\(\int \frac {-4 e^x x \log (5)+(2-20 x-2 x^2) \log (5)+(-2 x^2 \log (5)-4 e^x x^2 \log (5)) \log (x)}{x^2} \, dx\) [7077]

Optimal result

Integrand size = 47, antiderivative size = 26 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-2+\frac {\log (5) \left (-2-x \left (4 \left (5+e^x\right )+2 x\right ) \log (x)\right )}{x} \]

[Out]

ln(5)/x*(-2-x*ln(x)*(2*x+4*exp(x)+20))-2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {14, 2326, 2332} \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-4 e^x \log (5) \log (x)-2 x \log (5) \log (x)-20 \log (5) \log (x)-\frac {2 \log (5)}{x} \]

[In]

Int[(-4*E^x*x*Log[5] + (2 - 20*x - 2*x^2)*Log[5] + (-2*x^2*Log[5] - 4*E^x*x^2*Log[5])*Log[x])/x^2,x]

[Out]

(-2*Log[5])/x - 20*Log[5]*Log[x] - 4*E^x*Log[5]*Log[x] - 2*x*Log[5]*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4 e^x \log (5) (1+x \log (x))}{x}-\frac {2 \log (5) \left (-1+10 x+x^2+x^2 \log (x)\right )}{x^2}\right ) \, dx \\ & = -\left ((2 \log (5)) \int \frac {-1+10 x+x^2+x^2 \log (x)}{x^2} \, dx\right )-(4 \log (5)) \int \frac {e^x (1+x \log (x))}{x} \, dx \\ & = -4 e^x \log (5) \log (x)-(2 \log (5)) \int \left (\frac {-1+10 x+x^2}{x^2}+\log (x)\right ) \, dx \\ & = -4 e^x \log (5) \log (x)-(2 \log (5)) \int \frac {-1+10 x+x^2}{x^2} \, dx-(2 \log (5)) \int \log (x) \, dx \\ & = 2 x \log (5)-4 e^x \log (5) \log (x)-2 x \log (5) \log (x)-(2 \log (5)) \int \left (1-\frac {1}{x^2}+\frac {10}{x}\right ) \, dx \\ & = -\frac {2 \log (5)}{x}-20 \log (5) \log (x)-4 e^x \log (5) \log (x)-2 x \log (5) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-2 \log (5) \left (\frac {1}{x}+\left (10+2 e^x+x\right ) \log (x)\right ) \]

[In]

Integrate[(-4*E^x*x*Log[5] + (2 - 20*x - 2*x^2)*Log[5] + (-2*x^2*Log[5] - 4*E^x*x^2*Log[5])*Log[x])/x^2,x]

[Out]

-2*Log[5]*(x^(-1) + (10 + 2*E^x + x)*Log[x])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19

method	result	size
risch	\(\left (-2 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )\right ) \ln \left (x \right )-\frac {2 \ln \left (5\right ) \left (10 x \ln \left (x \right )+1\right )}{x}\)	\(31\)
default	\(-4 \ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )-2 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-2 \ln \left (5\right ) \left (x +10 \ln \left (x \right )+\frac {1}{x}\right )\)	\(35\)
norman	\(\frac {-20 x \ln \left (5\right ) \ln \left (x \right )-2 x^{2} \ln \left (5\right ) \ln \left (x \right )-4 x \ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )-2 \ln \left (5\right )}{x}\)	\(35\)
parts	\(-4 \ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )-2 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-2 \ln \left (5\right ) \left (x +10 \ln \left (x \right )+\frac {1}{x}\right )\)	\(35\)
parallelrisch	\(-\frac {2 x^{2} \ln \left (5\right ) \ln \left (x \right )+4 x \ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )+20 x \ln \left (5\right ) \ln \left (x \right )+2 \ln \left (5\right )}{x}\)	\(36\)

[In]

int(((-4*x^2*ln(5)*exp(x)-2*x^2*ln(5))*ln(x)-4*x*exp(x)*ln(5)+(-2*x^2-20*x+2)*ln(5))/x^2,x,method=_RETURNVERBO
SE)

[Out]

(-2*x*ln(5)-4*exp(x)*ln(5))*ln(x)-2*ln(5)*(10*x*ln(x)+1)/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-\frac {2 \, {\left ({\left (2 \, x e^{x} \log \left (5\right ) + {\left (x^{2} + 10 \, x\right )} \log \left (5\right )\right )} \log \left (x\right ) + \log \left (5\right )\right )}}{x} \]

[In]

integrate(((-4*x^2*log(5)*exp(x)-2*x^2*log(5))*log(x)-4*x*exp(x)*log(5)+(-2*x^2-20*x+2)*log(5))/x^2,x, algorit
hm="fricas")

[Out]

-2*((2*x*e^x*log(5) + (x^2 + 10*x)*log(5))*log(x) + log(5))/x

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=- 2 x \log {\left (5 \right )} \log {\left (x \right )} - 4 e^{x} \log {\left (5 \right )} \log {\left (x \right )} - 20 \log {\left (5 \right )} \log {\left (x \right )} - \frac {2 \log {\left (5 \right )}}{x} \]

[In]

integrate(((-4*x**2*ln(5)*exp(x)-2*x**2*ln(5))*ln(x)-4*x*exp(x)*ln(5)+(-2*x**2-20*x+2)*ln(5))/x**2,x)

[Out]

-2*x*log(5)*log(x) - 4*exp(x)*log(5)*log(x) - 20*log(5)*log(x) - 2*log(5)/x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-4 \, e^{x} \log \left (5\right ) \log \left (x\right ) - 2 \, {\left (x \log \left (x\right ) - x\right )} \log \left (5\right ) - 2 \, x \log \left (5\right ) - 20 \, \log \left (5\right ) \log \left (x\right ) - \frac {2 \, \log \left (5\right )}{x} \]

[In]

integrate(((-4*x^2*log(5)*exp(x)-2*x^2*log(5))*log(x)-4*x*exp(x)*log(5)+(-2*x^2-20*x+2)*log(5))/x^2,x, algorit
hm="maxima")

[Out]

-4*e^x*log(5)*log(x) - 2*(x*log(x) - x)*log(5) - 2*x*log(5) - 20*log(5)*log(x) - 2*log(5)/x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-\frac {2 \, {\left (x^{2} \log \left (5\right ) \log \left (x\right ) + 2 \, x e^{x} \log \left (5\right ) \log \left (x\right ) + 10 \, x \log \left (5\right ) \log \left (x\right ) + \log \left (5\right )\right )}}{x} \]

[In]

integrate(((-4*x^2*log(5)*exp(x)-2*x^2*log(5))*log(x)-4*x*exp(x)*log(5)+(-2*x^2-20*x+2)*log(5))/x^2,x, algorit
hm="giac")

[Out]

-2*(x^2*log(5)*log(x) + 2*x*e^x*log(5)*log(x) + 10*x*log(5)*log(x) + log(5))/x

Mupad [B] (verification not implemented)

Time = 15.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-4 e^x x \log (5)+\left (2-20 x-2 x^2\right ) \log (5)+\left (-2 x^2 \log (5)-4 e^x x^2 \log (5)\right ) \log (x)}{x^2} \, dx=-2\,\ln \left (5\right )\,\left (10\,\ln \left (x\right )+2\,{\mathrm {e}}^x\,\ln \left (x\right )\right )-\frac {2\,\ln \left (5\right )}{x}-2\,x\,\ln \left (5\right )\,\ln \left (x\right ) \]

[In]

int(-(log(5)*(20*x + 2*x^2 - 2) + log(x)*(2*x^2*log(5) + 4*x^2*exp(x)*log(5)) + 4*x*exp(x)*log(5))/x^2,x)

[Out]

- 2*log(5)*(10*log(x) + 2*exp(x)*log(x)) - (2*log(5))/x - 2*x*log(5)*log(x)