3.58.75 \(\int \frac {e (14-7 x)-21 x+56 x^2-21 x^3+(-14+7 x) \log (4-2 x)}{(-2 x^2+3 x^3-x^4+e (2 x-x^2)+(-2 x+x^2) \log (4-2 x)) \log ^2(e x-x^2+x^3-x \log (4-2 x))} \, dx\)

Optimal. Leaf size=23 \[ -\frac {7}{\log \left (x \left (e-x+x^2-\log (4-2 x)\right )\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 1, number of rules used = 1, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6686} \begin {gather*} -\frac {7}{\log \left (x^3-x^2+e x-x \log (4-2 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E*(14 - 7*x) - 21*x + 56*x^2 - 21*x^3 + (-14 + 7*x)*Log[4 - 2*x])/((-2*x^2 + 3*x^3 - x^4 + E*(2*x - x^2)
+ (-2*x + x^2)*Log[4 - 2*x])*Log[E*x - x^2 + x^3 - x*Log[4 - 2*x]]^2),x]

[Out]

-7/Log[E*x - x^2 + x^3 - x*Log[4 - 2*x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {7}{\log \left (e x-x^2+x^3-x \log (4-2 x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.96 \begin {gather*} -\frac {7}{\log (x (e+(-1+x) x-\log (4-2 x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(14 - 7*x) - 21*x + 56*x^2 - 21*x^3 + (-14 + 7*x)*Log[4 - 2*x])/((-2*x^2 + 3*x^3 - x^4 + E*(2*x -
x^2) + (-2*x + x^2)*Log[4 - 2*x])*Log[E*x - x^2 + x^3 - x*Log[4 - 2*x]]^2),x]

[Out]

-7/Log[x*(E + (-1 + x)*x - Log[4 - 2*x])]

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fricas [A]  time = 0.64, size = 27, normalized size = 1.17 \begin {gather*} -\frac {7}{\log \left (x^{3} - x^{2} + x e - x \log \left (-2 \, x + 4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x-14)*log(4-2*x)+(-7*x+14)*exp(1)-21*x^3+56*x^2-21*x)/((x^2-2*x)*log(4-2*x)+(-x^2+2*x)*exp(1)-x^
4+3*x^3-2*x^2)/log(-x*log(4-2*x)+x*exp(1)+x^3-x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 27, normalized size = 1.17 \begin {gather*} -\frac {7}{\log \left (x^{3} - x^{2} + x e - x \log \left (-2 \, x + 4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x-14)*log(4-2*x)+(-7*x+14)*exp(1)-21*x^3+56*x^2-21*x)/((x^2-2*x)*log(4-2*x)+(-x^2+2*x)*exp(1)-x^
4+3*x^3-2*x^2)/log(-x*log(4-2*x)+x*exp(1)+x^3-x^2)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.13, size = 189, normalized size = 8.22

method	result	size
risch	\(\frac {14 i}{\pi \,\mathrm {csgn}\left (i \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right ) \mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right ) \mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right ) \mathrm {csgn}\left (i x \right )-\pi \mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right )^{3}+\pi \mathrm {csgn}\left (i x \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )\right )^{2} \mathrm {csgn}\left (i x \right )-2 i \ln \relax (x )-2 i \ln \left (x^{2}+{\mathrm e}-\ln \left (4-2 x \right )-x \right )}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((7*x-14)*ln(4-2*x)+(-7*x+14)*exp(1)-21*x^3+56*x^2-21*x)/((x^2-2*x)*ln(4-2*x)+(-x^2+2*x)*exp(1)-x^4+3*x^3-
2*x^2)/ln(-x*ln(4-2*x)+x*exp(1)+x^3-x^2)^2,x,method=_RETURNVERBOSE)

[Out]

14*I/(Pi*csgn(I*(x^2+exp(1)-ln(4-2*x)-x))*csgn(I*x*(x^2+exp(1)-ln(4-2*x)-x))^2-Pi*csgn(I*(x^2+exp(1)-ln(4-2*x)
-x))*csgn(I*x*(x^2+exp(1)-ln(4-2*x)-x))*csgn(I*x)-Pi*csgn(I*x*(x^2+exp(1)-ln(4-2*x)-x))^3+Pi*csgn(I*x*(x^2+exp
(1)-ln(4-2*x)-x))^2*csgn(I*x)-2*I*ln(x)-2*I*ln(x^2+exp(1)-ln(4-2*x)-x))

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maxima [C]  time = 0.51, size = 30, normalized size = 1.30 \begin {gather*} -\frac {7}{\log \left (-i \, \pi + x^{2} - x + e - \log \relax (2) - \log \left (x - 2\right )\right ) + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x-14)*log(4-2*x)+(-7*x+14)*exp(1)-21*x^3+56*x^2-21*x)/((x^2-2*x)*log(4-2*x)+(-x^2+2*x)*exp(1)-x^
4+3*x^3-2*x^2)/log(-x*log(4-2*x)+x*exp(1)+x^3-x^2)^2,x, algorithm="maxima")

[Out]

-7/(log(-I*pi + x^2 - x + e - log(2) - log(x - 2)) + log(x))

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mupad [B]  time = 5.09, size = 27, normalized size = 1.17 \begin {gather*} -\frac {7}{\ln \left (x\,\mathrm {e}-x\,\ln \left (4-2\,x\right )-x^2+x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((21*x - log(4 - 2*x)*(7*x - 14) - 56*x^2 + 21*x^3 + exp(1)*(7*x - 14))/(log(x*exp(1) - x*log(4 - 2*x) - x^
2 + x^3)^2*(log(4 - 2*x)*(2*x - x^2) - exp(1)*(2*x - x^2) + 2*x^2 - 3*x^3 + x^4)),x)

[Out]

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

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sympy [A]  time = 0.46, size = 24, normalized size = 1.04 \begin {gather*} - \frac {7}{\log {\left (x^{3} - x^{2} - x \log {\left (4 - 2 x \right )} + e x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x-14)*ln(4-2*x)+(-7*x+14)*exp(1)-21*x**3+56*x**2-21*x)/((x**2-2*x)*ln(4-2*x)+(-x**2+2*x)*exp(1)-
x**4+3*x**3-2*x**2)/ln(-x*ln(4-2*x)+x*exp(1)+x**3-x**2)**2,x)

[Out]

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

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