3.59.42 \(\int \frac {e^x (-15-20 x)+e^x (15+55 x+20 x^2) \log (2 x)}{12 \log ^2(2 x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {5 e^x \left (4+\frac {3}{x}\right ) x^2}{12 \log (2 x)} \]

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Rubi [F]  time = 0.34, 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^x (-15-20 x)+e^x \left (15+55 x+20 x^2\right ) \log (2 x)}{12 \log ^2(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-15 - 20*x) + E^x*(15 + 55*x + 20*x^2)*Log[2*x])/(12*Log[2*x]^2),x]

[Out]

(-5*Defer[Int][E^x/Log[2*x]^2, x])/4 - (5*Defer[Int][(E^x*x)/Log[2*x]^2, x])/3 + (5*Defer[Int][E^x/Log[2*x], x
])/4 + (55*Defer[Int][(E^x*x)/Log[2*x], x])/12 + (5*Defer[Int][(E^x*x^2)/Log[2*x], x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{12} \int \frac {e^x (-15-20 x)+e^x \left (15+55 x+20 x^2\right ) \log (2 x)}{\log ^2(2 x)} \, dx\\ &=\frac {1}{12} \int \left (-\frac {15 e^x}{\log ^2(2 x)}-\frac {20 e^x x}{\log ^2(2 x)}+\frac {15 e^x}{\log (2 x)}+\frac {55 e^x x}{\log (2 x)}+\frac {20 e^x x^2}{\log (2 x)}\right ) \, dx\\ &=-\left (\frac {5}{4} \int \frac {e^x}{\log ^2(2 x)} \, dx\right )+\frac {5}{4} \int \frac {e^x}{\log (2 x)} \, dx-\frac {5}{3} \int \frac {e^x x}{\log ^2(2 x)} \, dx+\frac {5}{3} \int \frac {e^x x^2}{\log (2 x)} \, dx+\frac {55}{12} \int \frac {e^x x}{\log (2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 19, normalized size = 0.83 \begin {gather*} \frac {5 e^x x (3+4 x)}{12 \log (2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-15 - 20*x) + E^x*(15 + 55*x + 20*x^2)*Log[2*x])/(12*Log[2*x]^2),x]

[Out]

(5*E^x*x*(3 + 4*x))/(12*Log[2*x])

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fricas [A]  time = 0.62, size = 19, normalized size = 0.83 \begin {gather*} \frac {5 \, {\left (4 \, x^{2} + 3 \, x\right )} e^{x}}{12 \, \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((20*x^2+55*x+15)*exp(x)*log(2*x)+(-20*x-15)*exp(x))/log(2*x)^2,x, algorithm="fricas")

[Out]

5/12*(4*x^2 + 3*x)*e^x/log(2*x)

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giac [A]  time = 0.12, size = 21, normalized size = 0.91 \begin {gather*} \frac {5 \, {\left (4 \, x^{2} e^{x} + 3 \, x e^{x}\right )}}{12 \, \log \left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((20*x^2+55*x+15)*exp(x)*log(2*x)+(-20*x-15)*exp(x))/log(2*x)^2,x, algorithm="giac")

[Out]

5/12*(4*x^2*e^x + 3*x*e^x)/log(2*x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.74

method	result	size
risch	\(\frac {5 x \,{\mathrm e}^{x} \left (3+4 x \right )}{12 \ln \left (2 x \right )}\)	\(17\)
norman	\(\frac {\frac {5 \,{\mathrm e}^{x} x}{4}+\frac {5 \,{\mathrm e}^{x} x^{2}}{3}}{\ln \left (2 x \right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*((20*x^2+55*x+15)*exp(x)*ln(2*x)+(-20*x-15)*exp(x))/ln(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

5/12*x*exp(x)*(3+4*x)/ln(2*x)

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maxima [A]  time = 0.47, size = 20, normalized size = 0.87 \begin {gather*} \frac {5 \, {\left (4 \, x^{2} + 3 \, x\right )} e^{x}}{12 \, {\left (\log \relax (2) + \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((20*x^2+55*x+15)*exp(x)*log(2*x)+(-20*x-15)*exp(x))/log(2*x)^2,x, algorithm="maxima")

[Out]

5/12*(4*x^2 + 3*x)*e^x/(log(2) + log(x))

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mupad [B]  time = 4.15, size = 26, normalized size = 1.13 \begin {gather*} \frac {15\,x^2\,{\mathrm {e}}^x+20\,x^3\,{\mathrm {e}}^x}{12\,x\,\ln \left (2\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(20*x + 15))/12 - (log(2*x)*exp(x)*(55*x + 20*x^2 + 15))/12)/log(2*x)^2,x)

[Out]

(15*x^2*exp(x) + 20*x^3*exp(x))/(12*x*log(2*x))

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sympy [A]  time = 0.26, size = 17, normalized size = 0.74 \begin {gather*} \frac {\left (20 x^{2} + 15 x\right ) e^{x}}{12 \log {\left (2 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((20*x**2+55*x+15)*exp(x)*ln(2*x)+(-20*x-15)*exp(x))/ln(2*x)**2,x)

[Out]

(20*x**2 + 15*x)*exp(x)/(12*log(2*x))

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