Integrand size = 40, antiderivative size = 32 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=\frac {1}{5} \left (-2+\frac {\left (3-e^5\right ) x^2}{5 \log (7) \left (1+\log \left (x^2\right )\right )}\right ) \] Output:
1/25*x^2/(ln(x^2)+1)/ln(7)*(3-exp(5))-2/5
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=-\frac {\left (-3+e^5\right ) x^2}{25 \log (7) \left (1+\log \left (x^2\right )\right )} \] Input:
Integrate[((6*x - 2*E^5*x)*Log[x^2])/(25*Log[7] + 50*Log[7]*Log[x^2] + 25* Log[7]*Log[x^2]^2),x]
Output:
-1/25*((-3 + E^5)*x^2)/(Log[7]*(1 + Log[x^2]))
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6, 27, 27, 7239, 27, 2813, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7) \log ^2\left (x^2\right )+50 \log (7) \log \left (x^2\right )+25 \log (7)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (6-2 e^5\right ) x \log \left (x^2\right )}{25 \log (7) \log ^2\left (x^2\right )+50 \log (7) \log \left (x^2\right )+25 \log (7)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (3-e^5\right ) \int \frac {x \log \left (x^2\right )}{25 \left (\log (7) \log ^2\left (x^2\right )+2 \log (7) \log \left (x^2\right )+\log (7)\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{25} \left (3-e^5\right ) \int \frac {x \log \left (x^2\right )}{\log (7) \log ^2\left (x^2\right )+2 \log (7) \log \left (x^2\right )+\log (7)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {2}{25} \left (3-e^5\right ) \int \frac {x \log \left (x^2\right )}{\log (7) \left (\log \left (x^2\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \left (3-e^5\right ) \int \frac {x \log \left (x^2\right )}{\left (\log \left (x^2\right )+1\right )^2}dx}{25 \log (7)}\) |
\(\Big \downarrow \) 2813 |
\(\displaystyle \frac {2 \left (3-e^5\right ) \left (-2 \int \left (\frac {\operatorname {ExpIntegralEi}\left (\log \left (x^2\right )+1\right )}{2 e x}-\frac {x}{2 \left (\log \left (x^2\right )+1\right )}\right )dx+\frac {\log \left (x^2\right ) \operatorname {ExpIntegralEi}\left (\log \left (x^2\right )+1\right )}{2 e}-\frac {x^2 \log \left (x^2\right )}{2 \left (\log \left (x^2\right )+1\right )}\right )}{25 \log (7)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (3-e^5\right ) \left (\frac {\log \left (x^2\right ) \operatorname {ExpIntegralEi}\left (\log \left (x^2\right )+1\right )}{2 e}-2 \left (-\frac {\operatorname {ExpIntegralEi}\left (\log \left (x^2\right )+1\right )}{4 e}+\frac {\left (\log \left (x^2\right )+1\right ) \operatorname {ExpIntegralEi}\left (\log \left (x^2\right )+1\right )}{4 e}-\frac {x^2}{4}\right )-\frac {x^2 \log \left (x^2\right )}{2 \left (\log \left (x^2\right )+1\right )}\right )}{25 \log (7)}\) |
Input:
Int[((6*x - 2*E^5*x)*Log[x^2])/(25*Log[7] + 50*Log[7]*Log[x^2] + 25*Log[7] *Log[x^2]^2),x]
Output:
(2*(3 - E^5)*((ExpIntegralEi[1 + Log[x^2]]*Log[x^2])/(2*E) - (x^2*Log[x^2] )/(2*(1 + Log[x^2])) - 2*(-1/4*x^2 - ExpIntegralEi[1 + Log[x^2]]/(4*E) + ( ExpIntegralEi[1 + Log[x^2]]*(1 + Log[x^2]))/(4*E))))/(25*Log[7])
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ .)]*(e_.))*((g_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Simp[(d + e*Log[f*x^r]) u, x] - Simp[e*r Int[Simp lifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] && !(EqQ[p, 1] && EqQ[a, 0] && NeQ[d, 0])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69
method | result | size |
default | \(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \left (7\right ) \left (\ln \left (x^{2}\right )+1\right )}\) | \(22\) |
norman | \(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \left (7\right ) \left (\ln \left (x^{2}\right )+1\right )}\) | \(22\) |
risch | \(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \left (7\right ) \left (\ln \left (x^{2}\right )+1\right )}\) | \(22\) |
parallelrisch | \(-\frac {2 x^{2} {\mathrm e}^{5}-6 x^{2}}{50 \ln \left (7\right ) \left (\ln \left (x^{2}\right )+1\right )}\) | \(28\) |
Input:
int((-2*x*exp(5)+6*x)*ln(x^2)/(25*ln(7)*ln(x^2)^2+50*ln(7)*ln(x^2)+25*ln(7 )),x,method=_RETURNVERBOSE)
Output:
-1/25*(exp(5)-3)/ln(7)*x^2/(ln(x^2)+1)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=-\frac {x^{2} e^{5} - 3 \, x^{2}}{25 \, {\left (\log \left (7\right ) \log \left (x^{2}\right ) + \log \left (7\right )\right )}} \] Input:
integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x ^2)+25*log(7)),x, algorithm="fricas")
Output:
-1/25*(x^2*e^5 - 3*x^2)/(log(7)*log(x^2) + log(7))
Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=\frac {- x^{2} e^{5} + 3 x^{2}}{25 \log {\left (7 \right )} \log {\left (x^{2} \right )} + 25 \log {\left (7 \right )}} \] Input:
integrate((-2*x*exp(5)+6*x)*ln(x**2)/(25*ln(7)*ln(x**2)**2+50*ln(7)*ln(x** 2)+25*ln(7)),x)
Output:
(-x**2*exp(5) + 3*x**2)/(25*log(7)*log(x**2) + 25*log(7))
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=-\frac {x^{2} {\left (e^{5} - 3\right )}}{25 \, {\left (2 \, \log \left (7\right ) \log \left (x\right ) + \log \left (7\right )\right )}} \] Input:
integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x ^2)+25*log(7)),x, algorithm="maxima")
Output:
-1/25*x^2*(e^5 - 3)/(2*log(7)*log(x) + log(7))
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=-\frac {x^{2} e^{5}}{25 \, {\left (\log \left (7\right ) \log \left (x^{2}\right ) + \log \left (7\right )\right )}} + \frac {3 \, x^{2}}{25 \, {\left (\log \left (7\right ) \log \left (x^{2}\right ) + \log \left (7\right )\right )}} \] Input:
integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x ^2)+25*log(7)),x, algorithm="giac")
Output:
-1/25*x^2*e^5/(log(7)*log(x^2) + log(7)) + 3/25*x^2/(log(7)*log(x^2) + log (7))
Time = 2.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=-\frac {x^2\,\left ({\mathrm {e}}^5-3\right )}{25\,\ln \left (7\right )\,\left (\ln \left (x^2\right )+1\right )} \] Input:
int((log(x^2)*(6*x - 2*x*exp(5)))/(25*log(7) + 50*log(x^2)*log(7) + 25*log (x^2)^2*log(7)),x)
Output:
-(x^2*(exp(5) - 3))/(25*log(7)*(log(x^2) + 1))
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\left (6 x-2 e^5 x\right ) \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx=\frac {x^{2} \left (-e^{5}+3\right )}{25 \,\mathrm {log}\left (7\right ) \left (\mathrm {log}\left (x^{2}\right )+1\right )} \] Input:
int((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x^2)+25 *log(7)),x)
Output:
(x**2*( - e**5 + 3))/(25*log(7)*(log(x**2) + 1))