3.44.27 \(\int \frac {6 x+3 x^2+2 x \log (\frac {1+2 e^3+e^6}{4 e^6})}{9+18 x+9 x^2+(6+6 x) \log (\frac {1+2 e^3+e^6}{4 e^6})+\log ^2(\frac {1+2 e^3+e^6}{4 e^6})} \, dx\)

Optimal. Leaf size=23 \[ \frac {x^2}{3+3 x+\log \left (\frac {1}{4} \left (1+\frac {1}{e^3}\right )^2\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 5, number of rules used = 5, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6, 1593, 1983, 27, 74} \begin {gather*} -\frac {\left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )^2}{9 \left (-3 x+3-2 \log \left (1+e^3\right )+\log (4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*x + 3*x^2 + 2*x*Log[(1 + 2*E^3 + E^6)/(4*E^6)])/(9 + 18*x + 9*x^2 + (6 + 6*x)*Log[(1 + 2*E^3 + E^6)/(4*
E^6)] + Log[(1 + 2*E^3 + E^6)/(4*E^6)]^2),x]

[Out]

-1/9*(3*x - 2*(3 + Log[4] - 2*Log[1 + E^3]))^2/(3 - 3*x + Log[4] - 2*Log[1 + E^3])

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 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1983

Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n*ExpandToSum[w,
x]^p, x] /; FreeQ[{m, n, p}, x] && LinearQ[{u, v}, x] && QuadraticQ[w, x] &&  !(LinearMatchQ[{u, v}, x] && Qua
draticMatchQ[w, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x^2+x \left (6+2 \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx\\ &=\int \frac {x \left (6+3 x+2 \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx\\ &=\int \frac {x \left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )}{9 x^2-6 x \left (3+\log (4)-2 \log \left (1+e^3\right )\right )+\left (3+\log (4)-2 \log \left (1+e^3\right )\right )^2} \, dx\\ &=\int \frac {x \left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )}{\left (-3+3 x-\log (4)+2 \log \left (1+e^3\right )\right )^2} \, dx\\ &=-\frac {\left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )^2}{9 \left (3-3 x+\log (4)-2 \log \left (1+e^3\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 54, normalized size = 2.35 \begin {gather*} \frac {1}{9} \left (-3+3 x-\log (4)-\frac {\left (3+\log (4)-2 \log \left (1+e^3\right )\right )^2}{3-3 x+\log (4)-2 \log \left (1+e^3\right )}+2 \log \left (1+e^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*x + 3*x^2 + 2*x*Log[(1 + 2*E^3 + E^6)/(4*E^6)])/(9 + 18*x + 9*x^2 + (6 + 6*x)*Log[(1 + 2*E^3 + E^
6)/(4*E^6)] + Log[(1 + 2*E^3 + E^6)/(4*E^6)]^2),x]

[Out]

(-3 + 3*x - Log[4] - (3 + Log[4] - 2*Log[1 + E^3])^2/(3 - 3*x + Log[4] - 2*Log[1 + E^3]) + 2*Log[1 + E^3])/9

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fricas [B]  time = 0.63, size = 65, normalized size = 2.83 \begin {gather*} \frac {9 \, x^{2} + 3 \, {\left (x + 2\right )} \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 9 \, x + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(
6*x+6)*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+9*x^2+18*x+9),x, algorithm="fricas")

[Out]

1/9*(9*x^2 + 3*(x + 2)*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 9*x + 9)/(3*x
+ log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 3)

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giac [B]  time = 0.12, size = 58, normalized size = 2.52 \begin {gather*} \frac {1}{3} \, x + \frac {\log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 6 \, \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(
6*x+6)*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+9*x^2+18*x+9),x, algorithm="giac")

[Out]

1/3*x + 1/9*(log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 6*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 9)/(3*x + log(1/4*(e^
6 + 2*e^3 + 1)*e^(-6)) + 3)

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maple [A]  time = 0.14, size = 25, normalized size = 1.09

method	result	size
norman	\(\frac {x^{2}}{\ln \left (\frac {{\mathrm e}^{6}}{4}+\frac {{\mathrm e}^{3}}{2}+\frac {1}{4}\right )+3 x -3}\)	\(25\)
gosper	\(\frac {x^{2}}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3 x +3}\)	\(29\)
default	\(\frac {x}{3}-\frac {-\frac {\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )^{2}}{9}-\frac {2 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )}{3}-1}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3 x +3}\)	\(61\)
risch	\(\frac {x}{3}+\frac {2 \ln \left ({\mathrm e}^{3}+1\right )^{2}}{9 \left (-\ln \relax (2)+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}-\frac {4 \ln \relax (2) \ln \left ({\mathrm e}^{3}+1\right )}{9 \left (-\ln \relax (2)+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {2 \ln \relax (2)^{2}}{9 \left (-\ln \relax (2)+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{3}+1\right )}{3 \left (-\ln \relax (2)+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {2 \ln \relax (2)}{3 \left (-\ln \relax (2)+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {1}{-2 \ln \relax (2)+2 \ln \left ({\mathrm e}^{3}+1\right )-3+3 x}\)	\(138\)
meijerg	\(\frac {\left (2 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+6\right ) \left (-\frac {3 x}{\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right ) \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )}+\ln \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )\right )}{9}+\frac {\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right )^{2} \left (\frac {x \left (\frac {9 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}+6\right )}{\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right ) \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )}-2 \ln \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )\right )}{9 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+27}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(ln(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(6*x+6)*l
n(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+9*x^2+18*x+9),x,method=_RETURNVERBOSE)

[Out]

x^2/(ln(1/4*exp(3)^2+1/2*exp(3)+1/4)+3*x-3)

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maxima [B]  time = 0.35, size = 58, normalized size = 2.52 \begin {gather*} \frac {1}{3} \, x + \frac {\log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 6 \, \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(
6*x+6)*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+9*x^2+18*x+9),x, algorithm="maxima")

[Out]

1/3*x + 1/9*(log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 6*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 9)/(3*x + log(1/4*(e^
6 + 2*e^3 + 1)*e^(-6)) + 3)

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mupad [B]  time = 3.95, size = 196, normalized size = 8.52 \begin {gather*} \frac {x}{3}-\frac {2\,\mathrm {atan}\left (\frac {\frac {\left (\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )-18\right )\,\left (\ln \left (\frac {4096}{{\left ({\mathrm {e}}^3+1\right )}^{12}}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+9\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}}+\frac {x\,\left (\ln \left (\frac {105312291668557186697918027683670432318895095400549111254310977536}{{\left ({\mathrm {e}}^3+1\right )}^{216}}\right )+72\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+162\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}}}{\ln \left (\frac {16}{{\left ({\mathrm {e}}^3+1\right )}^4}\right )+\frac {4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2}{3}+3}\right )\,\left (\ln \left (\frac {4096}{{\left ({\mathrm {e}}^3+1\right )}^{12}}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+9\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + 3*x^2 + 2*x*log(exp(-6)*(exp(3)/2 + exp(6)/4 + 1/4)))/(18*x + log(exp(-6)*(exp(3)/2 + exp(6)/4 + 1/
4))^2 + log(exp(-6)*(exp(3)/2 + exp(6)/4 + 1/4))*(6*x + 6) + 9*x^2 + 9),x)

[Out]

x/3 - (2*atan((((log((exp(3) + 1)^12/4096) - 18)*(log(4096/(exp(3) + 1)^12) + 4*log(exp(3)/2 + 1/2)^2 + 9))/(3
*(144*log(exp(3)/2 + 1/2)^2 - log((exp(3) + 1)^12/4096)^2)^(1/2)) + (x*(log(1053122916685571866979180276836704
32318895095400549111254310977536/(exp(3) + 1)^216) + 72*log(exp(3)/2 + 1/2)^2 + 162))/(3*(144*log(exp(3)/2 + 1
/2)^2 - log((exp(3) + 1)^12/4096)^2)^(1/2)))/(log(16/(exp(3) + 1)^4) + (4*log(exp(3)/2 + 1/2)^2)/3 + 3))*(log(
4096/(exp(3) + 1)^12) + 4*log(exp(3)/2 + 1/2)^2 + 9))/(3*(144*log(exp(3)/2 + 1/2)^2 - log((exp(3) + 1)^12/4096
)^2)^(1/2))

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sympy [B]  time = 0.63, size = 83, normalized size = 3.61 \begin {gather*} \frac {x}{3} + \frac {- 6 \log {\left (1 + 2 e^{3} + e^{6} \right )} - 4 \log {\relax (2 )} \log {\left (1 + 2 e^{3} + e^{6} \right )} + 4 \log {\relax (2 )}^{2} + 12 \log {\relax (2 )} + 9 + \log {\left (1 + 2 e^{3} + e^{6} \right )}^{2}}{27 x - 27 - 18 \log {\relax (2 )} + 9 \log {\left (1 + 2 e^{3} + e^{6} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(1/4*(exp(3)**2+2*exp(3)+1)/exp(3)**2)+3*x**2+6*x)/(ln(1/4*(exp(3)**2+2*exp(3)+1)/exp(3)**2)*
*2+(6*x+6)*ln(1/4*(exp(3)**2+2*exp(3)+1)/exp(3)**2)+9*x**2+18*x+9),x)

[Out]

x/3 + (-6*log(1 + 2*exp(3) + exp(6)) - 4*log(2)*log(1 + 2*exp(3) + exp(6)) + 4*log(2)**2 + 12*log(2) + 9 + log
(1 + 2*exp(3) + exp(6))**2)/(27*x - 27 - 18*log(2) + 9*log(1 + 2*exp(3) + exp(6)))

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