3.6.87 \(\int \frac {1-2 e^3+6 x+3 x^2+(-6-6 x) \log (3)+3 \log ^2(3)}{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} \, dx\)

Optimal. Leaf size=23 \[ 5+3 x+\frac {2 \left (1+e^3\right )}{1+x-\log (3)}+\log (4) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {1984, 27, 683} \begin {gather*} 3 x+\frac {2 \left (1+e^3\right )}{x+1-\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*x + (2*(1 + E^3))/(1 + x - Log[3])

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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 26, normalized size = 1.13 \begin {gather*} \frac {2 \left (1+e^3\right )}{1+x-\log (3)}+3 (1+x-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 \, x^{2} - 3 \, x \log \relax (3) + 3 \, x + 2 \, e^{3} + 2}{x - \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.74, size = 19, normalized size = 0.83 \begin {gather*} 3 \, x + \frac {2 \, {\left (e^{3} + 1\right )}}{x - \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + 2*(e^3 + 1)/(x - log(3) + 1)

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maple [A]  time = 0.38, size = 22, normalized size = 0.96

method	result	size
default	\(3 x -\frac {-2-2 \,{\mathrm e}^{3}}{1+x -\ln \relax (3)}\)	\(22\)
risch	\(3 x -\frac {2}{\ln \relax (3)-x -1}-\frac {2 \,{\mathrm e}^{3}}{\ln \relax (3)-x -1}\)	\(29\)
norman	\(\frac {-3 x^{2}+1+3 \ln \relax (3)^{2}-2 \,{\mathrm e}^{3}-6 \ln \relax (3)}{\ln \relax (3)-x -1}\)	\(32\)
gosper	\(-\frac {3 x^{2}-1-3 \ln \relax (3)^{2}+2 \,{\mathrm e}^{3}+6 \ln \relax (3)}{\ln \relax (3)-x -1}\)	\(33\)
meijerg	\(\left (-6 \ln \relax (3)+6\right ) \left (\frac {x}{\left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}+\ln \left (1-\frac {x}{\ln \relax (3)-1}\right )\right )+\frac {3 \left (\ln \relax (3)-1\right )^{2} \left (-\frac {x \left (-\frac {3 x}{\ln \relax (3)-1}+6\right )}{3 \left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}-2 \ln \left (1-\frac {x}{\ln \relax (3)-1}\right )\right )}{1-\ln \relax (3)}-\frac {3 \ln \relax (3)^{2} x}{\left (1-\ln \relax (3)\right ) \left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}+\frac {2 \,{\mathrm e}^{3} x}{\left (1-\ln \relax (3)\right ) \left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}+\frac {6 \ln \relax (3) x}{\left (1-\ln \relax (3)\right ) \left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}-\frac {x}{\left (1-\ln \relax (3)\right ) \left (\ln \relax (3)-1\right ) \left (1-\frac {x}{\ln \relax (3)-1}\right )}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*ln(3)^2+(-6*x-6)*ln(3)-2*exp(3)+3*x^2+6*x+1)/(ln(3)^2+(-2*x-2)*ln(3)+x^2+2*x+1),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [A]  time = 0.57, size = 19, normalized size = 0.83 \begin {gather*} 3 \, x + \frac {2 \, {\left (e^{3} + 1\right )}}{x - \log \relax (3) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x + 2*(e^3 + 1)/(x - log(3) + 1)

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mupad [B]  time = 0.62, size = 61, normalized size = 2.65 \begin {gather*} 3\,x-\frac {\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}-\ln \relax (9)\,1{}\mathrm {i}+2{}\mathrm {i}}{\sqrt {2\,\ln \relax (3)+\ln \relax (9)}\,\sqrt {\ln \relax (9)-2\,\ln \relax (3)}}\right )\,\left ({\mathrm {e}}^3+1\right )\,4{}\mathrm {i}}{\sqrt {2\,\ln \relax (3)+\ln \relax (9)}\,\sqrt {\ln \relax (9)-2\,\ln \relax (3)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x - (atan((x*2i - log(9)*1i + 2i)/((2*log(3) + log(9))^(1/2)*(log(9) - 2*log(3))^(1/2)))*(exp(3) + 1)*4i)/((
2*log(3) + log(9))^(1/2)*(log(9) - 2*log(3))^(1/2))

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sympy [A]  time = 0.19, size = 15, normalized size = 0.65 \begin {gather*} 3 x + \frac {2 + 2 e^{3}}{x - \log {\relax (3 )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*ln(3)**2+(-6*x-6)*ln(3)-2*exp(3)+3*x**2+6*x+1)/(ln(3)**2+(-2*x-2)*ln(3)+x**2+2*x+1),x)

[Out]

3*x + (2 + 2*exp(3))/(x - log(3) + 1)

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