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

Optimal. Leaf size=20 \[ \frac {e^{2 x} \left (-e+\log ^2(x)\right )}{8 x} \]

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Rubi [F]  time = 0.41, 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^{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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 \text {Ei}(2 x)}{4}-\frac {e^{2 x} \log (x)}{4 x}+\frac {1}{2} \text {Ei}(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 \text {Ei}(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} \text {Ei}(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 \text {Ei}(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} \text {Ei}(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 {\text {Ei}(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} \text {Ei}(2 x) \log (x)-\frac {1}{2} (E_1(-2 x)+\text {Ei}(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 {E_1(-2 x)}{x} \, dx\\ &=-\frac {e^{2 x}}{4 x}-\frac {e^{1+2 x}}{8 x}+\frac {\text {Ei}(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} \text {Ei}(2 x) \log (x)-\frac {1}{2} (E_1(-2 x)+\text {Ei}(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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 20, normalized size = 1.00 \begin {gather*} -\frac {e^{2 x} \left (e-\log ^2(x)\right )}{8 x} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.29, size = 27, normalized size = 1.35 \begin {gather*} \frac {{\left (e^{\left (2 \, x + 1\right )} \log \relax (x)^{2} - e^{\left (2 \, x + 2\right )}\right )} e^{\left (-1\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((2*x-1)*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

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giac [A]  time = 0.38, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{\left (2 \, x\right )} \log \relax (x)^{2} - e^{\left (2 \, x + 1\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((2*x-1)*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

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maple [A]  time = 0.06, size = 26, normalized size = 1.30

method	result	size
default	\(\frac {{\mathrm e}^{2 x} \ln \relax (x )^{2}}{8 x}-\frac {{\mathrm e}^{2 x +1}}{8 x}\)	\(26\)
risch	\(\frac {{\mathrm e}^{2 x} \ln \relax (x )^{2}}{8 x}-\frac {{\mathrm e}^{2 x +1}}{8 x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [C]  time = 0.64, size = 31, normalized size = 1.55 \begin {gather*} -\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 \relax (x)^{2}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((2*x-1)*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

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mupad [B]  time = 0.62, size = 18, normalized size = 0.90 \begin {gather*} -\frac {{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}-{\ln \relax (x)}^2\right )}{8\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.28, size = 15, normalized size = 0.75 \begin {gather*} \frac {\left (\log {\relax (x )}^{2} - e\right ) e^{2 x}}{8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((2*x-1)*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)

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