\(\int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log (x^2)}{\log (4)}} (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log (x^2))}{x^2 \log (4)} \, dx\) [3959]

Optimal result

Integrand size = 90, antiderivative size = 31 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} \left (5+e^{3-e^x-\frac {2 x \log \left (x^2\right )}{\log (4)}}\right )}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx \]

[In]

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

[Out]

(5*E^(2*x))/x - Defer[Int][E^(3 - E^x + 2*x)*(x^2)^(-1 - (2*x)/Log[4]), x] - (4*(1 - Log[2])*Defer[Int][E^(3 -
E^x + 2*x)*x*(x^2)^(-1 - (2*x)/Log[4]), x])/Log[4] - (2*Log[x^2]*Defer[Int][E^(3 - E^x + 2*x)*x*(x^2)^(-1 - (
2*x)/Log[4]), x])/Log[4] - Defer[Int][E^(3 - E^x + 3*x)*x*(x^2)^(-1 - (2*x)/Log[4]), x] + (4*Defer[Int][Defer[
Int][E^(3 - E^x + 2*x)/(x*(x^2)^((2*x)/Log[4])), x]/x, x])/Log[4]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2} \, dx}{\log (4)} \\ & = \frac {\int \left (\frac {5 e^{2 x} (-1+2 x) \log (4)}{x^2}+e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \left (-4 x (1-\log (2))-\log (4)-e^x x \log (4)-2 x \log \left (x^2\right )\right )\right ) \, dx}{\log (4)} \\ & = 5 \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx+\frac {\int e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \left (-4 x (1-\log (2))-\log (4)-e^x x \log (4)-2 x \log \left (x^2\right )\right ) \, dx}{\log (4)} \\ & = \frac {5 e^{2 x}}{x}+\frac {\int \left (4 e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} (-1+\log (2))-e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \log (4)-e^{3-e^x+3 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \log (4)-2 e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \log \left (x^2\right )\right ) \, dx}{\log (4)} \\ & = \frac {5 e^{2 x}}{x}-\frac {2 \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \log \left (x^2\right ) \, dx}{\log (4)}-\frac {(4 (1-\log (2))) \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx}{\log (4)}-\int e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx-\int e^{3-e^x+3 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx \\ & = \frac {5 e^{2 x}}{x}+\frac {2 \int \frac {2 \int \frac {e^{3-e^x+2 x} \left (x^2\right )^{-\frac {2 x}{\log (4)}}}{x} \, dx}{x} \, dx}{\log (4)}-\frac {(4 (1-\log (2))) \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx}{\log (4)}-\frac {\left (2 \log \left (x^2\right )\right ) \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx}{\log (4)}-\int e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx-\int e^{3-e^x+3 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx \\ & = \frac {5 e^{2 x}}{x}+\frac {4 \int \frac {\int \frac {e^{3-e^x+2 x} \left (x^2\right )^{-\frac {2 x}{\log (4)}}}{x} \, dx}{x} \, dx}{\log (4)}-\frac {(4 (1-\log (2))) \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx}{\log (4)}-\frac {\left (2 \log \left (x^2\right )\right ) \int e^{3-e^x+2 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx}{\log (4)}-\int e^{3-e^x+2 x} \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx-\int e^{3-e^x+3 x} x \left (x^2\right )^{-1-\frac {2 x}{\log (4)}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} \left (5 x \log (4)+\frac {e^{3-e^x} x \left (x^2\right )^{-\frac {2 x}{\log (4)}} \left (-4 \log (2)+e^x \log (4)\right )}{-2+e^x}\right )}{x^2 \log (4)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61

method	result	size
parallelrisch	\(\frac {4 \ln \left (2\right ) {\mathrm e}^{-\frac {{\mathrm e}^{x} \ln \left (2\right )+x \ln \left (x^{2}\right )-3 \ln \left (2\right )}{\ln \left (2\right )}} {\mathrm e}^{2 x}+20 \ln \left (2\right ) {\mathrm e}^{2 x}}{4 \ln \left (2\right ) x}\)	\(50\)
risch	\(\frac {5 \,{\mathrm e}^{2 x}}{x}+\frac {x^{-\frac {2 x}{\ln \left (2\right )}} {\mathrm e}^{-\frac {-i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 \,{\mathrm e}^{x} \ln \left (2\right )-4 x \ln \left (2\right )-6 \ln \left (2\right )}{2 \ln \left (2\right )}}}{x}\)	\(99\)

[In]

int(1/2*((-2*x*exp(x)^2*ln(x^2)-2*x*ln(2)*exp(x)^3+(2*(-1+2*x)*ln(2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*ln(x^2)-2*ex
p(x)*ln(2)+6*ln(2))/ln(2))+2*(10*x-5)*ln(2)*exp(x)^2)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/4/ln(2)/x*(4*ln(2)*exp(-(exp(x)*ln(2)+x*ln(x^2)-3*ln(2))/ln(2))*exp(x)^2+20*ln(2)*exp(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x - \frac {e^{x} \log \left (2\right ) + x \log \left (x^{2}\right ) - 3 \, \log \left (2\right )}{\log \left (2\right )}\right )}}{x} \]

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log(2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log
(x^2)-2*exp(x)*log(2)+6*log(2))/log(2))+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} e^{\frac {- x \log {\left (x^{2} \right )} - e^{x} \log {\left (2 \right )} + 3 \log {\left (2 \right )}}{\log {\left (2 \right )}}}}{x} + \frac {5 e^{2 x}}{x} \]

[In]

integrate(1/2*((-2*x*exp(x)**2*ln(x**2)-2*x*ln(2)*exp(x)**3+(2*(-1+2*x)*ln(2)-4*x)*exp(x)**2)*exp(1/2*(-2*x*ln
(x**2)-2*exp(x)*ln(2)+6*ln(2))/ln(2))+2*(10*x-5)*ln(2)*exp(x)**2)/x**2/ln(2),x)

[Out]

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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {10 \, {\rm Ei}\left (2 \, x\right ) \log \left (2\right ) - 10 \, \Gamma \left (-1, -2 \, x\right ) \log \left (2\right ) + \frac {e^{\left (2 \, x - \frac {2 \, x \log \left (x\right )}{\log \left (2\right )} - e^{x} + 3\right )} \log \left (2\right )}{x}}{\log \left (2\right )} \]

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log(2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log
(x^2)-2*exp(x)*log(2)+6*log(2))/log(2))+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm="maxima")

[Out]

(10*Ei(2*x)*log(2) - 10*gamma(-1, -2*x)*log(2) + e^(2*x - 2*x*log(x)/log(2) - e^x + 3)*log(2)/x)/log(2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {5 \, e^{\left (2 \, x\right )} \log \left (2\right ) + e^{\left (\frac {2 \, x \log \left (2\right ) - e^{x} \log \left (2\right ) - x \log \left (x^{2}\right )}{\log \left (2\right )} + 3\right )} \log \left (2\right )}{x \log \left (2\right )} \]

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log(2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log
(x^2)-2*exp(x)*log(2)+6*log(2))/log(2))+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\int -\frac {\frac {{\mathrm {e}}^{-\frac {x\,\ln \left (x^2\right )-3\,\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )}{\ln \left (2\right )}}\,\left ({\mathrm {e}}^{2\,x}\,\left (4\,x-2\,\ln \left (2\right )\,\left (2\,x-1\right )\right )+2\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{3\,x}\,\ln \left (2\right )\right )}{2}-{\mathrm {e}}^{2\,x}\,\ln \left (2\right )\,\left (10\,x-5\right )}{x^2\,\ln \left (2\right )} \,d x \]

[In]

int(-((exp(-(x*log(x^2) - 3*log(2) + exp(x)*log(2))/log(2))*(exp(2*x)*(4*x - 2*log(2)*(2*x - 1)) + 2*x*log(x^2
)*exp(2*x) + 2*x*exp(3*x)*log(2)))/2 - exp(2*x)*log(2)*(10*x - 5))/(x^2*log(2)),x)

[Out]

int(-((exp(-(x*log(x^2) - 3*log(2) + exp(x)*log(2))/log(2))*(exp(2*x)*(4*x - 2*log(2)*(2*x - 1)) + 2*x*log(x^2
)*exp(2*x) + 2*x*exp(3*x)*log(2)))/2 - exp(2*x)*log(2)*(10*x - 5))/(x^2*log(2)), x)