x + \log \relax (3)} + \frac {2 \, e^{6}}{x + \log \relax (3)} - \frac {\log \left (x + \log \relax (3)\right )}{\log \relax (3)} - \frac {\log \relax (x)}{x + \log \relax (3)} + \frac {\log \relax (x)}{\log \relax (3)} + \frac {1}{x + \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)^4+2*x*exp(3)^2+x)*log(x)+((2*x^2-1)*exp(3)^4+(4*x^2-2)*exp(3)^2+2*x^2-1)*log(3)+(x^3-x)*e
xp(3)^4+(2*x^3-2*x)*exp(3)^2+x^3-x)/(x*log(3)^2+2*x^2*log(3)+x^3),x, algorithm="maxima")

[Out]

2*(log(3)/(x + log(3)) + log(x + log(3)))*e^12*log(3) - (1/(x*log(3) + log(3)^2) - log(x + log(3))/log(3)^2 +
log(x)/log(3)^2)*e^12*log(3) + 4*(log(3)/(x + log(3)) + log(x + log(3)))*e^6*log(3) - 2*(1/(x*log(3) + log(3)^
2) - log(x + log(3))/log(3)^2 + log(x)/log(3)^2)*e^6*log(3) - (2*log(3)*log(x + log(3)) - x + log(3)^2/(x + lo
g(3)))*e^12 - (log(x + log(3))/log(3) - log(x)/log(3))*e^12 - 2*(2*log(3)*log(x + log(3)) - x + log(3)^2/(x +
log(3)))*e^6 - 2*(log(x + log(3))/log(3) - log(x)/log(3))*e^6 + 2*(log(3)/(x + log(3)) + log(x + log(3)))*log(
3) - (1/(x*log(3) + log(3)^2) - log(x + log(3))/log(3)^2 + log(x)/log(3)^2)*log(3) - 2*log(3)*log(x + log(3))
+ x - log(3)^2/(x + log(3)) - e^12*log(x)/(x + log(3)) - 2*e^6*log(x)/(x + log(3)) + e^12/(x + log(3)) + 2*e^6
/(x + log(3)) - log(x + log(3))/log(3) - log(x)/(x + log(3)) + log(x)/log(3) + 1/(x + log(3))

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mupad [B]  time = 3.51, size = 22, normalized size = 0.85 \begin {gather*} -\frac {\left (\ln \relax (x)-x^2\right )\,{\left ({\mathrm {e}}^6+1\right )}^2}{x+\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.31, size = 60, normalized size = 2.31 \begin {gather*} x \left (1 + 2 e^{6} + e^{12}\right ) + \frac {\left (- e^{12} - 2 e^{6} - 1\right ) \log {\relax (x )}}{x + \log {\relax (3 )}} + \frac {\log {\relax (3 )}^{2} + 2 e^{6} \log {\relax (3 )}^{2} + e^{12} \log {\relax (3 )}^{2}}{x + \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)**4+2*x*exp(3)**2+x)*ln(x)+((2*x**2-1)*exp(3)**4+(4*x**2-2)*exp(3)**2+2*x**2-1)*ln(3)+(x**
3-x)*exp(3)**4+(2*x**3-2*x)*exp(3)**2+x**3-x)/(x*ln(3)**2+2*x**2*ln(3)+x**3),x)

[Out]

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

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