3.10.2 \(\int \frac {(e^{-3+x} (-2 e^5-2 x)-10 e^5 x-10 x^2+(10 e^5 x+e^{-3+x} (-2 x+2 e^5 x+2 x^2)) \log (x)) \log (\frac {e^{-3+x}+5 x}{(e^5+x) \log (x)})}{(5 e^5 x^2+5 x^3+e^{-3+x} (e^5 x+x^2)) \log (x)} \, dx\) [902]

Optimal. Leaf size=24 \[ \log ^2\left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \]

[Out]

ln((exp(-3+x)+5*x)/(exp(5)+x)/ln(x))^2

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Rubi [F]
time = 13.45, 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 {\left (e^{-3+x} \left (-2 e^5-2 x\right )-10 e^5 x-10 x^2+\left (10 e^5 x+e^{-3+x} \left (-2 x+2 e^5 x+2 x^2\right )\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (5 e^5 x^2+5 x^3+e^{-3+x} \left (e^5 x+x^2\right )\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

2*x - x^2 - 2*E^5*Log[E^5 + x] + 2*x*Log[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])] + 2*LogIntegral[x] + 10*E^3*Lo
g[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])]*Defer[Int][(E^x + 5*E^3*x)^(-1), x] - 10*E^3*Defer[Int][x/(E^x + 5*E^
3*x), x] - 10*E^3*Log[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])]*Defer[Int][x/(E^x + 5*E^3*x), x] + 10*E^3*Defer[I
nt][x^2/(E^x + 5*E^3*x), x] - 2*E^5*Defer[Int][Log[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])]/(E^5 + x), x] - 2*(1
- E^5)*Defer[Int][Log[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])]/(E^5 + x), x] - 2*Defer[Int][Log[(E^(-3 + x) + 5
*x)/((E^5 + x)*Log[x])]/(x*Log[x]), x] - 10*E^3*Defer[Int][Defer[Int][(E^x + 5*E^3*x)^(-1), x], x] + 10*E^8*De
fer[Int][Defer[Int][(E^x + 5*E^3*x)^(-1), x]/(E^5 + x), x] + 10*E^3*(1 - E^5)*Defer[Int][Defer[Int][(E^x + 5*E
^3*x)^(-1), x]/(E^5 + x), x] - 50*E^6*Defer[Int][Defer[Int][(E^x + 5*E^3*x)^(-1), x]/(E^x + 5*E^3*x), x] + 50*
E^6*Defer[Int][(x*Defer[Int][(E^x + 5*E^3*x)^(-1), x])/(E^x + 5*E^3*x), x] + 10*E^3*Defer[Int][Defer[Int][(E^x
+ 5*E^3*x)^(-1), x]/(x*Log[x]), x] + 10*E^3*Defer[Int][Defer[Int][x/(E^x + 5*E^3*x), x], x] - 10*E^8*Defer[In
t][Defer[Int][x/(E^x + 5*E^3*x), x]/(E^5 + x), x] - 10*E^3*(1 - E^5)*Defer[Int][Defer[Int][x/(E^x + 5*E^3*x),
x]/(E^5 + x), x] + 50*E^6*Defer[Int][Defer[Int][x/(E^x + 5*E^3*x), x]/(E^x + 5*E^3*x), x] - 50*E^6*Defer[Int][
(x*Defer[Int][x/(E^x + 5*E^3*x), x])/(E^x + 5*E^3*x), x] - 10*E^3*Defer[Int][Defer[Int][x/(E^x + 5*E^3*x), x]/
(x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-\left (\left (e^5+x\right ) \left (e^x+5 e^3 x\right )\right )+\left (5 e^8+e^{5+x}+e^x (-1+x)\right ) x \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \left (e^5+x\right ) \left (e^x+5 e^3 x\right ) \log (x)} \, dx\\ &=2 \int \frac {\left (-\left (\left (e^5+x\right ) \left (e^x+5 e^3 x\right )\right )+\left (5 e^8+e^{5+x}+e^x (-1+x)\right ) x \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \left (e^5+x\right ) \left (e^x+5 e^3 x\right ) \log (x)} \, dx\\ &=2 \int \left (-\frac {5 e^3 (-1+x) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^x+5 e^3 x}+\frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \left (e^5+x\right ) \log (x)}\right ) \, dx\\ &=2 \int \frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \left (e^5+x\right ) \log (x)} \, dx-\left (10 e^3\right ) \int \frac {(-1+x) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^x+5 e^3 x} \, dx\\ &=2 \int \frac {\left (-e^5-x+x \left (-1+e^5+x\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \left (e^5+x\right ) \log (x)} \, dx+\left (10 e^3\right ) \int \left (\frac {5 e^8+e^{5+x}+e^x (-1+x)}{\left (e^5+x\right ) \left (e^x+5 e^3 x\right )}-\frac {1}{x \log (x)}\right ) \left (-\int \frac {1}{e^x+5 e^3 x} \, dx+\int \frac {x}{e^x+5 e^3 x} \, dx\right ) \, dx+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=2 \int \left (\frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5 \left (e^5+x\right ) \log (x)}+\frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5 x \log (x)}\right ) \, dx+\left (10 e^3\right ) \int \left (\frac {5 e^3 (-1+x) \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{e^x+5 e^3 x}+\frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{x \left (e^5+x\right ) \log (x)}\right ) \, dx+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=\frac {2 \int \frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)} \, dx}{e^5}+\frac {2 \int \frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \log (x)} \, dx}{e^5}+\left (10 e^3\right ) \int \frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{x \left (e^5+x\right ) \log (x)} \, dx+\left (50 e^6\right ) \int \frac {(-1+x) \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{e^x+5 e^3 x} \, dx+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=\frac {2 \int \frac {\left (e^5+x-x \left (-1+e^5+x\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)} \, dx}{e^5}+\frac {2 \int \frac {\left (-e^5-x+x \left (-1+e^5+x\right ) \log (x)\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \log (x)} \, dx}{e^5}+\left (10 e^3\right ) \int \frac {\left (e^5+x-x \left (-1+e^5+x\right ) \log (x)\right ) \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{x \left (e^5+x\right ) \log (x)} \, dx+\left (50 e^6\right ) \int \left (-\frac {\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x}+\frac {x \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{e^x+5 e^3 x}\right ) \, dx+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=\frac {2 \int \left (-\left (\left (1-e^5\right ) \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right )+x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )-\frac {\log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\log (x)}-\frac {e^5 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \log (x)}\right ) \, dx}{e^5}+\frac {2 \int \left (\frac {\left (1-e^5\right ) x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5+x}-\frac {x^2 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5+x}+\frac {e^5 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)}+\frac {x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)}\right ) \, dx}{e^5}+\left (10 e^3\right ) \int \left (\frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \int \frac {1}{e^x+5 e^3 x} \, dx}{x \left (e^5+x\right ) \log (x)}+\frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \int \frac {x}{e^x+5 e^3 x} \, dx}{x \left (e^5+x\right ) \log (x)}\right ) \, dx-\left (50 e^6\right ) \int \frac {\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x} \, dx+\left (50 e^6\right ) \int \frac {x \left (\int \frac {1}{e^x+5 e^3 x} \, dx-\int \frac {x}{e^x+5 e^3 x} \, dx\right )}{e^x+5 e^3 x} \, dx+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=-\left (2 \int \frac {\log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{x \log (x)} \, dx\right )+2 \int \frac {\log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)} \, dx-\left (2 \left (1-\frac {1}{e^5}\right )\right ) \int \frac {x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5+x} \, dx+\frac {2 \int x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \, dx}{e^5}-\frac {2 \int \frac {x^2 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{e^5+x} \, dx}{e^5}-\frac {2 \int \frac {\log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\log (x)} \, dx}{e^5}+\frac {2 \int \frac {x \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )}{\left (e^5+x\right ) \log (x)} \, dx}{e^5}+\left (10 e^3\right ) \int \frac {\left (e^5+x+\left (1-e^5\right ) x \log (x)-x^2 \log (x)\right ) \int \frac {1}{e^x+5 e^3 x} \, dx}{x \left (e^5+x\right ) \log (x)} \, dx+\left (10 e^3\right ) \int \frac {\left (-e^5-x-\left (1-e^5\right ) x \log (x)+x^2 \log (x)\right ) \int \frac {x}{e^x+5 e^3 x} \, dx}{x \left (e^5+x\right ) \log (x)} \, dx-\left (50 e^6\right ) \int \left (\frac {\int \frac {1}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x}-\frac {\int \frac {x}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x}\right ) \, dx+\left (50 e^6\right ) \int \left (\frac {x \int \frac {1}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x}-\frac {x \int \frac {x}{e^x+5 e^3 x} \, dx}{e^x+5 e^3 x}\right ) \, dx+\frac {\left (2 \left (-1+e^5\right )\right ) \int \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \, dx}{e^5}+\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {1}{e^x+5 e^3 x} \, dx-\left (10 e^3 \log \left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right )\right ) \int \frac {x}{e^x+5 e^3 x} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 24, normalized size = 1.00 \begin {gather*} \log ^2\left (\frac {e^{-3+x}+5 x}{\left (e^5+x\right ) \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^(-3 + x)*(-2*E^5 - 2*x) - 10*E^5*x - 10*x^2 + (10*E^5*x + E^(-3 + x)*(-2*x + 2*E^5*x + 2*x^2))*L
og[x])*Log[(E^(-3 + x) + 5*x)/((E^5 + x)*Log[x])])/((5*E^5*x^2 + 5*x^3 + E^(-3 + x)*(E^5*x + x^2))*Log[x]),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.45, size = 1440, normalized size = 60.00

method	result	size
risch	\(\text {Expression too large to display}\)	\(1440\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-18-2*ln(ln(x))*ln(5)-I*Pi*ln(exp(x-3)+5*x)*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^3+3*I*Pi*csgn(I/ln(x))*c
sgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^2+3*I*Pi*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(x)/(exp(5)+x)*
(1/5*exp(x-3)+x))^2+I*Pi*ln(ln(x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^3+I*Pi*ln(exp(5)+x)*csgn(I*(1/5*e
xp(x-3)+x)/(exp(5)+x))^3+I*Pi*ln(exp(5)+x)*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^3+I*Pi*ln(ln(x))*csgn(I*(
1/5*exp(x-3)+x)/(exp(5)+x))^3+3*I*Pi*csgn(I/(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^2+3*I*Pi*csgn(I*(1
/5*exp(x-3)+x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^2-I*Pi*ln(exp(x-3)+5*x)*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x)
)^3+6*ln(5)+ln(ln(x))^2-6*ln(exp(x-3)+5*x)-3*I*Pi*csgn(I/ln(x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(
x)/(exp(5)+x)*(1/5*exp(x-3)+x))-3*I*Pi*csgn(I/(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1/5*exp(x-3)+x)/(ex
p(5)+x))-I*Pi*ln(exp(5)+x)*csgn(I/ln(x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^2-I*Pi*ln(exp(5)+x)*csgn(I/
(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^2-I*Pi*ln(exp(5)+x)*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1/5*exp(x
-3)+x)/(exp(5)+x))^2-I*Pi*ln(exp(5)+x)*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-
3)+x))^2-I*Pi*ln(ln(x))*csgn(I/ln(x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^2-I*Pi*ln(ln(x))*csgn(I/(exp(5
)+x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^2-I*Pi*ln(ln(x))*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1/5*exp(x-3)+x)/(e
xp(5)+x))^2-I*Pi*ln(ln(x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^2+I*P
i*ln(exp(x-3)+5*x)*csgn(I/ln(x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^2+I*Pi*ln(exp(x-3)+5*x)*csgn(I/(exp
(5)+x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^2+I*Pi*ln(exp(x-3)+5*x)*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1/5*exp(x
-3)+x)/(exp(5)+x))^2+I*Pi*ln(exp(x-3)+5*x)*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*ex
p(x-3)+x))^2+ln(1/5*exp(x-3)+x)^2+ln(exp(5)+x)^2+I*Pi*ln(ln(x))*csgn(I/ln(x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+
x))*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))+I*Pi*ln(ln(x))*csgn(I/(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x))*csgn(
I*(1/5*exp(x-3)+x)/(exp(5)+x))-I*Pi*ln(exp(x-3)+5*x)*csgn(I/ln(x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/
ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))-I*Pi*ln(exp(x-3)+5*x)*csgn(I/(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1
/5*exp(x-3)+x)/(exp(5)+x))+I*Pi*ln(exp(5)+x)*csgn(I/ln(x))*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))*csgn(I/ln(x)/(e
xp(5)+x)*(1/5*exp(x-3)+x))+I*Pi*ln(exp(5)+x)*csgn(I/(exp(5)+x))*csgn(I*(1/5*exp(x-3)+x))*csgn(I*(1/5*exp(x-3)+
x)/(exp(5)+x))-3*I*Pi*csgn(I/ln(x)/(exp(5)+x)*(1/5*exp(x-3)+x))^3-3*I*Pi*csgn(I*(1/5*exp(x-3)+x)/(exp(5)+x))^3
-2*ln(5)*ln(exp(5)+x)+(-2*ln(exp(5)+x)+6-2*ln(ln(x)))*ln(1/5*exp(x-3)+x)+2*ln(ln(x))*ln(exp(5)+x)+2*ln(5)*ln(e
xp(x-3)+5*x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (22) = 44\).
time = 0.34, size = 104, normalized size = 4.33 \begin {gather*} 2 \, {\left (\log \left (x + e^{5}\right ) + \log \left (\log \left (x\right )\right )\right )} \log \left (5 \, x e^{3} + e^{x}\right ) - \log \left (5 \, x e^{3} + e^{x}\right )^{2} - \log \left (x + e^{5}\right )^{2} + 2 \, {\left (\log \left (5 \, x e^{3} + e^{x}\right ) - \log \left (x + e^{5}\right ) - \log \left (\log \left (x\right )\right )\right )} \log \left (\frac {5 \, x + e^{\left (x - 3\right )}}{{\left (x + e^{5}\right )} \log \left (x\right )}\right ) - 2 \, \log \left (x + e^{5}\right ) \log \left (\log \left (x\right )\right ) - \log \left (\log \left (x\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(5)+2*x^2-2*x)*exp(-3+x)+10*x*exp(5))*log(x)+(-2*exp(5)-2*x)*exp(-3+x)-10*x*exp(5)-10*x^2)
*log((exp(-3+x)+5*x)/(exp(5)+x)/log(x))/((x*exp(5)+x^2)*exp(-3+x)+5*x^2*exp(5)+5*x^3)/log(x),x, algorithm="max
ima")

[Out]

2*(log(x + e^5) + log(log(x)))*log(5*x*e^3 + e^x) - log(5*x*e^3 + e^x)^2 - log(x + e^5)^2 + 2*(log(5*x*e^3 + e
^x) - log(x + e^5) - log(log(x)))*log((5*x + e^(x - 3))/((x + e^5)*log(x))) - 2*log(x + e^5)*log(log(x)) - log
(log(x))^2

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Fricas [A]
time = 0.38, size = 22, normalized size = 0.92 \begin {gather*} \log \left (\frac {5 \, x + e^{\left (x - 3\right )}}{{\left (x + e^{5}\right )} \log \left (x\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(5)+2*x^2-2*x)*exp(-3+x)+10*x*exp(5))*log(x)+(-2*exp(5)-2*x)*exp(-3+x)-10*x*exp(5)-10*x^2)
*log((exp(-3+x)+5*x)/(exp(5)+x)/log(x))/((x*exp(5)+x^2)*exp(-3+x)+5*x^2*exp(5)+5*x^3)/log(x),x, algorithm="fri
cas")

[Out]

log((5*x + e^(x - 3))/((x + e^5)*log(x)))^2

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Sympy [A]
time = 0.86, size = 19, normalized size = 0.79 \begin {gather*} \log {\left (\frac {5 x + e^{x - 3}}{\left (x + e^{5}\right ) \log {\left (x \right )}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((5*x + exp(x - 3))/((x + exp(5))*log(x)))**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(5)+2*x^2-2*x)*exp(-3+x)+10*x*exp(5))*log(x)+(-2*exp(5)-2*x)*exp(-3+x)-10*x*exp(5)-10*x^2)
*log((exp(-3+x)+5*x)/(exp(5)+x)/log(x))/((x*exp(5)+x^2)*exp(-3+x)+5*x^2*exp(5)+5*x^3)/log(x),x, algorithm="gia
c")

[Out]

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

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Mupad [B]
time = 2.14, size = 23, normalized size = 0.96 \begin {gather*} {\ln \left (\frac {5\,x+{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{\ln \left (x\right )\,\left (x+{\mathrm {e}}^5\right )}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((5*x + exp(-3)*exp(x))/(log(x)*(x + exp(5))))^2

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