\(\int \frac {e^x x \log (x)+(2 e^{2 x} x+e^x (1+2 x^2)) \log ^2(x)+(-e^x x+e^x (x+x^2) \log (x)+(e^{2 x} (2 x+4 x^2)+e^x (x+4 x^2+2 x^3)) \log ^2(x)) \log (25 x)}{x \log ^2(x)} \, dx\) [8776]

Optimal result

Integrand size = 101, antiderivative size = 24 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=e^x \left (1+2 x \left (e^x+x\right )+\frac {x}{\log (x)}\right ) \log (25 x) \]

[Out]

exp(x)*ln(25*x)*(x/ln(x)+1+2*(exp(x)+x)*x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(24)=48\).

Time = 1.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6820, 6874, 2326} \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\frac {e^x \left (2 x^3 \log ^2(x) \log (25 x)+x^2 \log (x) \log (25 x)+x \log ^2(x) \log (25 x)\right )}{x \log ^2(x)}+2 e^{2 x} x \log (25 x) \]

[In]

Int[(E^x*x*Log[x] + (2*E^(2*x)*x + E^x*(1 + 2*x^2))*Log[x]^2 + (-(E^x*x) + E^x*(x + x^2)*Log[x] + (E^(2*x)*(2*
x + 4*x^2) + E^x*(x + 4*x^2 + 2*x^3))*Log[x]^2)*Log[25*x])/(x*Log[x]^2),x]

[Out]

2*E^(2*x)*x*Log[25*x] + (E^x*(x^2*Log[x]*Log[25*x] + x*Log[x]^2*Log[25*x] + 2*x^3*Log[x]^2*Log[25*x]))/(x*Log[
x]^2)

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 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-x \log (25 x)+\log (x) (x+x (1+x) \log (25 x))+\log ^2(x) \left (1+2 e^x x+2 x^2+x \left (1+4 x+2 x^2+e^x (2+4 x)\right ) \log (25 x)\right )\right )}{x \log ^2(x)} \, dx \\ & = \int \left (2 e^{2 x} (1+\log (25 x)+2 x \log (25 x))+\frac {e^x \left (x \log (x)+\log ^2(x)+2 x^2 \log ^2(x)-x \log (25 x)+x \log (x) \log (25 x)+x^2 \log (x) \log (25 x)+x \log ^2(x) \log (25 x)+4 x^2 \log ^2(x) \log (25 x)+2 x^3 \log ^2(x) \log (25 x)\right )}{x \log ^2(x)}\right ) \, dx \\ & = 2 \int e^{2 x} (1+\log (25 x)+2 x \log (25 x)) \, dx+\int \frac {e^x \left (x \log (x)+\log ^2(x)+2 x^2 \log ^2(x)-x \log (25 x)+x \log (x) \log (25 x)+x^2 \log (x) \log (25 x)+x \log ^2(x) \log (25 x)+4 x^2 \log ^2(x) \log (25 x)+2 x^3 \log ^2(x) \log (25 x)\right )}{x \log ^2(x)} \, dx \\ & = 2 e^{2 x} x \log (25 x)+\frac {e^x \left (x^2 \log (x) \log (25 x)+x \log ^2(x) \log (25 x)+2 x^3 \log ^2(x) \log (25 x)\right )}{x \log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\frac {e^x \left (x+\left (1+2 e^x x+2 x^2\right ) \log (x)\right ) \log (25 x)}{\log (x)} \]

[In]

Integrate[(E^x*x*Log[x] + (2*E^(2*x)*x + E^x*(1 + 2*x^2))*Log[x]^2 + (-(E^x*x) + E^x*(x + x^2)*Log[x] + (E^(2*
x)*(2*x + 4*x^2) + E^x*(x + 4*x^2 + 2*x^3))*Log[x]^2)*Log[25*x])/(x*Log[x]^2),x]

[Out]

(E^x*(x + (1 + 2*E^x*x + 2*x^2)*Log[x])*Log[25*x])/Log[x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(52\) vs. \(2(22)=44\).

Time = 0.44 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21

method	result	size
parallelrisch	\(\frac {120 \ln \left (x \right ) {\mathrm e}^{x} \ln \left (25 x \right ) x^{2}+60 \,{\mathrm e}^{x} \ln \left (25 x \right ) x +120 \ln \left (25 x \right ) {\mathrm e}^{2 x} \ln \left (x \right ) x +60 \ln \left (x \right ) {\mathrm e}^{x} \ln \left (25 x \right )}{60 \ln \left (x \right )}\)	\(53\)
risch	\(\left (2 \,{\mathrm e}^{x} x^{2}+2 x \,{\mathrm e}^{2 x}+{\mathrm e}^{x}\right ) \ln \left (x \right )+4 x \ln \left (5\right ) {\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} \ln \left (5\right )+{\mathrm e}^{x} x +4 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+\frac {2 x \,{\mathrm e}^{x} \ln \left (5\right )}{\ln \left (x \right )}\)	\(61\)

[In]

int(((((4*x^2+2*x)*exp(x)^2+(2*x^3+4*x^2+x)*exp(x))*ln(x)^2+(x^2+x)*exp(x)*ln(x)-exp(x)*x)*ln(25*x)+(2*x*exp(x
)^2+(2*x^2+1)*exp(x))*ln(x)^2+x*exp(x)*ln(x))/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/60*(120*ln(x)*exp(x)*ln(25*x)*x^2+60*exp(x)*ln(25*x)*x+120*ln(25*x)*exp(x)^2*ln(x)*x+60*ln(x)*exp(x)*ln(25*x
))/ln(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (22) = 44\).

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\frac {2 \, x e^{x} \log \left (5\right ) + {\left (2 \, x e^{\left (2 \, x\right )} + {\left (2 \, x^{2} + 1\right )} e^{x}\right )} \log \left (x\right )^{2} + {\left (4 \, x e^{\left (2 \, x\right )} \log \left (5\right ) + {\left (2 \, {\left (2 \, x^{2} + 1\right )} \log \left (5\right ) + x\right )} e^{x}\right )} \log \left (x\right )}{\log \left (x\right )} \]

[In]

integrate(((((4*x^2+2*x)*exp(x)^2+(2*x^3+4*x^2+x)*exp(x))*log(x)^2+(x^2+x)*exp(x)*log(x)-exp(x)*x)*log(25*x)+(
2*x*exp(x)^2+(2*x^2+1)*exp(x))*log(x)^2+x*exp(x)*log(x))/x/log(x)^2,x, algorithm="fricas")

[Out]

(2*x*e^x*log(5) + (2*x*e^(2*x) + (2*x^2 + 1)*e^x)*log(x)^2 + (4*x*e^(2*x)*log(5) + (2*(2*x^2 + 1)*log(5) + x)*
e^x)*log(x))/log(x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).

Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.17 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\frac {\left (2 x \log {\left (x \right )}^{2} + 4 x \log {\left (5 \right )} \log {\left (x \right )}\right ) e^{2 x} + \left (2 x^{2} \log {\left (x \right )}^{2} + 4 x^{2} \log {\left (5 \right )} \log {\left (x \right )} + x \log {\left (x \right )} + 2 x \log {\left (5 \right )} + \log {\left (x \right )}^{2} + 2 \log {\left (5 \right )} \log {\left (x \right )}\right ) e^{x}}{\log {\left (x \right )}} \]

[In]

integrate(((((4*x**2+2*x)*exp(x)**2+(2*x**3+4*x**2+x)*exp(x))*ln(x)**2+(x**2+x)*exp(x)*ln(x)-exp(x)*x)*ln(25*x
)+(2*x*exp(x)**2+(2*x**2+1)*exp(x))*ln(x)**2+x*exp(x)*ln(x))/x/ln(x)**2,x)

[Out]

((2*x*log(x)**2 + 4*x*log(5)*log(x))*exp(2*x) + (2*x**2*log(x)**2 + 4*x**2*log(5)*log(x) + x*log(x) + 2*x*log(
5) + log(x)**2 + 2*log(5)*log(x))*exp(x))/log(x)

Maxima [F]

\[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\int { \frac {x e^{x} \log \left (x\right ) + {\left (2 \, x e^{\left (2 \, x\right )} + {\left (2 \, x^{2} + 1\right )} e^{x}\right )} \log \left (x\right )^{2} + {\left ({\left (x^{2} + x\right )} e^{x} \log \left (x\right ) + {\left (2 \, {\left (2 \, x^{2} + x\right )} e^{\left (2 \, x\right )} + {\left (2 \, x^{3} + 4 \, x^{2} + x\right )} e^{x}\right )} \log \left (x\right )^{2} - x e^{x}\right )} \log \left (25 \, x\right )}{x \log \left (x\right )^{2}} \,d x } \]

[In]

integrate(((((4*x^2+2*x)*exp(x)^2+(2*x^3+4*x^2+x)*exp(x))*log(x)^2+(x^2+x)*exp(x)*log(x)-exp(x)*x)*log(25*x)+(
2*x*exp(x)^2+(2*x^2+1)*exp(x))*log(x)^2+x*exp(x)*log(x))/x/log(x)^2,x, algorithm="maxima")

[Out]

(2*x^2 + 1)*e^x*log(x) + (4*x*log(5) + 2*x*log(x) - 1)*e^(2*x) + 2*(x - 1)*e^x + Ei(x) + e^(2*x) + integrate((
(4*x^3*log(5) + x^2*(8*log(5) - 1) + x*(2*log(5) + 1) - 1)*log(x)^2 - 2*x*log(5) + 2*(x^2*log(5) + x*log(5))*l
og(x))*e^x/(x*log(x)^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.25 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx=\frac {4 \, x^{2} e^{x} \log \left (5\right ) \log \left (x\right ) + 2 \, x^{2} e^{x} \log \left (x\right )^{2} + 4 \, x e^{\left (2 \, x\right )} \log \left (5\right ) \log \left (x\right ) + 2 \, x e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 2 \, x e^{x} \log \left (5\right ) + x e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (5\right ) \log \left (x\right ) + e^{x} \log \left (x\right )^{2}}{\log \left (x\right )} \]

[In]

integrate(((((4*x^2+2*x)*exp(x)^2+(2*x^3+4*x^2+x)*exp(x))*log(x)^2+(x^2+x)*exp(x)*log(x)-exp(x)*x)*log(25*x)+(
2*x*exp(x)^2+(2*x^2+1)*exp(x))*log(x)^2+x*exp(x)*log(x))/x/log(x)^2,x, algorithm="giac")

[Out]

(4*x^2*e^x*log(5)*log(x) + 2*x^2*e^x*log(x)^2 + 4*x*e^(2*x)*log(5)*log(x) + 2*x*e^(2*x)*log(x)^2 + 2*x*e^x*log
(5) + x*e^x*log(x) + 2*e^x*log(5)*log(x) + e^x*log(x)^2)/log(x)

Mupad [B] (verification not implemented)

Time = 13.83 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.62 \[ \int \frac {e^x x \log (x)+\left (2 e^{2 x} x+e^x \left (1+2 x^2\right )\right ) \log ^2(x)+\left (-e^x x+e^x \left (x+x^2\right ) \log (x)+\left (e^{2 x} \left (2 x+4 x^2\right )+e^x \left (x+4 x^2+2 x^3\right )\right ) \log ^2(x)\right ) \log (25 x)}{x \log ^2(x)} \, dx={\mathrm {e}}^x\,\ln \left (x\right )+2\,{\mathrm {e}}^x\,\ln \left (5\right )+x\,{\mathrm {e}}^x+4\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (5\right )+4\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )+2\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right )+2\,x^2\,{\mathrm {e}}^x\,\ln \left (x\right )+\frac {2\,x\,{\mathrm {e}}^x\,\ln \left (5\right )}{\ln \left (x\right )} \]

[In]

int((log(25*x)*(log(x)^2*(exp(2*x)*(2*x + 4*x^2) + exp(x)*(x + 4*x^2 + 2*x^3)) - x*exp(x) + exp(x)*log(x)*(x +
x^2)) + log(x)^2*(2*x*exp(2*x) + exp(x)*(2*x^2 + 1)) + x*exp(x)*log(x))/(x*log(x)^2),x)

[Out]

exp(x)*log(x) + 2*exp(x)*log(5) + x*exp(x) + 4*x*exp(2*x)*log(5) + 4*x^2*exp(x)*log(5) + 2*x*exp(2*x)*log(x) +
2*x^2*exp(x)*log(x) + (2*x*exp(x)*log(5))/log(x)