3.23.40 \(\int \frac {e^{\frac {135 x}{-5 x^2+\log (x)}} (-270 x+1350 x^3+50 x^4+(270 x-20 x^2) \log (x)+2 \log ^2(x))}{375 x^4-150 x^2 \log (x)+15 \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {2}{15} e^{\frac {27}{-x+\frac {\log (x)}{5 x}}} x \]

________________________________________________________________________________________

Rubi [B]  time = 0.58, antiderivative size = 81, normalized size of antiderivative = 3.38, number of steps used = 3, number of rules used = 3, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6741, 12, 2288} \begin {gather*} \frac {2 e^{-\frac {135 x}{5 x^2-\log (x)}} \left (-5 x^3+x-x \log (x)\right )}{15 \left (\frac {\left (\frac {1}{x}-10 x\right ) x}{\left (5 x^2-\log (x)\right )^2}+\frac {1}{5 x^2-\log (x)}\right ) \left (5 x^2-\log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((135*x)/(-5*x^2 + Log[x]))*(-270*x + 1350*x^3 + 50*x^4 + (270*x - 20*x^2)*Log[x] + 2*Log[x]^2))/(375*x
^4 - 150*x^2*Log[x] + 15*Log[x]^2),x]

[Out]

(2*(x - 5*x^3 - x*Log[x]))/(15*E^((135*x)/(5*x^2 - Log[x]))*(((x^(-1) - 10*x)*x)/(5*x^2 - Log[x])^2 + (5*x^2 -
Log[x])^(-1))*(5*x^2 - Log[x])^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 2288

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{\frac {135 x}{-5 x^2+\log (x)}} \left (-135 x+675 x^3+25 x^4+135 x \log (x)-10 x^2 \log (x)+\log ^2(x)\right )}{15 \left (5 x^2-\log (x)\right )^2} \, dx\\ &=\frac {2}{15} \int \frac {e^{\frac {135 x}{-5 x^2+\log (x)}} \left (-135 x+675 x^3+25 x^4+135 x \log (x)-10 x^2 \log (x)+\log ^2(x)\right )}{\left (5 x^2-\log (x)\right )^2} \, dx\\ &=\frac {2 e^{-\frac {135 x}{5 x^2-\log (x)}} \left (x-5 x^3-x \log (x)\right )}{15 \left (\frac {\left (\frac {1}{x}-10 x\right ) x}{\left (5 x^2-\log (x)\right )^2}+\frac {1}{5 x^2-\log (x)}\right ) \left (5 x^2-\log (x)\right )^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 20, normalized size = 0.83 \begin {gather*} \frac {2}{15} e^{\frac {135 x}{-5 x^2+\log (x)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((135*x)/(-5*x^2 + Log[x]))*(-270*x + 1350*x^3 + 50*x^4 + (270*x - 20*x^2)*Log[x] + 2*Log[x]^2))/
(375*x^4 - 150*x^2*Log[x] + 15*Log[x]^2),x]

[Out]

(2*E^((135*x)/(-5*x^2 + Log[x]))*x)/15

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 19, normalized size = 0.79 \begin {gather*} \frac {2}{15} \, x e^{\left (-\frac {135 \, x}{5 \, x^{2} - \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+(-20*x^2+270*x)*log(x)+50*x^4+1350*x^3-270*x)*exp(135*x/(log(x)-5*x^2))/(15*log(x)^2-150
*x^2*log(x)+375*x^4),x, algorithm="fricas")

[Out]

2/15*x*e^(-135*x/(5*x^2 - log(x)))

________________________________________________________________________________________

giac [A]  time = 0.28, size = 19, normalized size = 0.79 \begin {gather*} \frac {2}{15} \, x e^{\left (-\frac {135 \, x}{5 \, x^{2} - \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+(-20*x^2+270*x)*log(x)+50*x^4+1350*x^3-270*x)*exp(135*x/(log(x)-5*x^2))/(15*log(x)^2-150
*x^2*log(x)+375*x^4),x, algorithm="giac")

[Out]

2/15*x*e^(-135*x/(5*x^2 - log(x)))

________________________________________________________________________________________

maple [A]  time = 0.02, size = 18, normalized size = 0.75

method	result	size
risch	\(\frac {2 x \,{\mathrm e}^{\frac {135 x}{\ln \relax (x )-5 x^{2}}}}{15}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)^2+(-20*x^2+270*x)*ln(x)+50*x^4+1350*x^3-270*x)*exp(135*x/(ln(x)-5*x^2))/(15*ln(x)^2-150*x^2*ln(x)
+375*x^4),x,method=_RETURNVERBOSE)

[Out]

2/15*x*exp(135*x/(ln(x)-5*x^2))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+(-20*x^2+270*x)*log(x)+50*x^4+1350*x^3-270*x)*exp(135*x/(log(x)-5*x^2))/(15*log(x)^2-150
*x^2*log(x)+375*x^4),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

________________________________________________________________________________________

mupad [B]  time = 1.43, size = 17, normalized size = 0.71 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^{\frac {135\,x}{\ln \relax (x)-5\,x^2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((135*x)/(log(x) - 5*x^2))*(2*log(x)^2 - 270*x + log(x)*(270*x - 20*x^2) + 1350*x^3 + 50*x^4))/(15*log
(x)^2 - 150*x^2*log(x) + 375*x^4),x)

[Out]

(2*x*exp((135*x)/(log(x) - 5*x^2)))/15

________________________________________________________________________________________

sympy [A]  time = 7.84, size = 17, normalized size = 0.71 \begin {gather*} \frac {2 x e^{\frac {135 x}{- 5 x^{2} + \log {\relax (x )}}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)**2+(-20*x**2+270*x)*ln(x)+50*x**4+1350*x**3-270*x)*exp(135*x/(ln(x)-5*x**2))/(15*ln(x)**2-1
50*x**2*ln(x)+375*x**4),x)

[Out]

2*x*exp(135*x/(-5*x**2 + log(x)))/15

________________________________________________________________________________________