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

Optimal. Leaf size=31 \[ \frac {e^{2 x} \left (5+e^{3-e^x-\frac {2 x \log \left (x^2\right )}{\log (4)}}\right )}{x} \]

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Rubi [F]  time = 1.28, 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 {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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 1.02, size = 32, normalized size = 1.03 \begin {gather*} \frac {e^{2 x} \left (5+e^{3-e^x} \left (x^2\right )^{-\frac {2 x}{\log (4)}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

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

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fricas [A]  time = 0.62, size = 38, normalized size = 1.23 \begin {gather*} \frac {5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x - \frac {e^{x} \log \relax (2) + x \log \left (x^{2}\right ) - 3 \, \log \relax (2)}{\log \relax (2)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(2*x-1)*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

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giac [A]  time = 0.29, size = 47, normalized size = 1.52 \begin {gather*} \frac {5 \, e^{\left (2 \, x\right )} \log \relax (2) + e^{\left (\frac {2 \, x \log \relax (2) - e^{x} \log \relax (2) - x \log \left (x^{2}\right )}{\log \relax (2)} + 3\right )} \log \relax (2)}{x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(2*x-1)*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))

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maple [C]  time = 0.54, size = 95, normalized size = 3.06

method	result	size
risch	\(\frac {5 \,{\mathrm e}^{2 x}}{x}+\frac {{\mathrm e}^{-\frac {-i x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i x \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+4 x \ln \relax (x )+2 \,{\mathrm e}^{x} \ln \relax (2)-4 x \ln \relax (2)-6 \ln \relax (2)}{2 \ln \relax (2)}}}{x}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-2*x*exp(x)^2*ln(x^2)-2*x*ln(2)*exp(x)^3+(2*(2*x-1)*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,method=_RETURNVERBOSE)

[Out]

5*exp(2*x)/x+1/x*exp(-1/2*(-I*x*Pi*csgn(I*x^2)^3+2*I*x*Pi*csgn(I*x^2)^2*csgn(I*x)-I*x*Pi*csgn(I*x^2)*csgn(I*x)
^2+4*x*ln(x)+2*exp(x)*ln(2)-4*x*ln(2)-6*ln(2))/ln(2))

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maxima [C]  time = 0.71, size = 48, normalized size = 1.55 \begin {gather*} \frac {10 \, {\rm Ei}\left (2 \, x\right ) \log \relax (2) - 10 \, \Gamma \left (-1, -2 \, x\right ) \log \relax (2) + \frac {e^{\left (2 \, x - \frac {2 \, x \log \relax (x)}{\log \relax (2)} - e^{x} + 3\right )} \log \relax (2)}{x}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(2*x-1)*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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\frac {{\mathrm {e}}^{-\frac {x\,\ln \left (x^2\right )-3\,\ln \relax (2)+{\mathrm {e}}^x\,\ln \relax (2)}{\ln \relax (2)}}\,\left ({\mathrm {e}}^{2\,x}\,\left (4\,x-2\,\ln \relax (2)\,\left (2\,x-1\right )\right )+2\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{3\,x}\,\ln \relax (2)\right )}{2}-{\mathrm {e}}^{2\,x}\,\ln \relax (2)\,\left (10\,x-5\right )}{x^2\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.42, size = 37, normalized size = 1.19 \begin {gather*} \frac {e^{2 x} e^{\frac {- x \log {\left (x^{2} \right )} - e^{x} \log {\relax (2 )} + 3 \log {\relax (2 )}}{\log {\relax (2 )}}}}{x} + \frac {5 e^{2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-2*x*exp(x)**2*ln(x**2)-2*x*ln(2)*exp(x)**3+(2*(2*x-1)*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

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