3.46.8 \(\int \frac {3-3 x+x^2+(-3+2 x) \log (49)+\log ^2(49)}{-3 x+x^2+(-3+2 x) \log (49)+\log ^2(49)} \, dx\)

Optimal. Leaf size=19 \[ 2+e^{10}+x+\log \left (\frac {2}{3}-\frac {2}{x+\log (49)}\right ) \]

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Rubi [B]  time = 0.09, antiderivative size = 47, normalized size of antiderivative = 2.47, number of steps used = 5, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1984, 1657, 618, 206} \begin {gather*} x+\frac {6 \tanh ^{-1}\left (\frac {-2 x+3-\log (2401)}{\sqrt {9+\log ^2(2401)-\log (49) \log (5764801)}}\right )}{\sqrt {9+\log ^2(2401)-\log (49) \log (5764801)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 3*x + x^2 + (-3 + 2*x)*Log[49] + Log[49]^2)/(-3*x + x^2 + (-3 + 2*x)*Log[49] + Log[49]^2),x]

[Out]

x + (6*ArcTanh[(3 - 2*x - Log[2401])/Sqrt[9 + Log[2401]^2 - Log[49]*Log[5764801]]])/Sqrt[9 + Log[2401]^2 - Log
[49]*Log[5764801]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 {3+x^2-3 \log (49)+\log ^2(49)-x (3-\log (2401))}{x^2-(3-\log (49)) \log (49)+x (-3+\log (2401))} \, dx\\ &=\int \left (1+\frac {3}{x^2-(3-\log (49)) \log (49)+x (-3+\log (2401))}\right ) \, dx\\ &=x+3 \int \frac {1}{x^2-(3-\log (49)) \log (49)+x (-3+\log (2401))} \, dx\\ &=x-6 \operatorname {Subst}\left (\int \frac {1}{9-x^2+\log ^2(2401)-\log (49) \log (5764801)} \, dx,x,-3+2 x+\log (2401)\right )\\ &=x+\frac {6 \tanh ^{-1}\left (\frac {3-2 x-\log (2401)}{\sqrt {9+\log ^2(2401)-\log (49) \log (5764801)}}\right )}{\sqrt {9+\log ^2(2401)-\log (49) \log (5764801)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} x+\log (3-x-\log (49))-\log (x+\log (49)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 3*x + x^2 + (-3 + 2*x)*Log[49] + Log[49]^2)/(-3*x + x^2 + (-3 + 2*x)*Log[49] + Log[49]^2),x]

[Out]

x + Log[3 - x - Log[49]] - Log[x + Log[49]]

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fricas [A]  time = 0.49, size = 19, normalized size = 1.00 \begin {gather*} x - \log \left (x + 2 \, \log \relax (7)\right ) + \log \left (x + 2 \, \log \relax (7) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x+3)/(4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x),x, algorithm="fricas")

[Out]

x - log(x + 2*log(7)) + log(x + 2*log(7) - 3)

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giac [A]  time = 0.12, size = 21, normalized size = 1.11 \begin {gather*} x - \log \left ({\left | x + 2 \, \log \relax (7) \right |}\right ) + \log \left ({\left | x + 2 \, \log \relax (7) - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x+3)/(4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x),x, algorithm="giac")

[Out]

x - log(abs(x + 2*log(7))) + log(abs(x + 2*log(7) - 3))

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maple [A]  time = 0.19, size = 20, normalized size = 1.05

method	result	size
default	\(x +\ln \left (2 \ln \relax (7)+x -3\right )-\ln \left (x +2 \ln \relax (7)\right )\)	\(20\)
norman	\(x +\ln \left (2 \ln \relax (7)+x -3\right )-\ln \left (x +2 \ln \relax (7)\right )\)	\(20\)
risch	\(x +\ln \left (2 \ln \relax (7)+x -3\right )-\ln \left (x +2 \ln \relax (7)\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(7)^2+2*(2*x-3)*ln(7)+x^2-3*x+3)/(4*ln(7)^2+2*(2*x-3)*ln(7)+x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

x+ln(2*ln(7)+x-3)-ln(x+2*ln(7))

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maxima [A]  time = 0.35, size = 19, normalized size = 1.00 \begin {gather*} x - \log \left (x + 2 \, \log \relax (7)\right ) + \log \left (x + 2 \, \log \relax (7) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x+3)/(4*log(7)^2+2*(2*x-3)*log(7)+x^2-3*x),x, algorithm="maxima")

[Out]

x - log(x + 2*log(7)) + log(x + 2*log(7) - 3)

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mupad [B]  time = 0.09, size = 14, normalized size = 0.74 \begin {gather*} x-2\,\mathrm {atanh}\left (\frac {2\,x}{3}+\frac {4\,\ln \relax (7)}{3}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(7)*(2*x - 3) - 3*x + 4*log(7)^2 + x^2 + 3)/(2*log(7)*(2*x - 3) - 3*x + 4*log(7)^2 + x^2),x)

[Out]

x - 2*atanh((2*x)/3 + (4*log(7))/3 - 1)

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sympy [A]  time = 0.16, size = 19, normalized size = 1.00 \begin {gather*} x - \log {\left (x + 2 \log {\relax (7 )} \right )} + \log {\left (x - 3 + 2 \log {\relax (7 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(7)**2+2*(2*x-3)*ln(7)+x**2-3*x+3)/(4*ln(7)**2+2*(2*x-3)*ln(7)+x**2-3*x),x)

[Out]

x - log(x + 2*log(7)) + log(x - 3 + 2*log(7))

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