3.84.19 \(\int \frac {20+e^{2 x} (-25+50 x) \log (4)}{4 x^2} \, dx\)

Optimal. Leaf size=24 \[ -\frac {5 \left (4-\frac {2 x}{e}-5 e^{2 x} \log (4)\right )}{4 x} \]

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Rubi [A]  time = 0.04, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2197} \begin {gather*} \frac {25 e^{2 x} \log (4)}{4 x}-\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + E^(2*x)*(-25 + 50*x)*Log[4])/(4*x^2),x]

[Out]

-5/x + (25*E^(2*x)*Log[4])/(4*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {20+e^{2 x} (-25+50 x) \log (4)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {20}{x^2}+\frac {25 e^{2 x} (-1+2 x) \log (4)}{x^2}\right ) \, dx\\ &=-\frac {5}{x}+\frac {1}{4} (25 \log (4)) \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx\\ &=-\frac {5}{x}+\frac {25 e^{2 x} \log (4)}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.75 \begin {gather*} \frac {5 \left (-4+5 e^{2 x} \log (4)\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + E^(2*x)*(-25 + 50*x)*Log[4])/(4*x^2),x]

[Out]

(5*(-4 + 5*E^(2*x)*Log[4]))/(4*x)

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fricas [A]  time = 0.59, size = 15, normalized size = 0.62 \begin {gather*} \frac {5 \, {\left (5 \, e^{\left (2 \, x\right )} \log \relax (2) - 2\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*(50*x-25)*log(2)*exp(x)^2+20)/x^2,x, algorithm="fricas")

[Out]

5/2*(5*e^(2*x)*log(2) - 2)/x

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giac [A]  time = 0.15, size = 15, normalized size = 0.62 \begin {gather*} \frac {5 \, {\left (5 \, e^{\left (2 \, x\right )} \log \relax (2) - 2\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*(50*x-25)*log(2)*exp(x)^2+20)/x^2,x, algorithm="giac")

[Out]

5/2*(5*e^(2*x)*log(2) - 2)/x

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maple [A]  time = 0.04, size = 15, normalized size = 0.62

method	result	size
norman	\(\frac {-5+\frac {25 \ln \relax (2) {\mathrm e}^{2 x}}{2}}{x}\)	\(15\)
default	\(-\frac {5}{x}+\frac {25 \ln \relax (2) {\mathrm e}^{2 x}}{2 x}\)	\(18\)
risch	\(-\frac {5}{x}+\frac {25 \ln \relax (2) {\mathrm e}^{2 x}}{2 x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(2*(50*x-25)*ln(2)*exp(x)^2+20)/x^2,x,method=_RETURNVERBOSE)

[Out]

(-5+25/2*ln(2)*exp(x)^2)/x

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maxima [C]  time = 0.38, size = 23, normalized size = 0.96 \begin {gather*} 25 \, {\rm Ei}\left (2 \, x\right ) \log \relax (2) - 25 \, \Gamma \left (-1, -2 \, x\right ) \log \relax (2) - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*(50*x-25)*log(2)*exp(x)^2+20)/x^2,x, algorithm="maxima")

[Out]

25*Ei(2*x)*log(2) - 25*gamma(-1, -2*x)*log(2) - 5/x

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mupad [B]  time = 0.07, size = 15, normalized size = 0.62 \begin {gather*} \frac {25\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)-10}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2*x)*log(2)*(50*x - 25))/2 + 5)/x^2,x)

[Out]

(25*exp(2*x)*log(2) - 10)/(2*x)

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sympy [A]  time = 0.12, size = 15, normalized size = 0.62 \begin {gather*} \frac {25 e^{2 x} \log {\relax (2 )}}{2 x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*(50*x-25)*ln(2)*exp(x)**2+20)/x**2,x)

[Out]

25*exp(2*x)*log(2)/(2*x) - 5/x

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