3.67.18 \(\int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} (2 x^2+x^3+(-4 x-2 x^2) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} (-x^2+(2 x+x^2) \log (10+5 x)))}{2 x^2+x^3+(-4 x-2 x^2) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx\)

Optimal. Leaf size=24 \[ e^{7-e^{\frac {x}{x-\log (10+5 x)}}} x \]

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Rubi [B]  time = 0.50, antiderivative size = 121, normalized size of antiderivative = 5.04, number of steps used = 1, number of rules used = 1, integrand size = 138, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used = {2288} \begin {gather*} -\frac {e^{7-e^{\frac {x}{x-\log (5 x+10)}}} \left (x^2-\left (x^2+2 x\right ) \log (5 x+10)\right )}{\left (\frac {x \left (1-\frac {1}{x+2}\right )}{(x-\log (5 x+10))^2}-\frac {1}{x-\log (5 x+10)}\right ) \left (x^3+2 x^2-2 \left (x^2+2 x\right ) \log (5 x+10)+(x+2) \log ^2(5 x+10)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(7 - E^(-(x/(-x + Log[10 + 5*x]))))*(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]
^2 + (-x^2 + (2*x + x^2)*Log[10 + 5*x])/E^(x/(-x + Log[10 + 5*x]))))/(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[10 + 5*
x] + (2 + x)*Log[10 + 5*x]^2),x]

[Out]

-((E^(7 - E^(x/(x - Log[10 + 5*x])))*(x^2 - (2*x + x^2)*Log[10 + 5*x]))/(((x*(1 - (2 + x)^(-1)))/(x - Log[10 +
5*x])^2 - (x - Log[10 + 5*x])^(-1))*(2*x^2 + x^3 - 2*(2*x + x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2)))

Rule 2288

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 {gather*} \begin {aligned} \text {integral} &=-\frac {e^{7-e^{\frac {x}{x-\log (10+5 x)}}} \left (x^2-\left (2 x+x^2\right ) \log (10+5 x)\right )}{\left (\frac {x \left (1-\frac {1}{2+x}\right )}{(x-\log (10+5 x))^2}-\frac {1}{x-\log (10+5 x)}\right ) \left (2 x^2+x^3-2 \left (2 x+x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 25, normalized size = 1.04 \begin {gather*} e^{7-e^{-\frac {x}{-x+\log (5 (2+x))}}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(7 - E^(-(x/(-x + Log[10 + 5*x]))))*(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[10 + 5*x] + (2 + x)*Log[10
+ 5*x]^2 + (-x^2 + (2*x + x^2)*Log[10 + 5*x])/E^(x/(-x + Log[10 + 5*x]))))/(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[1
0 + 5*x] + (2 + x)*Log[10 + 5*x]^2),x]

[Out]

E^(7 - E^(-(x/(-x + Log[5*(2 + x)]))))*x

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fricas [A]  time = 0.50, size = 22, normalized size = 0.92 \begin {gather*} x e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^
3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm=
"fricas")

[Out]

x*e^(-e^(x/(x - log(5*x + 10))) + 7)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^
3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm=
"giac")

[Out]

integrate((x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - (x^2 - (x^2 + 2*x)*log(5*x + 10))*e^(x/(x - log(5*x + 10)))
- 2*(x^2 + 2*x)*log(5*x + 10))*e^(-e^(x/(x - log(5*x + 10))) + 7)/(x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - 2*
(x^2 + 2*x)*log(5*x + 10)), x)

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maple [A]  time = 0.18, size = 23, normalized size = 0.96

method	result	size
risch	\(x \,{\mathrm e}^{-{\mathrm e}^{\frac {x}{x -\ln \left (5 x +10\right )}}+7}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+2*x)*ln(5*x+10)-x^2)*exp(-x/(ln(5*x+10)-x))+(2+x)*ln(5*x+10)^2+(-2*x^2-4*x)*ln(5*x+10)+x^3+2*x^2)/(
(2+x)*ln(5*x+10)^2+(-2*x^2-4*x)*ln(5*x+10)+x^3+2*x^2)/exp(exp(-x/(ln(5*x+10)-x))-7),x,method=_RETURNVERBOSE)

[Out]

x*exp(-exp(x/(x-ln(5*x+10)))+7)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^
3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm=
"maxima")

[Out]

integrate((x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - (x^2 - (x^2 + 2*x)*log(5*x + 10))*e^(x/(x - log(5*x + 10)))
- 2*(x^2 + 2*x)*log(5*x + 10))*e^(-e^(x/(x - log(5*x + 10))) + 7)/(x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - 2*
(x^2 + 2*x)*log(5*x + 10)), x)

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mupad [B]  time = 4.73, size = 22, normalized size = 0.92 \begin {gather*} x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {x}{x-\ln \left (5\,x+10\right )}}}\,{\mathrm {e}}^7 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(7 - exp(x/(x - log(5*x + 10))))*(log(5*x + 10)^2*(x + 2) - log(5*x + 10)*(4*x + 2*x^2) + exp(x/(x - l
og(5*x + 10)))*(log(5*x + 10)*(2*x + x^2) - x^2) + 2*x^2 + x^3))/(log(5*x + 10)^2*(x + 2) - log(5*x + 10)*(4*x
+ 2*x^2) + 2*x^2 + x^3),x)

[Out]

x*exp(-exp(x/(x - log(5*x + 10))))*exp(7)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+2*x)*ln(5*x+10)-x**2)*exp(-x/(ln(5*x+10)-x))+(2+x)*ln(5*x+10)**2+(-2*x**2-4*x)*ln(5*x+10)+x*
*3+2*x**2)/((2+x)*ln(5*x+10)**2+(-2*x**2-4*x)*ln(5*x+10)+x**3+2*x**2)/exp(exp(-x/(ln(5*x+10)-x))-7),x)

[Out]

Timed out

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