\(\int \frac {e^{6-2 e^x x+2 x \log (25-x)} ((50-2 x+2 x^2) \log (2)+e^x (50 x+48 x^2-2 x^3) \log (2)+(-50 x+2 x^2) \log (2) \log (25-x))}{-25 x^3+x^4} \, dx\) [339]

Optimal result

Integrand size = 83, antiderivative size = 33 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{6+2 x^2+2 x \left (-e^x-x+\log (25-x)\right )} \log (2)}{x^2} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(33)=66\).

Time = 7.80 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1607, 6873, 12, 2326} \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{2 \left (3-e^x x\right )} (25-x)^{2 x-1} \log (2) \left (-e^x x^3+24 e^x x^2+x^2+x^2 \log (25-x)+25 e^x x-25 x \log (25-x)\right )}{x^3 \left (e^x x+\frac {x}{25-x}+e^x-\log (25-x)\right )} \]

[In]

Int[(E^(6 - 2*E^x*x + 2*x*Log[25 - x])*((50 - 2*x + 2*x^2)*Log[2] + E^x*(50*x + 48*x^2 - 2*x^3)*Log[2] + (-50*
x + 2*x^2)*Log[2]*Log[25 - x]))/(-25*x^3 + x^4),x]

[Out]

(E^(2*(3 - E^x*x))*(25 - x)^(-1 + 2*x)*Log[2]*(25*E^x*x + x^2 + 24*E^x*x^2 - E^x*x^3 - 25*x*Log[25 - x] + x^2*
Log[25 - x]))/(x^3*(E^x + E^x*x + x/(25 - x) - Log[25 - 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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{(-25+x) x^3} \, dx \\ & = \int \frac {2 e^{-2 \left (-3+e^x x-x \log (25-x)\right )} \log (2) \left (-25+x-25 e^x x-x^2-24 e^x x^2+e^x x^3+25 x \log (25-x)-x^2 \log (25-x)\right )}{(25-x) x^3} \, dx \\ & = (2 \log (2)) \int \frac {e^{-2 \left (-3+e^x x-x \log (25-x)\right )} \left (-25+x-25 e^x x-x^2-24 e^x x^2+e^x x^3+25 x \log (25-x)-x^2 \log (25-x)\right )}{(25-x) x^3} \, dx \\ & = \frac {e^{2 \left (3-e^x x\right )} (25-x)^{-1+2 x} \log (2) \left (25 e^x x+x^2+24 e^x x^2-e^x x^3-25 x \log (25-x)+x^2 \log (25-x)\right )}{x^3 \left (e^x+e^x x+\frac {x}{25-x}-\log (25-x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{6-2 e^x x} (25-x)^{2 x} \log (2)}{x^2} \]

[In]

Integrate[(E^(6 - 2*E^x*x + 2*x*Log[25 - x])*((50 - 2*x + 2*x^2)*Log[2] + E^x*(50*x + 48*x^2 - 2*x^3)*Log[2] +
(-50*x + 2*x^2)*Log[2]*Log[25 - x]))/(-25*x^3 + x^4),x]

[Out]

(E^(6 - 2*E^x*x)*(25 - x)^(2*x)*Log[2])/x^2

Maple [A] (verified)

Time = 3.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\ln \left (2\right ) \left (-x +25\right )^{2 x} {\mathrm e}^{-2 \,{\mathrm e}^{x} x +6}}{x^{2}}\)	\(24\)
parallelrisch	\(\frac {\ln \left (2\right ) {\mathrm e}^{2 x \ln \left (-x +25\right )-2 \,{\mathrm e}^{x} x +6}}{x^{2}}\)	\(25\)

[In]

int(((2*x^2-50*x)*ln(2)*ln(-x+25)+(-2*x^3+48*x^2+50*x)*ln(2)*exp(x)+(2*x^2-2*x+50)*ln(2))*exp(x*ln(-x+25)-exp(
x)*x+3)^2/(x^4-25*x^3),x,method=_RETURNVERBOSE)

[Out]

1/x^2*ln(2)*((-x+25)^x)^2*exp(-2*exp(x)*x+6)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{\left (-2 \, x e^{x} + 2 \, x \log \left (-x + 25\right ) + 6\right )} \log \left (2\right )}{x^{2}} \]

[In]

integrate(((2*x^2-50*x)*log(2)*log(-x+25)+(-2*x^3+48*x^2+50*x)*log(2)*exp(x)+(2*x^2-2*x+50)*log(2))*exp(x*log(
-x+25)-exp(x)*x+3)^2/(x^4-25*x^3),x, algorithm="fricas")

[Out]

e^(-2*x*e^x + 2*x*log(-x + 25) + 6)*log(2)/x^2

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{- 2 x e^{x} + 2 x \log {\left (25 - x \right )} + 6} \log {\left (2 \right )}}{x^{2}} \]

[In]

integrate(((2*x**2-50*x)*ln(2)*ln(-x+25)+(-2*x**3+48*x**2+50*x)*ln(2)*exp(x)+(2*x**2-2*x+50)*ln(2))*exp(x*ln(-
x+25)-exp(x)*x+3)**2/(x**4-25*x**3),x)

[Out]

exp(-2*x*exp(x) + 2*x*log(25 - x) + 6)*log(2)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {e^{\left (-2 \, x e^{x} + 2 \, x \log \left (-x + 25\right ) + 6\right )} \log \left (2\right )}{x^{2}} \]

[In]

integrate(((2*x^2-50*x)*log(2)*log(-x+25)+(-2*x^3+48*x^2+50*x)*log(2)*exp(x)+(2*x^2-2*x+50)*log(2))*exp(x*log(
-x+25)-exp(x)*x+3)^2/(x^4-25*x^3),x, algorithm="maxima")

[Out]

e^(-2*x*e^x + 2*x*log(-x + 25) + 6)*log(2)/x^2

Giac [F]

\[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\int { -\frac {2 \, {\left ({\left (x^{3} - 24 \, x^{2} - 25 \, x\right )} e^{x} \log \left (2\right ) - {\left (x^{2} - 25 \, x\right )} \log \left (2\right ) \log \left (-x + 25\right ) - {\left (x^{2} - x + 25\right )} \log \left (2\right )\right )} e^{\left (-2 \, x e^{x} + 2 \, x \log \left (-x + 25\right ) + 6\right )}}{x^{4} - 25 \, x^{3}} \,d x } \]

[In]

integrate(((2*x^2-50*x)*log(2)*log(-x+25)+(-2*x^3+48*x^2+50*x)*log(2)*exp(x)+(2*x^2-2*x+50)*log(2))*exp(x*log(
-x+25)-exp(x)*x+3)^2/(x^4-25*x^3),x, algorithm="giac")

[Out]

integrate(-2*((x^3 - 24*x^2 - 25*x)*e^x*log(2) - (x^2 - 25*x)*log(2)*log(-x + 25) - (x^2 - x + 25)*log(2))*e^(
-2*x*e^x + 2*x*log(-x + 25) + 6)/(x^4 - 25*x^3), x)

Mupad [B] (verification not implemented)

Time = 8.86 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{6-2 e^x x+2 x \log (25-x)} \left (\left (50-2 x+2 x^2\right ) \log (2)+e^x \left (50 x+48 x^2-2 x^3\right ) \log (2)+\left (-50 x+2 x^2\right ) \log (2) \log (25-x)\right )}{-25 x^3+x^4} \, dx=\frac {{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^6\,\ln \left (2\right )\,{\left (25-x\right )}^{2\,x}}{x^2} \]

[In]

int(-(exp(2*x*log(25 - x) - 2*x*exp(x) + 6)*(log(2)*(2*x^2 - 2*x + 50) + exp(x)*log(2)*(50*x + 48*x^2 - 2*x^3)
- log(2)*log(25 - x)*(50*x - 2*x^2)))/(25*x^3 - x^4),x)

[Out]

(exp(-2*x*exp(x))*exp(6)*log(2)*(25 - x)^(2*x))/x^2