\(\int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} (-30 x+10 x^2+10 x^3) \log (6561-4374 x-3645 x^2+1458 x^3+729 x^4)}{-3+x+x^2} \, dx\) [1273]

Optimal result

Integrand size = 66, antiderivative size = 19 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^2} \log \left (729 \left (-3+x+x^2\right )^2\right ) \]

[Out]

ln(27*(x^2+x-3)*(27*x^2+27*x-81))*exp(5*x^2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6820, 12, 2326} \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^2} \log \left (729 \left (-x^2-x+3\right )^2\right ) \]

[In]

Int[(E^(5*x^2)*(2 + 4*x) + E^(5*x^2)*(-30*x + 10*x^2 + 10*x^3)*Log[6561 - 4374*x - 3645*x^2 + 1458*x^3 + 729*x
^4])/(-3 + x + x^2),x]

[Out]

E^(5*x^2)*Log[729*(3 - x - x^2)^2]

Rule 12

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{5 x^2} \left (-1-2 x-5 x \left (-3+x+x^2\right ) \log \left (729 \left (-3+x+x^2\right )^2\right )\right )}{3-x-x^2} \, dx \\ & = 2 \int \frac {e^{5 x^2} \left (-1-2 x-5 x \left (-3+x+x^2\right ) \log \left (729 \left (-3+x+x^2\right )^2\right )\right )}{3-x-x^2} \, dx \\ & = e^{5 x^2} \log \left (729 \left (3-x-x^2\right )^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^2} \log \left (729 \left (-3+x+x^2\right )^2\right ) \]

[In]

Integrate[(E^(5*x^2)*(2 + 4*x) + E^(5*x^2)*(-30*x + 10*x^2 + 10*x^3)*Log[6561 - 4374*x - 3645*x^2 + 1458*x^3 +
729*x^4])/(-3 + x + x^2),x]

[Out]

E^(5*x^2)*Log[729*(-3 + x + x^2)^2]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53

method	result	size
norman	\({\mathrm e}^{5 x^{2}} \ln \left (729 x^{4}+1458 x^{3}-3645 x^{2}-4374 x +6561\right )\)	\(29\)
parallelrisch	\({\mathrm e}^{5 x^{2}} \ln \left (729 x^{4}+1458 x^{3}-3645 x^{2}-4374 x +6561\right )\)	\(29\)
default	\(\left (6 \ln \left (3\right )+\ln \left (\left (x^{2}+x -3\right )^{2}\right )-2 \ln \left (x^{2}+x -3\right )\right ) {\mathrm e}^{5 x^{2}}+2 \,{\mathrm e}^{5 x^{2}} \ln \left (x^{2}+x -3\right )\)	\(47\)
risch	\(2 \,{\mathrm e}^{5 x^{2}} \ln \left (x^{2}+x -3\right )+\frac {\left (-i \pi {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left (x^{2}+x -3\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )}^{3}+12 \ln \left (3\right )\right ) {\mathrm e}^{5 x^{2}}}{2}\)	\(104\)

[In]

int(((10*x^3+10*x^2-30*x)*exp(5*x^2)*ln(729*x^4+1458*x^3-3645*x^2-4374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2+x-3),x
,method=_RETURNVERBOSE)

[Out]

exp(5*x^2)*ln(729*x^4+1458*x^3-3645*x^2-4374*x+6561)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{\left (5 \, x^{2}\right )} \log \left (729 \, x^{4} + 1458 \, x^{3} - 3645 \, x^{2} - 4374 \, x + 6561\right ) \]

[In]

integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2
+x-3),x, algorithm="fricas")

[Out]

e^(5*x^2)*log(729*x^4 + 1458*x^3 - 3645*x^2 - 4374*x + 6561)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^{2}} \log {\left (729 x^{4} + 1458 x^{3} - 3645 x^{2} - 4374 x + 6561 \right )} \]

[In]

integrate(((10*x**3+10*x**2-30*x)*exp(5*x**2)*ln(729*x**4+1458*x**3-3645*x**2-4374*x+6561)+(4*x+2)*exp(5*x**2)
)/(x**2+x-3),x)

[Out]

exp(5*x**2)*log(729*x**4 + 1458*x**3 - 3645*x**2 - 4374*x + 6561)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=6 \, e^{\left (5 \, x^{2}\right )} \log \left (3\right ) + 2 \, e^{\left (5 \, x^{2}\right )} \log \left (x^{2} + x - 3\right ) \]

[In]

integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2
+x-3),x, algorithm="maxima")

[Out]

6*e^(5*x^2)*log(3) + 2*e^(5*x^2)*log(x^2 + x - 3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{\left (5 \, x^{2}\right )} \log \left (729 \, x^{4} + 1458 \, x^{3} - 3645 \, x^{2} - 4374 \, x + 6561\right ) \]

[In]

integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2
+x-3),x, algorithm="giac")

[Out]

e^(5*x^2)*log(729*x^4 + 1458*x^3 - 3645*x^2 - 4374*x + 6561)

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx={\mathrm {e}}^{5\,x^2}\,\left (\ln \left (729\right )+\ln \left (x^4+2\,x^3-5\,x^2-6\,x+9\right )\right ) \]

[In]

int((exp(5*x^2)*(4*x + 2) + log(1458*x^3 - 3645*x^2 - 4374*x + 729*x^4 + 6561)*exp(5*x^2)*(10*x^2 - 30*x + 10*
x^3))/(x + x^2 - 3),x)

[Out]

exp(5*x^2)*(log(729) + log(2*x^3 - 5*x^2 - 6*x + x^4 + 9))