3.16.24 \(\int \frac {16 x-8 x^2+(16 x-8 x^2+4 x^3-3 x^4) \log (7)}{(4-8 x+4 x^2) \log (7)} \, dx\)

Optimal. Leaf size=25 \[ \frac {2 x \left (x+\frac {x^3}{8}+\frac {x}{\log (7)}\right )}{1-x} \]

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Rubi [B]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 5, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 27, 1850} \begin {gather*} -\frac {x^3}{4}-\frac {x^2}{4}-\frac {x (8+9 \log (7))}{4 \log (7)}+\frac {8+9 \log (7)}{4 (1-x) \log (7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x - 8*x^2 + (16*x - 8*x^2 + 4*x^3 - 3*x^4)*Log[7])/((4 - 8*x + 4*x^2)*Log[7]),x]

[Out]

-1/4*x^2 - x^3/4 + (8 + 9*Log[7])/(4*(1 - x)*Log[7]) - (x*(8 + 9*Log[7]))/(4*Log[7])

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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{4-8 x+4 x^2} \, dx}{\log (7)}\\ &=\frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{4 (-1+x)^2} \, dx}{\log (7)}\\ &=\frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{(-1+x)^2} \, dx}{4 \log (7)}\\ &=\frac {\int \left (-2 x \log (7)-3 x^2 \log (7)-8 \left (1+\frac {9 \log (7)}{8}\right )+\frac {8+9 \log (7)}{(-1+x)^2}\right ) \, dx}{4 \log (7)}\\ &=-\frac {x^2}{4}-\frac {x^3}{4}+\frac {8+9 \log (7)}{4 (1-x) \log (7)}-\frac {x (8+9 \log (7))}{4 \log (7)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 2.00 \begin {gather*} \frac {\frac {-8-9 \log (7)}{-1+x}-4 (-1+x)^2 \log (7)-(-1+x)^3 \log (7)-2 (-1+x) (4+7 \log (7))}{4 \log (7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x - 8*x^2 + (16*x - 8*x^2 + 4*x^3 - 3*x^4)*Log[7])/((4 - 8*x + 4*x^2)*Log[7]),x]

[Out]

((-8 - 9*Log[7])/(-1 + x) - 4*(-1 + x)^2*Log[7] - (-1 + x)^3*Log[7] - 2*(-1 + x)*(4 + 7*Log[7]))/(4*Log[7])

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fricas [A]  time = 0.62, size = 37, normalized size = 1.48 \begin {gather*} -\frac {8 \, x^{2} + {\left (x^{4} + 8 \, x^{2} - 9 \, x + 9\right )} \log \relax (7) - 8 \, x + 8}{4 \, {\left (x - 1\right )} \log \relax (7)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+4*x^3-8*x^2+16*x)*log(7)-8*x^2+16*x)/(4*x^2-8*x+4)/log(7),x, algorithm="fricas")

[Out]

-1/4*(8*x^2 + (x^4 + 8*x^2 - 9*x + 9)*log(7) - 8*x + 8)/((x - 1)*log(7))

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giac [A]  time = 0.24, size = 39, normalized size = 1.56 \begin {gather*} -\frac {x^{3} \log \relax (7) + x^{2} \log \relax (7) + 9 \, x \log \relax (7) + 8 \, x + \frac {9 \, \log \relax (7) + 8}{x - 1}}{4 \, \log \relax (7)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+4*x^3-8*x^2+16*x)*log(7)-8*x^2+16*x)/(4*x^2-8*x+4)/log(7),x, algorithm="giac")

[Out]

-1/4*(x^3*log(7) + x^2*log(7) + 9*x*log(7) + 8*x + (9*log(7) + 8)/(x - 1))/log(7)

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maple [A]  time = 0.79, size = 26, normalized size = 1.04

method	result	size
norman	\(\frac {-\frac {x^{4}}{4}-\frac {2 \left (\ln \relax (7)+1\right ) x^{2}}{\ln \relax (7)}}{x -1}\)	\(26\)
gosper	\(-\frac {x^{2} \left (x^{2} \ln \relax (7)+8 \ln \relax (7)+8\right )}{4 \ln \relax (7) \left (x -1\right )}\)	\(27\)
risch	\(-\frac {x^{3}}{4}-\frac {x^{2}}{4}-\frac {9 x}{4}-\frac {2 x}{\ln \relax (7)}-\frac {9}{4 \left (x -1\right )}-\frac {2}{\ln \relax (7) \left (x -1\right )}\)	\(40\)
default	\(\frac {-\ln \relax (7) x^{3}-x^{2} \ln \relax (7)-9 x \ln \relax (7)-8 x -\frac {9 \ln \relax (7)+8}{x -1}}{4 \ln \relax (7)}\)	\(43\)
meijerg	\(-\frac {x \left (-5 x^{3}-10 x^{2}-30 x +60\right )}{20 \left (1-x \right )}+\frac {x \left (-2 x^{2}-6 x +12\right )}{-4 x +4}-\frac {\left (-2 \ln \relax (7)-2\right ) \left (-\frac {x \left (-3 x +6\right )}{3 \left (1-x \right )}-2 \ln \left (1-x \right )\right )}{\ln \relax (7)}+\frac {\left (4 \ln \relax (7)+4\right ) \left (\frac {x}{1-x}+\ln \left (1-x \right )\right )}{\ln \relax (7)}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^4+4*x^3-8*x^2+16*x)*ln(7)-8*x^2+16*x)/(4*x^2-8*x+4)/ln(7),x,method=_RETURNVERBOSE)

[Out]

(-1/4*x^4-2*(ln(7)+1)/ln(7)*x^2)/(x-1)

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maxima [A]  time = 0.53, size = 39, normalized size = 1.56 \begin {gather*} -\frac {x^{3} \log \relax (7) + x^{2} \log \relax (7) + x {\left (9 \, \log \relax (7) + 8\right )} + \frac {9 \, \log \relax (7) + 8}{x - 1}}{4 \, \log \relax (7)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^4+4*x^3-8*x^2+16*x)*log(7)-8*x^2+16*x)/(4*x^2-8*x+4)/log(7),x, algorithm="maxima")

[Out]

-1/4*(x^3*log(7) + x^2*log(7) + x*(9*log(7) + 8) + (9*log(7) + 8)/(x - 1))/log(7)

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mupad [B]  time = 1.04, size = 47, normalized size = 1.88 \begin {gather*} \frac {9\,\ln \relax (7)+8}{4\,\ln \relax (7)-4\,x\,\ln \relax (7)}-x\,\left (\frac {8\,\ln \relax (7)+8}{4\,\ln \relax (7)}+\frac {1}{4}\right )-\frac {x^2}{4}-\frac {x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + log(7)*(16*x - 8*x^2 + 4*x^3 - 3*x^4) - 8*x^2)/(log(7)*(4*x^2 - 8*x + 4)),x)

[Out]

(9*log(7) + 8)/(4*log(7) - 4*x*log(7)) - x*((8*log(7) + 8)/(4*log(7)) + 1/4) - x^2/4 - x^3/4

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sympy [B]  time = 0.23, size = 39, normalized size = 1.56 \begin {gather*} - \frac {x^{3}}{4} - \frac {x^{2}}{4} - x \left (\frac {2}{\log {\relax (7 )}} + \frac {9}{4}\right ) - \frac {8 + 9 \log {\relax (7 )}}{4 x \log {\relax (7 )} - 4 \log {\relax (7 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**4+4*x**3-8*x**2+16*x)*ln(7)-8*x**2+16*x)/(4*x**2-8*x+4)/ln(7),x)

[Out]

-x**3/4 - x**2/4 - x*(2/log(7) + 9/4) - (8 + 9*log(7))/(4*x*log(7) - 4*log(7))

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