\(\int \frac {1}{2} (2+(2 e^2 x+e (-4 x+3 x^2)) \log (625)) \, dx\) [1145]

Optimal result

Integrand size = 27, antiderivative size = 16 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=x+\frac {1}{2} e x^2 (-2+e+x) \log (625) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12} \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=\frac {1}{2} e x^3 \log (625)+\frac {1}{2} e^2 x^2 \log (625)-e x^2 \log (625)+x \]

[In]

Int[(2 + (2*E^2*x + E*(-4*x + 3*x^2))*Log[625])/2,x]

[Out]

x - E*x^2*Log[625] + (E^2*x^2*Log[625])/2 + (E*x^3*Log[625])/2

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx \\ & = x+\frac {1}{2} \log (625) \int \left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \, dx \\ & = x+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} (e \log (625)) \int \left (-4 x+3 x^2\right ) \, dx \\ & = x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625) \]

[In]

Integrate[(2 + (2*E^2*x + E*(-4*x + 3*x^2))*Log[625])/2,x]

[Out]

x - E*x^2*Log[625] + (E^2*x^2*Log[625])/2 + (E*x^3*Log[625])/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81

method	result	size
parallelrisch	\(2 \ln \left (5\right ) \left (x^{2} {\mathrm e}^{2}+x^{3} {\mathrm e}-2 x^{2} {\mathrm e}\right )+x\)	\(29\)
gosper	\(x \left (2 x \,{\mathrm e}^{2} \ln \left (5\right )+2 \ln \left (5\right ) {\mathrm e} x^{2}-4 x \,{\mathrm e} \ln \left (5\right )+1\right )\)	\(30\)
norman	\(2 \ln \left (5\right ) {\mathrm e} x^{3}+\left (2 \,{\mathrm e}^{2} \ln \left (5\right )-4 \,{\mathrm e} \ln \left (5\right )\right ) x^{2}+x\)	\(31\)
default	\(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\)	\(32\)
risch	\(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\)	\(32\)
parts	\(2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+2 \ln \left (5\right ) {\mathrm e} x^{3}-4 \ln \left (5\right ) {\mathrm e} x^{2}+x\)	\(32\)

[In]

int(2*(2*x*exp(1)^2+(3*x^2-4*x)*exp(1))*ln(5)+1,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]

[In]

integrate(2*(2*x*exp(1)^2+(3*x^2-4*x)*exp(1))*log(5)+1,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 e x^{3} \log {\left (5 \right )} + x^{2} \left (- 4 e \log {\left (5 \right )} + 2 e^{2} \log {\left (5 \right )}\right ) + x \]

[In]

integrate(2*(2*x*exp(1)**2+(3*x**2-4*x)*exp(1))*ln(5)+1,x)

[Out]

2*E*x**3*log(5) + x**2*(-4*E*log(5) + 2*exp(2)*log(5)) + x

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]

[In]

integrate(2*(2*x*exp(1)^2+(3*x^2-4*x)*exp(1))*log(5)+1,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (5\right ) + x \]

[In]

integrate(2*(2*x*exp(1)^2+(3*x^2-4*x)*exp(1))*log(5)+1,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {1}{2} \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx=2\,\mathrm {e}\,\ln \left (5\right )\,x^3-\ln \left (5\right )\,\left (4\,\mathrm {e}-2\,{\mathrm {e}}^2\right )\,x^2+x \]

[In]

int(1 - 2*log(5)*(exp(1)*(4*x - 3*x^2) - 2*x*exp(2)),x)

[Out]

x + 2*x^3*exp(1)*log(5) - x^2*log(5)*(4*exp(1) - 2*exp(2))