3.61.4 \(\int \frac {25^{-\frac {1}{\log (\frac {2+x}{2})}} (25 x+e^{2 x} x)^{\frac {1}{\log (\frac {2+x}{2})}} ((50+25 x+e^{2 x} (2+5 x+2 x^2)) \log (\frac {2+x}{2})+(-25 x-e^{2 x} x) \log (\frac {1}{25} (25 x+e^{2 x} x)))}{(50 x+25 x^2+e^{2 x} (2 x+x^2)) \log ^2(\frac {2+x}{2})} \, dx\)

Optimal. Leaf size=23 \[ \left (x+\frac {1}{25} e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \]

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Rubi [F]  time = 13.18, 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^{-\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (50 x+25 x^2+e^{2 x} \left (2 x+x^2\right )\right ) \log ^2\left (\frac {2+x}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((25*x + E^(2*x)*x)^Log[(2 + x)/2]^(-1)*((50 + 25*x + E^(2*x)*(2 + 5*x + 2*x^2))*Log[(2 + x)/2] + (-25*x -
E^(2*x)*x)*Log[(25*x + E^(2*x)*x)/25]))/(25^Log[(2 + x)/2]^(-1)*(50*x + 25*x^2 + E^(2*x)*(2*x + x^2))*Log[(2
+ x)/2]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (\left (50+25 x+e^{2 x} \left (2+5 x+2 x^2\right )\right ) \log \left (\frac {2+x}{2}\right )+\left (-25 x-e^{2 x} x\right ) \log \left (\frac {1}{25} \left (25 x+e^{2 x} x\right )\right )\right )}{\left (25+e^{2 x}\right ) x (2+x) \log ^2\left (1+\frac {x}{2}\right )} \, dx\\ &=\int \left (\frac {2\ 25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )}+\frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (2 \log \left (\frac {2+x}{2}\right )+5 x \log \left (\frac {2+x}{2}\right )+2 x^2 \log \left (\frac {2+x}{2}\right )-x \log \left (x+\frac {1}{25} e^{2 x} x\right )\right )}{x (2+x) \log ^2\left (1+\frac {x}{2}\right )}\right ) \, dx\\ &=2 \int \frac {25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )} \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (2 \log \left (\frac {2+x}{2}\right )+5 x \log \left (\frac {2+x}{2}\right )+2 x^2 \log \left (\frac {2+x}{2}\right )-x \log \left (x+\frac {1}{25} e^{2 x} x\right )\right )}{x (2+x) \log ^2\left (1+\frac {x}{2}\right )} \, dx\\ &=2 \int \frac {25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )} \, dx+\int \left (\frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} (1+2 x) \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{x \log \left (1+\frac {x}{2}\right )}+\frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \log \left (\frac {1}{25} \left (25+e^{2 x}\right ) x\right )}{(-2-x) \log ^2\left (1+\frac {x}{2}\right )}\right ) \, dx\\ &=2 \int \frac {25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )} \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} (1+2 x) \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{x \log \left (1+\frac {x}{2}\right )} \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \log \left (\frac {1}{25} \left (25+e^{2 x}\right ) x\right )}{(-2-x) \log ^2\left (1+\frac {x}{2}\right )} \, dx\\ &=2 \int \frac {25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )} \, dx+\int \left (\frac {2\ 25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\log \left (1+\frac {x}{2}\right )}+\frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{x \log \left (1+\frac {x}{2}\right )}\right ) \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \log \left (\frac {1}{25} \left (25+e^{2 x}\right ) x\right )}{(-2-x) \log ^2\left (1+\frac {x}{2}\right )} \, dx\\ &=2 \int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\log \left (1+\frac {x}{2}\right )} \, dx+2 \int \frac {25^{1-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{\left (-25-e^{2 x}\right ) \log \left (1+\frac {x}{2}\right )} \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}}}{x \log \left (1+\frac {x}{2}\right )} \, dx+\int \frac {25^{-\frac {1}{\log \left (1+\frac {x}{2}\right )}} \left (25 x+e^{2 x} x\right )^{\frac {1}{\log \left (1+\frac {x}{2}\right )}} \log \left (\frac {1}{25} \left (25+e^{2 x}\right ) x\right )}{(-2-x) \log ^2\left (1+\frac {x}{2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 23, normalized size = 1.00 \begin {gather*} \left (x+\frac {1}{25} e^{2 x} x\right )^{\frac {1}{\log \left (\frac {2+x}{2}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((25*x + E^(2*x)*x)^Log[(2 + x)/2]^(-1)*((50 + 25*x + E^(2*x)*(2 + 5*x + 2*x^2))*Log[(2 + x)/2] + (-
25*x - E^(2*x)*x)*Log[(25*x + E^(2*x)*x)/25]))/(25^Log[(2 + x)/2]^(-1)*(50*x + 25*x^2 + E^(2*x)*(2*x + x^2))*L
og[(2 + x)/2]^2),x]

[Out]

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

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fricas [A]  time = 0.88, size = 18, normalized size = 0.78 \begin {gather*} {\left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right )}^{\left (\frac {1}{\log \left (\frac {1}{2} \, x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x)^2+25*x+50)*log(1+1/2*x))*exp(log(1/
25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm="fricas")

[Out]

(1/25*x*e^(2*x) + x)^(1/log(1/2*x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x e^{\left (2 \, x\right )} + 25 \, x\right )} \log \left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right ) - {\left ({\left (2 \, x^{2} + 5 \, x + 2\right )} e^{\left (2 \, x\right )} + 25 \, x + 50\right )} \log \left (\frac {1}{2} \, x + 1\right )\right )} {\left (\frac {1}{25} \, x e^{\left (2 \, x\right )} + x\right )}^{\left (\frac {1}{\log \left (\frac {1}{2} \, x + 1\right )}\right )}}{{\left (25 \, x^{2} + {\left (x^{2} + 2 \, x\right )} e^{\left (2 \, x\right )} + 50 \, x\right )} \log \left (\frac {1}{2} \, x + 1\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x)^2+25*x+50)*log(1+1/2*x))*exp(log(1/
25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm="giac")

[Out]

integrate(-((x*e^(2*x) + 25*x)*log(1/25*x*e^(2*x) + x) - ((2*x^2 + 5*x + 2)*e^(2*x) + 25*x + 50)*log(1/2*x + 1
))*(1/25*x*e^(2*x) + x)^(1/log(1/2*x + 1))/((25*x^2 + (x^2 + 2*x)*e^(2*x) + 50*x)*log(1/2*x + 1)^2), x)

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maple [C]  time = 0.32, size = 126, normalized size = 5.48

method	result	size
risch	\({\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )^{3}-i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+25\right )\right )+i \pi \,\mathrm {csgn}\left (i x \left ({\mathrm e}^{2 x}+25\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}+25\right )\right )-2 \ln \relax (x )+4 \ln \relax (5)-2 \ln \left ({\mathrm e}^{2 x}+25\right )}{2 \ln \left (1+\frac {x}{2}\right )}}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(x)^2-25*x)*ln(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x)^2+25*x+50)*ln(1+1/2*x))*exp(ln(1/25*x*exp(
x)^2+x)/ln(1+1/2*x))/((x^2+2*x)*exp(x)^2+25*x^2+50*x)/ln(1+1/2*x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-1/2*(I*Pi*csgn(I*x*(exp(2*x)+25))^3-I*Pi*csgn(I*x*(exp(2*x)+25))^2*csgn(I*x)-I*Pi*csgn(I*x*(exp(2*x)+25))
^2*csgn(I*(exp(2*x)+25))+I*Pi*csgn(I*x*(exp(2*x)+25))*csgn(I*x)*csgn(I*(exp(2*x)+25))-2*ln(x)+4*ln(5)-2*ln(exp
(2*x)+25))/ln(1+1/2*x))

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maxima [B]  time = 0.60, size = 52, normalized size = 2.26 \begin {gather*} e^{\left (\frac {2 \, \log \relax (5)}{\log \relax (2) - \log \left (x + 2\right )} - \frac {\log \relax (x)}{\log \relax (2) - \log \left (x + 2\right )} - \frac {\log \left (e^{\left (2 \, x\right )} + 25\right )}{\log \relax (2) - \log \left (x + 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^2-25*x)*log(1/25*x*exp(x)^2+x)+((2*x^2+5*x+2)*exp(x)^2+25*x+50)*log(1+1/2*x))*exp(log(1/
25*x*exp(x)^2+x)/log(1+1/2*x))/((x^2+2*x)*exp(x)^2+25*x^2+50*x)/log(1+1/2*x)^2,x, algorithm="maxima")

[Out]

e^(2*log(5)/(log(2) - log(x + 2)) - log(x)/(log(2) - log(x + 2)) - log(e^(2*x) + 25)/(log(2) - log(x + 2)))

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mupad [B]  time = 4.54, size = 18, normalized size = 0.78 \begin {gather*} {\left (x+\frac {x\,{\mathrm {e}}^{2\,x}}{25}\right )}^{\frac {1}{\ln \left (\frac {x}{2}+1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x + (x*exp(2*x))/25)/log(x/2 + 1))*(log(x + (x*exp(2*x))/25)*(25*x + x*exp(2*x)) - log(x/2 + 1)*
(25*x + exp(2*x)*(5*x + 2*x^2 + 2) + 50)))/(log(x/2 + 1)^2*(50*x + exp(2*x)*(2*x + x^2) + 25*x^2)),x)

[Out]

(x + (x*exp(2*x))/25)^(1/log(x/2 + 1))

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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*exp(x)**2-25*x)*ln(1/25*x*exp(x)**2+x)+((2*x**2+5*x+2)*exp(x)**2+25*x+50)*ln(1+1/2*x))*exp(ln(1
/25*x*exp(x)**2+x)/ln(1+1/2*x))/((x**2+2*x)*exp(x)**2+25*x**2+50*x)/ln(1+1/2*x)**2,x)

[Out]

Timed out

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