\(\int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} (e^x (-1-x)-6 x^2 \log (5))}{25 \log (5)} \, dx\) [7629]

Optimal result

Integrand size = 51, antiderivative size = 26 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=2-\frac {1}{25} e^{2 x^3+\frac {e^x x}{\log (5)}}-x \]

[Out]

2-x-1/25*exp(exp(x)/ln(5)*x+2*x^3)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 6838} \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-\frac {1}{25} e^{2 x^3+\frac {e^x x}{\log (5)}}-x \]

[In]

Int[(-25*Log[5] + E^((E^x*x + 2*x^3*Log[5])/Log[5])*(E^x*(-1 - x) - 6*x^2*Log[5]))/(25*Log[5]),x]

[Out]

-1/25*E^(2*x^3 + (E^x*x)/Log[5]) - x

Rule 12

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )\right ) \, dx}{25 \log (5)} \\ & = -x+\frac {\int e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right ) \, dx}{25 \log (5)} \\ & = -\frac {1}{25} e^{2 x^3+\frac {e^x x}{\log (5)}}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-\frac {1}{25} e^{2 x^3+\frac {e^x x}{\log (5)}}-x \]

[In]

Integrate[(-25*Log[5] + E^((E^x*x + 2*x^3*Log[5])/Log[5])*(E^x*(-1 - x) - 6*x^2*Log[5]))/(25*Log[5]),x]

[Out]

-1/25*E^(2*x^3 + (E^x*x)/Log[5]) - x

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(-x -\frac {{\mathrm e}^{\frac {x \left (2 x^{2} \ln \left (5\right )+{\mathrm e}^{x}\right )}{\ln \left (5\right )}}}{25}\)	\(24\)
norman	\(-x -\frac {{\mathrm e}^{\frac {{\mathrm e}^{x} x +2 x^{3} \ln \left (5\right )}{\ln \left (5\right )}}}{25}\)	\(25\)
parallelrisch	\(\frac {-\ln \left (5\right ) {\mathrm e}^{\frac {x \left (2 x^{2} \ln \left (5\right )+{\mathrm e}^{x}\right )}{\ln \left (5\right )}}-25 x \ln \left (5\right )}{25 \ln \left (5\right )}\)	\(34\)
default	\(\frac {-\ln \left (5\right ) {\mathrm e}^{\frac {{\mathrm e}^{x} x +2 x^{3} \ln \left (5\right )}{\ln \left (5\right )}}-25 x \ln \left (5\right )}{25 \ln \left (5\right )}\)	\(35\)

[In]

int(1/25*(((-1-x)*exp(x)-6*x^2*ln(5))*exp((exp(x)*x+2*x^3*ln(5))/ln(5))-25*ln(5))/ln(5),x,method=_RETURNVERBOS
E)

[Out]

-x-1/25*exp(x*(2*x^2*ln(5)+exp(x))/ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-x - \frac {1}{25} \, e^{\left (\frac {2 \, x^{3} \log \left (5\right ) + x e^{x}}{\log \left (5\right )}\right )} \]

[In]

integrate(1/25*(((-1-x)*exp(x)-6*x^2*log(5))*exp((exp(x)*x+2*x^3*log(5))/log(5))-25*log(5))/log(5),x, algorith
m="fricas")

[Out]

-x - 1/25*e^((2*x^3*log(5) + x*e^x)/log(5))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=- x - \frac {e^{\frac {2 x^{3} \log {\left (5 \right )} + x e^{x}}{\log {\left (5 \right )}}}}{25} \]

[In]

integrate(1/25*(((-1-x)*exp(x)-6*x**2*ln(5))*exp((exp(x)*x+2*x**3*ln(5))/ln(5))-25*ln(5))/ln(5),x)

[Out]

-x - exp((2*x**3*log(5) + x*exp(x))/log(5))/25

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-\frac {25 \, x \log \left (5\right ) + e^{\left (2 \, x^{3} + \frac {x e^{x}}{\log \left (5\right )}\right )} \log \left (5\right )}{25 \, \log \left (5\right )} \]

[In]

integrate(1/25*(((-1-x)*exp(x)-6*x^2*log(5))*exp((exp(x)*x+2*x^3*log(5))/log(5))-25*log(5))/log(5),x, algorith
m="maxima")

[Out]

-1/25*(25*x*log(5) + e^(2*x^3 + x*e^x/log(5))*log(5))/log(5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-\frac {25 \, x \log \left (5\right ) + e^{\left (2 \, x^{3} + \frac {x e^{x}}{\log \left (5\right )}\right )} \log \left (5\right )}{25 \, \log \left (5\right )} \]

[In]

integrate(1/25*(((-1-x)*exp(x)-6*x^2*log(5))*exp((exp(x)*x+2*x^3*log(5))/log(5))-25*log(5))/log(5),x, algorith
m="giac")

[Out]

-1/25*(25*x*log(5) + e^(2*x^3 + x*e^x/log(5))*log(5))/log(5)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-25 \log (5)+e^{\frac {e^x x+2 x^3 \log (5)}{\log (5)}} \left (e^x (-1-x)-6 x^2 \log (5)\right )}{25 \log (5)} \, dx=-x-\frac {{\mathrm {e}}^{2\,x^3}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{\ln \left (5\right )}}}{25} \]

[In]

int(-(log(5) + (exp((2*x^3*log(5) + x*exp(x))/log(5))*(exp(x)*(x + 1) + 6*x^2*log(5)))/25)/log(5),x)

[Out]

- x - (exp(2*x^3)*exp((x*exp(x))/log(5)))/25