3.22.7 \(\int (e^{2 x} (-200+40 e^5)+e^{2 x} (50+e^5 (-10-18 x)+90 x) \log (x)+e^{2 x} (-10 x-10 x^2+e^5 (2 x+2 x^2)) \log ^2(x)) \, dx\)

Optimal. Leaf size=19 \[ e^{2 x} \left (-5+e^5\right ) (-5+x \log (x))^2 \]

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Rubi [F]  time = 0.42, 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 \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[E^(2*x)*(-200 + 40*E^5) + E^(2*x)*(50 + E^5*(-10 - 18*x) + 90*x)*Log[x] + E^(2*x)*(-10*x - 10*x^2 + E^5*(2
*x + 2*x^2))*Log[x]^2,x]

[Out]

(-49*E^(2*x)*(5 - E^5))/2 - ((5 - E^5)*ExpIntegralEi[2*x])/2 - (9*E^(2*x)*(5 - E^5)*Log[x])/2 + E^(2*x)*(5*(5
- E^5) + 9*(5 - E^5)*x)*Log[x] - 2*(5 - E^5)*Defer[Int][E^(2*x)*x*Log[x]^2, x] - 2*(5 - E^5)*Defer[Int][E^(2*x
)*x^2*Log[x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (40 \left (5-e^5\right )\right ) \int e^{2 x} \, dx\right )+\int e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x) \, dx+\int e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x) \, dx\\ &=-20 e^{2 x} \left (5-e^5\right )-\frac {9}{2} e^{2 x} \left (5-e^5\right ) \log (x)+e^{2 x} \left (5 \left (5-e^5\right )+9 \left (5-e^5\right ) x\right ) \log (x)-\int \frac {e^{2 x} \left (5-e^5\right ) (1+18 x)}{2 x} \, dx+\int 2 e^{2 x} \left (5-e^5\right ) (-1-x) x \log ^2(x) \, dx\\ &=-20 e^{2 x} \left (5-e^5\right )-\frac {9}{2} e^{2 x} \left (5-e^5\right ) \log (x)+e^{2 x} \left (5 \left (5-e^5\right )+9 \left (5-e^5\right ) x\right ) \log (x)-\frac {1}{2} \left (5-e^5\right ) \int \frac {e^{2 x} (1+18 x)}{x} \, dx+\left (2 \left (5-e^5\right )\right ) \int e^{2 x} (-1-x) x \log ^2(x) \, dx\\ &=-20 e^{2 x} \left (5-e^5\right )-\frac {9}{2} e^{2 x} \left (5-e^5\right ) \log (x)+e^{2 x} \left (5 \left (5-e^5\right )+9 \left (5-e^5\right ) x\right ) \log (x)-\frac {1}{2} \left (5-e^5\right ) \int \left (18 e^{2 x}+\frac {e^{2 x}}{x}\right ) \, dx+\left (2 \left (5-e^5\right )\right ) \int \left (-e^{2 x} x \log ^2(x)-e^{2 x} x^2 \log ^2(x)\right ) \, dx\\ &=-20 e^{2 x} \left (5-e^5\right )-\frac {9}{2} e^{2 x} \left (5-e^5\right ) \log (x)+e^{2 x} \left (5 \left (5-e^5\right )+9 \left (5-e^5\right ) x\right ) \log (x)-\frac {1}{2} \left (5-e^5\right ) \int \frac {e^{2 x}}{x} \, dx-\left (2 \left (5-e^5\right )\right ) \int e^{2 x} x \log ^2(x) \, dx-\left (2 \left (5-e^5\right )\right ) \int e^{2 x} x^2 \log ^2(x) \, dx-\left (9 \left (5-e^5\right )\right ) \int e^{2 x} \, dx\\ &=-\frac {49}{2} e^{2 x} \left (5-e^5\right )-\frac {1}{2} \left (5-e^5\right ) \text {Ei}(2 x)-\frac {9}{2} e^{2 x} \left (5-e^5\right ) \log (x)+e^{2 x} \left (5 \left (5-e^5\right )+9 \left (5-e^5\right ) x\right ) \log (x)-\left (2 \left (5-e^5\right )\right ) \int e^{2 x} x \log ^2(x) \, dx-\left (2 \left (5-e^5\right )\right ) \int e^{2 x} x^2 \log ^2(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 19, normalized size = 1.00 \begin {gather*} e^{2 x} \left (-5+e^5\right ) (-5+x \log (x))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)*(-200 + 40*E^5) + E^(2*x)*(50 + E^5*(-10 - 18*x) + 90*x)*Log[x] + E^(2*x)*(-10*x - 10*x^2 +
E^5*(2*x + 2*x^2))*Log[x]^2,x]

[Out]

E^(2*x)*(-5 + E^5)*(-5 + x*Log[x])^2

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fricas [B]  time = 0.65, size = 48, normalized size = 2.53 \begin {gather*} {\left (x^{2} e^{5} - 5 \, x^{2}\right )} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 10 \, {\left (x e^{5} - 5 \, x\right )} e^{\left (2 \, x\right )} \log \relax (x) + 25 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*exp(5)+90*x+50)*exp(x)^2*log(x)+(40*e
xp(5)-200)*exp(x)^2,x, algorithm="fricas")

[Out]

(x^2*e^5 - 5*x^2)*e^(2*x)*log(x)^2 - 10*(x*e^5 - 5*x)*e^(2*x)*log(x) + 25*(e^5 - 5)*e^(2*x)

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giac [B]  time = 0.41, size = 109, normalized size = 5.74 \begin {gather*} \frac {5}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \relax (x) - \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x + 5\right )} \log \relax (x) - {\left (5 \, x^{2} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + 5\right )}\right )} \log \relax (x)^{2} + 20 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, {\left (5 \, {\left (18 \, x + 1\right )} e^{\left (2 \, x\right )} - {\left (18 \, x + 1\right )} e^{\left (2 \, x + 5\right )}\right )} \log \relax (x) - 25 \, e^{\left (2 \, x\right )} + 5 \, e^{\left (2 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*exp(5)+90*x+50)*exp(x)^2*log(x)+(40*e
xp(5)-200)*exp(x)^2,x, algorithm="giac")

[Out]

5/2*(2*x - 1)*e^(2*x)*log(x) - 1/2*(2*x - 1)*e^(2*x + 5)*log(x) - (5*x^2*e^(2*x) - x^2*e^(2*x + 5))*log(x)^2 +
20*(e^5 - 5)*e^(2*x) + 1/2*(5*(18*x + 1)*e^(2*x) - (18*x + 1)*e^(2*x + 5))*log(x) - 25*e^(2*x) + 5*e^(2*x + 5
)

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maple [B]  time = 0.10, size = 63, normalized size = 3.32

method	result	size
risch	\(x^{2} \ln \relax (x )^{2} {\mathrm e}^{5+2 x}-5 \ln \relax (x )^{2} {\mathrm e}^{2 x} x^{2}-10 x \ln \relax (x ) {\mathrm e}^{5+2 x}+50 \ln \relax (x ) {\mathrm e}^{2 x} x +25 \,{\mathrm e}^{5+2 x}-125 \,{\mathrm e}^{2 x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*ln(x)^2+((-18*x-10)*exp(5)+90*x+50)*exp(x)^2*ln(x)+(40*exp(5)-20
0)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

x^2*ln(x)^2*exp(5+2*x)-5*ln(x)^2*exp(2*x)*x^2-10*x*ln(x)*exp(5+2*x)+50*ln(x)*exp(2*x)*x+25*exp(5+2*x)-125*exp(
2*x)

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maxima [B]  time = 0.83, size = 43, normalized size = 2.26 \begin {gather*} {\left (x^{2} {\left (e^{5} - 5\right )} \log \relax (x)^{2} - 10 \, x {\left (e^{5} - 5\right )} \log \relax (x) + 5 \, e^{5} - 25\right )} e^{\left (2 \, x\right )} + 20 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*exp(5)+90*x+50)*exp(x)^2*log(x)+(40*e
xp(5)-200)*exp(x)^2,x, algorithm="maxima")

[Out]

(x^2*(e^5 - 5)*log(x)^2 - 10*x*(e^5 - 5)*log(x) + 5*e^5 - 25)*e^(2*x) + 20*(e^5 - 5)*e^(2*x)

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mupad [B]  time = 1.24, size = 17, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{2\,x}\,{\left (x\,\ln \relax (x)-5\right )}^2\,\left ({\mathrm {e}}^5-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*(40*exp(5) - 200) - exp(2*x)*log(x)^2*(10*x - exp(5)*(2*x + 2*x^2) + 10*x^2) + exp(2*x)*log(x)*(9
0*x - exp(5)*(18*x + 10) + 50),x)

[Out]

exp(2*x)*(x*log(x) - 5)^2*(exp(5) - 5)

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sympy [B]  time = 0.47, size = 49, normalized size = 2.58 \begin {gather*} \left (- 5 x^{2} \log {\relax (x )}^{2} + x^{2} e^{5} \log {\relax (x )}^{2} - 10 x e^{5} \log {\relax (x )} + 50 x \log {\relax (x )} - 125 + 25 e^{5}\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+2*x)*exp(5)-10*x**2-10*x)*exp(x)**2*ln(x)**2+((-18*x-10)*exp(5)+90*x+50)*exp(x)**2*ln(x)+(4
0*exp(5)-200)*exp(x)**2,x)

[Out]

(-5*x**2*log(x)**2 + x**2*exp(5)*log(x)**2 - 10*x*exp(5)*log(x) + 50*x*log(x) - 125 + 25*exp(5))*exp(2*x)

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