\(\int \frac {2 e^{2+2 x}+e^{5 x} (-10 e^{2+2 x}+4 e^{4+4 x})+e^{4+4 x} (8-4 x)+(-2+e^{5 x} (10-4 e^{2+2 x})+e^{2+2 x} (-8+4 x)) \log (-2-e^{5 x}+x)}{2+e^{5 x}-x} \, dx\) [3411]

Optimal result

Integrand size = 105, antiderivative size = 25 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-1+\left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6820, 12, 6818} \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\left (e^{2 x+2}-\log \left (x-e^{5 x}-2\right )\right )^2 \]

[In]

Int[(2*E^(2 + 2*x) + E^(5*x)*(-10*E^(2 + 2*x) + 4*E^(4 + 4*x)) + E^(4 + 4*x)*(8 - 4*x) + (-2 + E^(5*x)*(10 - 4
*E^(2 + 2*x)) + E^(2 + 2*x)*(-8 + 4*x))*Log[-2 - E^(5*x) + x])/(2 + E^(5*x) - x),x]

[Out]

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

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (1-5 e^{5 x}+2 e^{2+7 x}-2 e^{2+2 x} (-2+x)\right ) \left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )}{2+e^{5 x}-x} \, dx \\ & = 2 \int \frac {\left (1-5 e^{5 x}+2 e^{2+7 x}-2 e^{2+2 x} (-2+x)\right ) \left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )}{2+e^{5 x}-x} \, dx \\ & = \left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )^2 \]

[In]

Integrate[(2*E^(2 + 2*x) + E^(5*x)*(-10*E^(2 + 2*x) + 4*E^(4 + 4*x)) + E^(4 + 4*x)*(8 - 4*x) + (-2 + E^(5*x)*(
10 - 4*E^(2 + 2*x)) + E^(2 + 2*x)*(-8 + 4*x))*Log[-2 - E^(5*x) + x])/(2 + E^(5*x) - x),x]

[Out]

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

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52

method	result	size
risch	\({\mathrm e}^{4+4 x}-2 \,{\mathrm e}^{2+2 x} \ln \left (-{\mathrm e}^{5 x}+x -2\right )+\ln \left (-{\mathrm e}^{5 x}+x -2\right )^{2}\)	\(38\)
parallelrisch	\({\mathrm e}^{4+4 x}-2 \,{\mathrm e}^{2+2 x} \ln \left (-{\mathrm e}^{5 x}+x -2\right )+\ln \left (-{\mathrm e}^{5 x}+x -2\right )^{2}\)	\(38\)

[In]

int((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*ln(-exp(5*x)+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x
)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/(exp(5*x)+2-x),x,method=_RETURNVERBOSE)

[Out]

exp(4+4*x)-2*exp(2+2*x)*ln(-exp(5*x)+x-2)+ln(-exp(5*x)+x-2)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-2 \, e^{\left (2 \, x + 2\right )} \log \left ({\left ({\left (x - 2\right )} e^{5} - e^{\left (5 \, x + 5\right )}\right )} e^{\left (-5\right )}\right ) + \log \left ({\left ({\left (x - 2\right )} e^{5} - e^{\left (5 \, x + 5\right )}\right )} e^{\left (-5\right )}\right )^{2} + e^{\left (4 \, x + 4\right )} \]

[In]

integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x)+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*
exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/(exp(5*x)+2-x),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\text {Timed out} \]

[In]

integrate((((-4*exp(1+x)**2+10)*exp(5*x)+(4*x-8)*exp(1+x)**2-2)*ln(-exp(5*x)+x-2)+(4*exp(1+x)**4-10*exp(1+x)**
2)*exp(5*x)+(-4*x+8)*exp(1+x)**4+2*exp(1+x)**2)/(exp(5*x)+2-x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-2 \, e^{\left (2 \, x + 2\right )} \log \left (x - e^{\left (5 \, x\right )} - 2\right ) + \log \left (x - e^{\left (5 \, x\right )} - 2\right )^{2} + e^{\left (4 \, x + 4\right )} \]

[In]

integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x)+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*
exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/(exp(5*x)+2-x),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\int { -\frac {2 \, {\left ({\left (2 \, e^{\left (4 \, x + 4\right )} - 5 \, e^{\left (2 \, x + 2\right )}\right )} e^{\left (5 \, x\right )} - 2 \, {\left (x - 2\right )} e^{\left (4 \, x + 4\right )} - {\left ({\left (2 \, e^{\left (2 \, x + 2\right )} - 5\right )} e^{\left (5 \, x\right )} - 2 \, {\left (x - 2\right )} e^{\left (2 \, x + 2\right )} + 1\right )} \log \left (x - e^{\left (5 \, x\right )} - 2\right ) + e^{\left (2 \, x + 2\right )}\right )}}{x - e^{\left (5 \, x\right )} - 2} \,d x } \]

[In]

integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x)+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*
exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/(exp(5*x)+2-x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.84 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx={\ln \left (x-{\mathrm {e}}^{5\,x}-2\right )}^2-2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^2\,\ln \left (x-{\mathrm {e}}^{5\,x}-2\right )+{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^4 \]

[In]

int(-(exp(4*x + 4)*(4*x - 8) - 2*exp(2*x + 2) + log(x - exp(5*x) - 2)*(exp(5*x)*(4*exp(2*x + 2) - 10) - exp(2*
x + 2)*(4*x - 8) + 2) + exp(5*x)*(10*exp(2*x + 2) - 4*exp(4*x + 4)))/(exp(5*x) - x + 2),x)

[Out]

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