3.3.67 \(\int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+(-144 x-96 x^2) \log (5)+24 x^2 \log ^2(5)} \, dx\)

Optimal. Leaf size=27 \[ \frac {2-e^{2 x}}{24 x \left (-2-\frac {3}{x}+\log (5)\right )} \]

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Rubi [A]  time = 0.58, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 8, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 6741, 27, 12, 6742, 2197} \begin {gather*} \frac {e^{2 x}}{24 (x (2-\log (5))+3)}-\frac {1}{12 (x (2-\log (5))+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 2*Log[5] + E^(2*x)*(4 + 4*x + (1 - 2*x)*Log[5]))/(216 + 288*x + 96*x^2 + (-144*x - 96*x^2)*Log[5] + 2
4*x^2*Log[5]^2),x]

[Out]

-1/12*1/(3 + x*(2 - Log[5])) + E^(2*x)/(24*(3 + x*(2 - Log[5])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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
]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+\left (-144 x-96 x^2\right ) \log (5)+x^2 \left (96+24 \log ^2(5)\right )} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+144 x (2-\log (5))+24 x^2 (2-\log (5))^2} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{24 (-3-2 x+x \log (5))^2} \, dx\\ &=\int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{24 (-3+x (-2+\log (5)))^2} \, dx\\ &=\frac {1}{24} \int \frac {4 \left (1-\frac {\log (5)}{2}\right )+e^{2 x} (4+4 x+(1-2 x) \log (5))}{(-3+x (-2+\log (5)))^2} \, dx\\ &=\frac {1}{24} \int \left (\frac {2 (2-\log (5))}{(3+x (2-\log (5)))^2}+\frac {e^{2 x} (4+2 x (2-\log (5))+\log (5))}{(3+x (2-\log (5)))^2}\right ) \, dx\\ &=-\frac {1}{12 (3+x (2-\log (5)))}+\frac {1}{24} \int \frac {e^{2 x} (4+2 x (2-\log (5))+\log (5))}{(3+x (2-\log (5)))^2} \, dx\\ &=-\frac {1}{12 (3+x (2-\log (5)))}+\frac {e^{2 x}}{24 (3+x (2-\log (5)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 23, normalized size = 0.85 \begin {gather*} \frac {2-e^{2 x}}{24 (-3+x (-2+\log (5)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 2*Log[5] + E^(2*x)*(4 + 4*x + (1 - 2*x)*Log[5]))/(216 + 288*x + 96*x^2 + (-144*x - 96*x^2)*Log[
5] + 24*x^2*Log[5]^2),x]

[Out]

(2 - E^(2*x))/(24*(-3 + x*(-2 + Log[5])))

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fricas [A]  time = 0.71, size = 19, normalized size = 0.70 \begin {gather*} -\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \relax (5) - 2 \, x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(-96*x^2-144*x)*log(5)+96*x^2+288*x+21
6),x, algorithm="fricas")

[Out]

-1/24*(e^(2*x) - 2)/(x*log(5) - 2*x - 3)

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giac [A]  time = 0.90, size = 19, normalized size = 0.70 \begin {gather*} -\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \relax (5) - 2 \, x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(-96*x^2-144*x)*log(5)+96*x^2+288*x+21
6),x, algorithm="giac")

[Out]

-1/24*(e^(2*x) - 2)/(x*log(5) - 2*x - 3)

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maple [A]  time = 0.56, size = 28, normalized size = 1.04

method	result	size
norman	\(\frac {\left (-\frac {1}{18}+\frac {\ln \relax (5)}{36}\right ) x -\frac {{\mathrm e}^{2 x}}{24}}{x \ln \relax (5)-2 x -3}\)	\(28\)
risch	\(\frac {1}{12 x \ln \relax (5)-24 x -36}-\frac {{\mathrm e}^{2 x}}{24 \left (x \ln \relax (5)-2 x -3\right )}\)	\(32\)
derivativedivides	\(-\frac {\ln \relax (5) {\mathrm e}^{2 x}}{12 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {1}{3 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{3 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{2}}+\frac {\ln \relax (5)}{6 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{\left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{\left (\ln \relax (5)-2\right )^{3}}+\frac {\ln \relax (5) {\mathrm e}^{2 x}}{2 \left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}+\frac {\ln \relax (5) {\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{3}}\)	\(251\)
default	\(-\frac {\ln \relax (5) {\mathrm e}^{2 x}}{12 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {1}{3 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{3 \left (\ln \relax (5)-2\right )^{2} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{2}}+\frac {\ln \relax (5)}{6 \left (2-\ln \relax (5)\right ) \left (-2 x \ln \relax (5)+4 x +6\right )}-\frac {{\mathrm e}^{2 x}}{\left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{\left (\ln \relax (5)-2\right )^{3}}+\frac {\ln \relax (5) {\mathrm e}^{2 x}}{2 \left (\ln \relax (5)-2\right )^{3} \left (2 x -\frac {6}{\ln \relax (5)-2}\right )}+\frac {\ln \relax (5) {\mathrm e}^{\frac {6}{\ln \relax (5)-2}} \expIntegralEi \left (1, -2 x +\frac {6}{\ln \relax (5)-2}\right )}{2 \left (\ln \relax (5)-2\right )^{3}}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1-2*x)*ln(5)+4*x+4)*exp(2*x)-2*ln(5)+4)/(24*x^2*ln(5)^2+(-96*x^2-144*x)*ln(5)+96*x^2+288*x+216),x,metho
d=_RETURNVERBOSE)

[Out]

((-1/18+1/36*ln(5))*x-1/24*exp(2*x))/(x*ln(5)-2*x-3)

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maxima [B]  time = 0.59, size = 63, normalized size = 2.33 \begin {gather*} -\frac {e^{\left (2 \, x\right )}}{24 \, {\left (x {\left (\log \relax (5) - 2\right )} - 3\right )}} + \frac {\log \relax (5)}{12 \, {\left ({\left (\log \relax (5)^{2} - 4 \, \log \relax (5) + 4\right )} x - 3 \, \log \relax (5) + 6\right )}} - \frac {1}{6 \, {\left ({\left (\log \relax (5)^{2} - 4 \, \log \relax (5) + 4\right )} x - 3 \, \log \relax (5) + 6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(-96*x^2-144*x)*log(5)+96*x^2+288*x+21
6),x, algorithm="maxima")

[Out]

-1/24*e^(2*x)/(x*(log(5) - 2) - 3) + 1/12*log(5)/((log(5)^2 - 4*log(5) + 4)*x - 3*log(5) + 6) - 1/6/((log(5)^2
- 4*log(5) + 4)*x - 3*log(5) + 6)

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mupad [B]  time = 0.26, size = 20, normalized size = 0.74 \begin {gather*} -\frac {\frac {{\mathrm {e}}^{2\,x}}{24}-\frac {1}{12}}{x\,\left (\ln \relax (5)-2\right )-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(4*x - log(5)*(2*x - 1) + 4) - 2*log(5) + 4)/(288*x + 24*x^2*log(5)^2 - log(5)*(144*x + 96*x^2)
+ 96*x^2 + 216),x)

[Out]

-(exp(2*x)/24 - 1/12)/(x*(log(5) - 2) - 3)

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sympy [B]  time = 0.39, size = 44, normalized size = 1.63 \begin {gather*} - \frac {2 - \log {\relax (5 )}}{x \left (- 48 \log {\relax (5 )} + 12 \log {\relax (5 )}^{2} + 48\right ) - 36 \log {\relax (5 )} + 72} - \frac {e^{2 x}}{- 48 x + 24 x \log {\relax (5 )} - 72} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1-2*x)*ln(5)+4*x+4)*exp(2*x)-2*ln(5)+4)/(24*x**2*ln(5)**2+(-96*x**2-144*x)*ln(5)+96*x**2+288*x+21
6),x)

[Out]

-(2 - log(5))/(x*(-48*log(5) + 12*log(5)**2 + 48) - 36*log(5) + 72) - exp(2*x)/(-48*x + 24*x*log(5) - 72)

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