3.45.70 \(\int \frac {25 e^4+e^{4+x} (5-5 x)}{(25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)) \log ^2(\frac {10+2 e^x+2 x}{5+e^x+2 x})} \, dx\)

Optimal. Leaf size=24 \[ \frac {5 e^4}{\log \left (\frac {2}{1+\frac {x}{5+e^x+x}}\right )} \]

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Rubi [F]  time = 1.69, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {25 e^4+e^{4+x} (5-5 x)}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {10+2 e^x+2 x}{5+e^x+2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*E^4 + E^(4 + x)*(5 - 5*x))/((25 + E^(2*x) + 15*x + 2*x^2 + E^x*(10 + 3*x))*Log[(10 + 2*E^x + 2*x)/(5 +
E^x + 2*x)]^2),x]

[Out]

20*E^4*Defer[Int][1/((5 + E^x + x)*Log[(2*(5 + E^x + x))/(5 + E^x + 2*x)]^2), x] + 5*E^4*Defer[Int][x/((5 + E^
x + x)*Log[(2*(5 + E^x + x))/(5 + E^x + 2*x)]^2), x] - 15*E^4*Defer[Int][1/((5 + E^x + 2*x)*Log[(2*(5 + E^x +
x))/(5 + E^x + 2*x)]^2), x] - 10*E^4*Defer[Int][x/((5 + E^x + 2*x)*Log[(2*(5 + E^x + x))/(5 + E^x + 2*x)]^2),
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^4 \left (5+e^x-e^x x\right )}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx\\ &=\left (5 e^4\right ) \int \frac {5+e^x-e^x x}{\left (25+e^{2 x}+15 x+2 x^2+e^x (10+3 x)\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx\\ &=\left (5 e^4\right ) \int \left (\frac {4+x}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}-\frac {3+2 x}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}\right ) \, dx\\ &=\left (5 e^4\right ) \int \frac {4+x}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx-\left (5 e^4\right ) \int \frac {3+2 x}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx\\ &=\left (5 e^4\right ) \int \left (\frac {4}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}+\frac {x}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}\right ) \, dx-\left (5 e^4\right ) \int \left (\frac {3}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}+\frac {2 x}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )}\right ) \, dx\\ &=\left (5 e^4\right ) \int \frac {x}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx-\left (10 e^4\right ) \int \frac {x}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx-\left (15 e^4\right ) \int \frac {1}{\left (5+e^x+2 x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx+\left (20 e^4\right ) \int \frac {1}{\left (5+e^x+x\right ) \log ^2\left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 26, normalized size = 1.08 \begin {gather*} \frac {5 e^4}{\log \left (\frac {2 \left (5+e^x+x\right )}{5+e^x+2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^4 + E^(4 + x)*(5 - 5*x))/((25 + E^(2*x) + 15*x + 2*x^2 + E^x*(10 + 3*x))*Log[(10 + 2*E^x + 2*x
)/(5 + E^x + 2*x)]^2),x]

[Out]

(5*E^4)/Log[(2*(5 + E^x + x))/(5 + E^x + 2*x)]

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fricas [A]  time = 0.65, size = 35, normalized size = 1.46 \begin {gather*} \frac {5 \, e^{4}}{\log \left (\frac {2 \, {\left ({\left (x + 5\right )} e^{4} + e^{\left (x + 4\right )}\right )}}{{\left (2 \, x + 5\right )} e^{4} + e^{\left (x + 4\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15*x+25)/log((2*exp(x)+2*x+10)/(e
xp(x)+5+2*x))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 25, normalized size = 1.04 \begin {gather*} \frac {5 \, e^{4}}{\log \relax (2) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15*x+25)/log((2*exp(x)+2*x+10)/(e
xp(x)+5+2*x))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.10, size = 161, normalized size = 6.71

method	result	size
risch	\(\frac {10 i {\mathrm e}^{4}}{\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) \mathrm {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x}+5+x \right )}{\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x}\right )^{3}+2 i \ln \left ({\mathrm e}^{x}+5+x \right )-2 i \ln \left (\frac {{\mathrm e}^{x}}{2}+\frac {5}{2}+x \right )}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15*x+25)/ln((2*exp(x)+2*x+10)/(exp(x)+5
+2*x))^2,x,method=_RETURNVERBOSE)

[Out]

10*I*exp(4)/(Pi*csgn(I*(exp(x)+5+x))*csgn(I/(1/2*exp(x)+5/2+x))*csgn(I*(exp(x)+5+x)/(1/2*exp(x)+5/2+x))-Pi*csg
n(I*(exp(x)+5+x))*csgn(I*(exp(x)+5+x)/(1/2*exp(x)+5/2+x))^2-Pi*csgn(I/(1/2*exp(x)+5/2+x))*csgn(I*(exp(x)+5+x)/
(1/2*exp(x)+5/2+x))^2+Pi*csgn(I*(exp(x)+5+x)/(1/2*exp(x)+5/2+x))^3+2*I*ln(exp(x)+5+x)-2*I*ln(1/2*exp(x)+5/2+x)
)

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maxima [A]  time = 0.64, size = 25, normalized size = 1.04 \begin {gather*} \frac {5 \, e^{4}}{\log \relax (2) - \log \left (2 \, x + e^{x} + 5\right ) + \log \left (x + e^{x} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)^2+(3*x+10)*exp(x)+2*x^2+15*x+25)/log((2*exp(x)+2*x+10)/(e
xp(x)+5+2*x))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.24, size = 26, normalized size = 1.08 \begin {gather*} \frac {5\,{\mathrm {e}}^4}{\ln \left (\frac {2\,x+2\,{\mathrm {e}}^x+10}{2\,x+{\mathrm {e}}^x+5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(4) - exp(4)*exp(x)*(5*x - 5))/(log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))^2*(15*x + exp(2*x) +
exp(x)*(3*x + 10) + 2*x^2 + 25)),x)

[Out]

(5*exp(4))/log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))

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sympy [A]  time = 0.25, size = 24, normalized size = 1.00 \begin {gather*} \frac {5 e^{4}}{\log {\left (\frac {2 x + 2 e^{x} + 10}{2 x + e^{x} + 5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(4)*exp(x)+25*exp(4))/(exp(x)**2+(3*x+10)*exp(x)+2*x**2+15*x+25)/ln((2*exp(x)+2*x+10)/(
exp(x)+5+2*x))**2,x)

[Out]

5*exp(4)/log((2*x + 2*exp(x) + 10)/(2*x + exp(x) + 5))

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