\(\int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx\) [3306]

Optimal result

Integrand size = 36, antiderivative size = 21 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=13-x+x^2-5 e^x \left (5-\log ^2(x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14, 2326} \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=x^2-x-\frac {5 e^x \left (5 x-x \log ^2(x)\right )}{x} \]

[In]

Int[(-x - 25*E^x*x + 2*x^2 + 10*E^x*Log[x] + 5*E^x*x*Log[x]^2)/x,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=-25 e^x+(-1+x) x+5 e^x \log ^2(x) \]

[In]

Integrate[(-x - 25*E^x*x + 2*x^2 + 10*E^x*Log[x] + 5*E^x*x*Log[x]^2)/x,x]

[Out]

-25*E^x + (-1 + x)*x + 5*E^x*Log[x]^2

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
default	\(-x +5 \,{\mathrm e}^{x} \ln \left (x \right )^{2}-25 \,{\mathrm e}^{x}+x^{2}\)	\(20\)
norman	\(-x +5 \,{\mathrm e}^{x} \ln \left (x \right )^{2}-25 \,{\mathrm e}^{x}+x^{2}\)	\(20\)
risch	\(-x +5 \,{\mathrm e}^{x} \ln \left (x \right )^{2}-25 \,{\mathrm e}^{x}+x^{2}\)	\(20\)
parallelrisch	\(-x +5 \,{\mathrm e}^{x} \ln \left (x \right )^{2}-25 \,{\mathrm e}^{x}+x^{2}\)	\(20\)
parts	\(-x +5 \,{\mathrm e}^{x} \ln \left (x \right )^{2}-25 \,{\mathrm e}^{x}+x^{2}\)	\(20\)

[In]

int((5*x*exp(x)*ln(x)^2+10*exp(x)*ln(x)-25*exp(x)*x+2*x^2-x)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=5 \, e^{x} \log \left (x\right )^{2} + x^{2} - x - 25 \, e^{x} \]

[In]

integrate((5*x*exp(x)*log(x)^2+10*exp(x)*log(x)-25*exp(x)*x+2*x^2-x)/x,x, algorithm="fricas")

[Out]

5*e^x*log(x)^2 + x^2 - x - 25*e^x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=x^{2} - x + \left (5 \log {\left (x \right )}^{2} - 25\right ) e^{x} \]

[In]

integrate((5*x*exp(x)*ln(x)**2+10*exp(x)*ln(x)-25*exp(x)*x+2*x**2-x)/x,x)

[Out]

x**2 - x + (5*log(x)**2 - 25)*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=5 \, e^{x} \log \left (x\right )^{2} + x^{2} - x - 25 \, e^{x} \]

[In]

integrate((5*x*exp(x)*log(x)^2+10*exp(x)*log(x)-25*exp(x)*x+2*x^2-x)/x,x, algorithm="maxima")

[Out]

5*e^x*log(x)^2 + x^2 - x - 25*e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=5 \, e^{x} \log \left (x\right )^{2} + x^{2} - x - 25 \, e^{x} \]

[In]

integrate((5*x*exp(x)*log(x)^2+10*exp(x)*log(x)-25*exp(x)*x+2*x^2-x)/x,x, algorithm="giac")

[Out]

5*e^x*log(x)^2 + x^2 - x - 25*e^x

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-x-25 e^x x+2 x^2+10 e^x \log (x)+5 e^x x \log ^2(x)}{x} \, dx=5\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2-25\,{\mathrm {e}}^x-x+x^2 \]

[In]

int((10*exp(x)*log(x) - x - 25*x*exp(x) + 2*x^2 + 5*x*exp(x)*log(x)^2)/x,x)

[Out]

5*exp(x)*log(x)^2 - 25*exp(x) - x + x^2