\(\int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx\) [662]

Optimal result

Integrand size = 45, antiderivative size = 20 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {e^{2 x} \left (-e+\log ^2(x)\right )}{8 x} \]

[Out]

1/8/x*(ln(x)^2-exp(1))*exp(x)^2

Rubi [F]

\[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx \]

[In]

Int[(E^(1 + 2*x)*(1 - 2*x) + 2*E^(2*x)*Log[x] + E^(2*x)*(-1 + 2*x)*Log[x]^2)/(8*x^2),x]

[Out]

-1/4*E^(2*x)/x - E^(1 + 2*x)/(8*x) + ExpIntegralEi[2*x]/2 - x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, 2*x] - L
og[-2*x]^2/4 - (EulerGamma*Log[x])/2 - (E^(2*x)*Log[x])/(4*x) + (ExpIntegralEi[2*x]*Log[x])/2 - ((ExpIntegralE
[1, -2*x] + ExpIntegralEi[2*x])*Log[x])/2 - Defer[Int][(E^(2*x)*Log[x]^2)/x^2, x]/8 + Defer[Int][(E^(2*x)*Log[
x]^2)/x, x]/4

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{x^2} \, dx \\ & = \frac {1}{8} \int \left (\frac {e^{1+2 x}}{x^2}-\frac {2 e^{1+2 x}}{x}+\frac {2 e^{2 x} \log (x)}{x^2}-\frac {e^{2 x} \log ^2(x)}{x^2}+\frac {2 e^{2 x} \log ^2(x)}{x}\right ) \, dx \\ & = \frac {1}{8} \int \frac {e^{1+2 x}}{x^2} \, dx-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx-\frac {1}{4} \int \frac {e^{1+2 x}}{x} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log (x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e \operatorname {ExpIntegralEi}(2 x)}{4}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{1+2 x}}{x} \, dx-\frac {1}{4} \int \frac {-e^{2 x}+2 x \operatorname {ExpIntegralEi}(2 x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx-\frac {1}{4} \int \left (-\frac {e^{2 x}}{x^2}+\frac {2 \operatorname {ExpIntegralEi}(2 x)}{x}\right ) \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ & = -\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x}}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx-\frac {1}{2} \int \frac {\operatorname {ExpIntegralEi}(2 x)}{x} \, dx \\ & = -\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{2} (\operatorname {ExpIntegralE}(1,-2 x)+\operatorname {ExpIntegralEi}(2 x)) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx+\frac {1}{2} \int \frac {e^{2 x}}{x} \, dx+\frac {1}{2} \int \frac {\operatorname {ExpIntegralE}(1,-2 x)}{x} \, dx \\ & = -\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}+\frac {\operatorname {ExpIntegralEi}(2 x)}{2}-x \, _3F_3(1,1,1;2,2,2;2 x)-\frac {1}{4} \log ^2(-2 x)-\frac {1}{2} \gamma \log (x)-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \operatorname {ExpIntegralEi}(2 x) \log (x)-\frac {1}{2} (\operatorname {ExpIntegralE}(1,-2 x)+\operatorname {ExpIntegralEi}(2 x)) \log (x)-\frac {1}{8} \int \frac {e^{2 x} \log ^2(x)}{x^2} \, dx+\frac {1}{4} \int \frac {e^{2 x} \log ^2(x)}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {e^{2 x} \left (e-\log ^2(x)\right )}{8 x} \]

[In]

Integrate[(E^(1 + 2*x)*(1 - 2*x) + 2*E^(2*x)*Log[x] + E^(2*x)*(-1 + 2*x)*Log[x]^2)/(8*x^2),x]

[Out]

-1/8*(E^(2*x)*(E - Log[x]^2))/x

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(-\frac {-{\mathrm e}^{2 x} \ln \left (x \right )^{2}+{\mathrm e} \,{\mathrm e}^{2 x}}{8 x}\)	\(24\)
default	\(\frac {{\mathrm e}^{2 x} \ln \left (x \right )^{2}}{8 x}-\frac {{\mathrm e}^{1+2 x}}{8 x}\)	\(26\)
risch	\(\frac {{\mathrm e}^{2 x} \ln \left (x \right )^{2}}{8 x}-\frac {{\mathrm e}^{1+2 x}}{8 x}\)	\(26\)

[In]

int(1/8*((-1+2*x)*exp(x)^2*ln(x)^2+2*exp(x)^2*ln(x)+(1-2*x)*exp(1)*exp(x)^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/8/x*(-exp(x)^2*ln(x)^2+exp(1)*exp(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {{\left (e^{\left (2 \, x + 1\right )} \log \left (x\right )^{2} - e^{\left (2 \, x + 2\right )}\right )} e^{\left (-1\right )}}{8 \, x} \]

[In]

integrate(1/8*((-1+2*x)*exp(x)^2*log(x)^2+2*exp(x)^2*log(x)+(1-2*x)*exp(1)*exp(x)^2)/x^2,x, algorithm="fricas"
)

[Out]

1/8*(e^(2*x + 1)*log(x)^2 - e^(2*x + 2))*e^(-1)/x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {\left (\log {\left (x \right )}^{2} - e\right ) e^{2 x}}{8 x} \]

[In]

integrate(1/8*((-1+2*x)*exp(x)**2*ln(x)**2+2*exp(x)**2*ln(x)+(1-2*x)*exp(1)*exp(x)**2)/x**2,x)

[Out]

(log(x)**2 - E)*exp(2*x)/(8*x)

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {1}{4} \, {\rm Ei}\left (2 \, x\right ) e + \frac {1}{4} \, e \Gamma \left (-1, -2 \, x\right ) + \frac {e^{\left (2 \, x\right )} \log \left (x\right )^{2}}{8 \, x} \]

[In]

integrate(1/8*((-1+2*x)*exp(x)^2*log(x)^2+2*exp(x)^2*log(x)+(1-2*x)*exp(1)*exp(x)^2)/x^2,x, algorithm="maxima"
)

[Out]

-1/4*Ei(2*x)*e + 1/4*e*gamma(-1, -2*x) + 1/8*e^(2*x)*log(x)^2/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=\frac {e^{\left (2 \, x\right )} \log \left (x\right )^{2} - e^{\left (2 \, x + 1\right )}}{8 \, x} \]

[In]

integrate(1/8*((-1+2*x)*exp(x)^2*log(x)^2+2*exp(x)^2*log(x)+(1-2*x)*exp(1)*exp(x)^2)/x^2,x, algorithm="giac")

[Out]

1/8*(e^(2*x)*log(x)^2 - e^(2*x + 1))/x

Mupad [B] (verification not implemented)

Time = 8.44 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1+2 x} (1-2 x)+2 e^{2 x} \log (x)+e^{2 x} (-1+2 x) \log ^2(x)}{8 x^2} \, dx=-\frac {{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}-{\ln \left (x\right )}^2\right )}{8\,x} \]

[In]

int(((exp(2*x)*log(x))/4 + (exp(2*x)*log(x)^2*(2*x - 1))/8 - (exp(2*x)*exp(1)*(2*x - 1))/8)/x^2,x)

[Out]

-(exp(2*x)*(exp(1) - log(x)^2))/(8*x)