3.97.58 \(\int \frac {e^{2 x} (864+10368 x)+e^{2 x} (72+1728 x) \log (5 x)+72 e^{2 x} x \log ^2(5 x)}{x} \, dx\)

Optimal. Leaf size=18 \[ 36 \left (-3+e^{2 x} (12+\log (5 x))^2\right ) \]

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Rubi [F]  time = 0.18, 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} (864+10368 x)+e^{2 x} (72+1728 x) \log (5 x)+72 e^{2 x} x \log ^2(5 x)}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(864 + 10368*x) + E^(2*x)*(72 + 1728*x)*Log[5*x] + 72*E^(2*x)*x*Log[5*x]^2)/x,x]

[Out]

5184*E^(2*x) - 144*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, 2*x] - 36*Log[-2*x]^2 - 72*EulerGamma*Log[x] - 72
*(ExpIntegralE[1, -2*x] + ExpIntegralEi[2*x])*Log[x] + 864*E^(2*x)*Log[5*x] + 72*ExpIntegralEi[2*x]*Log[5*x] +
72*Defer[Int][E^(2*x)*Log[5*x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (10368 e^{2 x}+\frac {864 e^{2 x}}{x}+1728 e^{2 x} \log (5 x)+\frac {72 e^{2 x} \log (5 x)}{x}+72 e^{2 x} \log ^2(5 x)\right ) \, dx\\ &=72 \int \frac {e^{2 x} \log (5 x)}{x} \, dx+72 \int e^{2 x} \log ^2(5 x) \, dx+864 \int \frac {e^{2 x}}{x} \, dx+1728 \int e^{2 x} \log (5 x) \, dx+10368 \int e^{2 x} \, dx\\ &=5184 e^{2 x}+864 \text {Ei}(2 x)+864 e^{2 x} \log (5 x)+72 \text {Ei}(2 x) \log (5 x)-72 \int \frac {\text {Ei}(2 x)}{x} \, dx+72 \int e^{2 x} \log ^2(5 x) \, dx-1728 \int \frac {e^{2 x}}{2 x} \, dx\\ &=5184 e^{2 x}+864 \text {Ei}(2 x)-72 (E_1(-2 x)+\text {Ei}(2 x)) \log (x)+864 e^{2 x} \log (5 x)+72 \text {Ei}(2 x) \log (5 x)+72 \int \frac {E_1(-2 x)}{x} \, dx+72 \int e^{2 x} \log ^2(5 x) \, dx-864 \int \frac {e^{2 x}}{x} \, dx\\ &=5184 e^{2 x}-144 x \, _3F_3(1,1,1;2,2,2;2 x)-36 \log ^2(-2 x)-72 \gamma \log (x)-72 (E_1(-2 x)+\text {Ei}(2 x)) \log (x)+864 e^{2 x} \log (5 x)+72 \text {Ei}(2 x) \log (5 x)+72 \int e^{2 x} \log ^2(5 x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 0.83 \begin {gather*} 36 e^{2 x} (12+\log (5 x))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(864 + 10368*x) + E^(2*x)*(72 + 1728*x)*Log[5*x] + 72*E^(2*x)*x*Log[5*x]^2)/x,x]

[Out]

36*E^(2*x)*(12 + Log[5*x])^2

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fricas [A]  time = 0.75, size = 29, normalized size = 1.61 \begin {gather*} 36 \, e^{\left (2 \, x\right )} \log \left (5 \, x\right )^{2} + 864 \, e^{\left (2 \, x\right )} \log \left (5 \, x\right ) + 5184 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x*exp(x)^2*log(5*x)^2+(1728*x+72)*exp(x)^2*log(5*x)+(10368*x+864)*exp(x)^2)/x,x, algorithm="fric
as")

[Out]

36*e^(2*x)*log(5*x)^2 + 864*e^(2*x)*log(5*x) + 5184*e^(2*x)

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giac [A]  time = 0.23, size = 29, normalized size = 1.61 \begin {gather*} 36 \, e^{\left (2 \, x\right )} \log \left (5 \, x\right )^{2} + 864 \, e^{\left (2 \, x\right )} \log \left (5 \, x\right ) + 5184 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x*exp(x)^2*log(5*x)^2+(1728*x+72)*exp(x)^2*log(5*x)+(10368*x+864)*exp(x)^2)/x,x, algorithm="giac
")

[Out]

36*e^(2*x)*log(5*x)^2 + 864*e^(2*x)*log(5*x) + 5184*e^(2*x)

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maple [A]  time = 0.04, size = 30, normalized size = 1.67

method	result	size
norman	\(5184 \,{\mathrm e}^{2 x}+864 \,{\mathrm e}^{2 x} \ln \left (5 x \right )+36 \,{\mathrm e}^{2 x} \ln \left (5 x \right )^{2}\)	\(30\)
risch	\(5184 \,{\mathrm e}^{2 x}+864 \,{\mathrm e}^{2 x} \ln \left (5 x \right )+36 \,{\mathrm e}^{2 x} \ln \left (5 x \right )^{2}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((72*x*exp(x)^2*ln(5*x)^2+(1728*x+72)*exp(x)^2*ln(5*x)+(10368*x+864)*exp(x)^2)/x,x,method=_RETURNVERBOSE)

[Out]

5184*exp(x)^2+864*exp(x)^2*ln(5*x)+36*exp(x)^2*ln(5*x)^2

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maxima [B]  time = 0.50, size = 38, normalized size = 2.11 \begin {gather*} 36 \, {\left (\log \relax (5)^{2} + 2 \, \log \relax (5) \log \relax (x) + \log \relax (x)^{2}\right )} e^{\left (2 \, x\right )} + 864 \, e^{\left (2 \, x\right )} \log \left (5 \, x\right ) + 5184 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x*exp(x)^2*log(5*x)^2+(1728*x+72)*exp(x)^2*log(5*x)+(10368*x+864)*exp(x)^2)/x,x, algorithm="maxi
ma")

[Out]

36*(log(5)^2 + 2*log(5)*log(x) + log(x)^2)*e^(2*x) + 864*e^(2*x)*log(5*x) + 5184*e^(2*x)

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mupad [B]  time = 7.13, size = 14, normalized size = 0.78 \begin {gather*} 36\,{\mathrm {e}}^{2\,x}\,{\left (\ln \left (5\,x\right )+12\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(10368*x + 864) + log(5*x)*exp(2*x)*(1728*x + 72) + 72*x*log(5*x)^2*exp(2*x))/x,x)

[Out]

36*exp(2*x)*(log(5*x) + 12)^2

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sympy [A]  time = 0.34, size = 20, normalized size = 1.11 \begin {gather*} \left (36 \log {\left (5 x \right )}^{2} + 864 \log {\left (5 x \right )} + 5184\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((72*x*exp(x)**2*ln(5*x)**2+(1728*x+72)*exp(x)**2*ln(5*x)+(10368*x+864)*exp(x)**2)/x,x)

[Out]

(36*log(5*x)**2 + 864*log(5*x) + 5184)*exp(2*x)

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