3.8.20 \(\int \frac {-4-4 x+(-2 x^2+e^3 (1+x)) \log (400)}{-4 x+(e^3 x-2 x^2) \log (400)} \, dx\)

Optimal. Leaf size=23 \[ -3+x+\log \left (\frac {x}{-e^3+2 x+\frac {4}{\log (400)}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1984, 1593, 893} \begin {gather*} x+\log (x)-\log \left (2 x \log (400)+4-e^3 \log (400)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 4*x + (-2*x^2 + E^3*(1 + x))*Log[400])/(-4*x + (E^3*x - 2*x^2)*Log[400]),x]

[Out]

x + Log[x] - Log[4 - E^3*Log[400] + 2*x*Log[400]]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 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 {-4+e^3 \log (400)-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )}{-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )} \, dx\\ &=\int \frac {-4+e^3 \log (400)-2 x^2 \log (400)-x \left (4-e^3 \log (400)\right )}{x \left (-4+e^3 \log (400)-2 x \log (400)\right )} \, dx\\ &=\int \left (1+\frac {1}{x}+\frac {2 \log (400)}{-4+e^3 \log (400)-2 x \log (400)}\right ) \, dx\\ &=x+\log (x)-\log \left (4-e^3 \log (400)+2 x \log (400)\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 21, normalized size = 0.91 \begin {gather*} x+\log (x)-\log \left (4-e^3 \log (400)+2 x \log (400)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 4*x + (-2*x^2 + E^3*(1 + x))*Log[400])/(-4*x + (E^3*x - 2*x^2)*Log[400]),x]

[Out]

x + Log[x] - Log[4 - E^3*Log[400] + 2*x*Log[400]]

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 20, normalized size = 0.87 \begin {gather*} x - \log \left ({\left (2 \, x - e^{3}\right )} \log \left (20\right ) + 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x+1)*exp(3)-2*x^2)*log(20)-4*x-4)/(2*(x*exp(3)-2*x^2)*log(20)-4*x),x, algorithm="fricas")

[Out]

x - log((2*x - e^3)*log(20) + 2) + log(x)

________________________________________________________________________________________

giac [A]  time = 0.39, size = 22, normalized size = 0.96 \begin {gather*} x - \log \left ({\left | 2 \, x \log \left (20\right ) - e^{3} \log \left (20\right ) + 2 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x+1)*exp(3)-2*x^2)*log(20)-4*x-4)/(2*(x*exp(3)-2*x^2)*log(20)-4*x),x, algorithm="giac")

[Out]

x - log(abs(2*x*log(20) - e^3*log(20) + 2)) + log(abs(x))

________________________________________________________________________________________

maple [A]  time = 0.28, size = 20, normalized size = 0.87

method	result	size
norman	\(x -\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2 x \ln \left (20\right )-2\right )+\ln \relax (x )\)	\(20\)
default	\(x -\ln \left (-\ln \left (20\right ) {\mathrm e}^{3}+2 x \ln \left (20\right )+2\right )+\ln \relax (x )\)	\(21\)
risch	\(x -\ln \left (\left (4 \ln \relax (2)+2 \ln \relax (5)\right ) x -{\mathrm e}^{3} \ln \relax (5)-2 \,{\mathrm e}^{3} \ln \relax (2)+2\right )+\ln \left (-x \right )\)	\(35\)
meijerg	\(-\frac {\left (\ln \left (20\right ) {\mathrm e}^{3}-2\right ) \ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )}{2 \ln \left (20\right )}-\frac {\left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )^{2} \left (-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}-\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )\right )}{\left (2 \ln \left (20\right ) {\mathrm e}^{3}-4\right ) \ln \left (20\right )}+\frac {2 \ln \left (20\right ) {\mathrm e}^{3} \left (-\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )+\ln \relax (x )+\ln \relax (2)+\ln \left (\ln \left (20\right )\right )-\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )+i \pi \right )}{2 \ln \left (20\right ) {\mathrm e}^{3}-4}-\frac {4 \left (-\ln \left (1-\frac {2 x \ln \left (20\right )}{\ln \left (20\right ) {\mathrm e}^{3}-2}\right )+\ln \relax (x )+\ln \relax (2)+\ln \left (\ln \left (20\right )\right )-\ln \left (\ln \left (20\right ) {\mathrm e}^{3}-2\right )+i \pi \right )}{2 \ln \left (20\right ) {\mathrm e}^{3}-4}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*((x+1)*exp(3)-2*x^2)*ln(20)-4*x-4)/(2*(x*exp(3)-2*x^2)*ln(20)-4*x),x,method=_RETURNVERBOSE)

[Out]

x-ln(ln(20)*exp(3)-2*x*ln(20)-2)+ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.68, size = 20, normalized size = 0.87 \begin {gather*} x - \log \left (2 \, x \log \left (20\right ) - e^{3} \log \left (20\right ) + 2\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x+1)*exp(3)-2*x^2)*log(20)-4*x-4)/(2*(x*exp(3)-2*x^2)*log(20)-4*x),x, algorithm="maxima")

[Out]

x - log(2*x*log(20) - e^3*log(20) + 2) + log(x)

________________________________________________________________________________________

mupad [B]  time = 0.18, size = 33, normalized size = 1.43 \begin {gather*} x-\mathrm {atan}\left (\frac {-{\mathrm {e}}^3\,\ln \left (20\right )\,2{}\mathrm {i}+x\,\ln \left (20\right )\,8{}\mathrm {i}+4{}\mathrm {i}}{2\,{\mathrm {e}}^3\,\ln \left (20\right )-4}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - 2*log(20)*(exp(3)*(x + 1) - 2*x^2) + 4)/(4*x - 2*log(20)*(x*exp(3) - 2*x^2)),x)

[Out]

x - atan((x*log(20)*8i - exp(3)*log(20)*2i + 4i)/(2*exp(3)*log(20) - 4))*2i

________________________________________________________________________________________

sympy [A]  time = 0.47, size = 22, normalized size = 0.96 \begin {gather*} x + \log {\relax (x )} - \log {\left (x + \frac {- 2 e^{3} \log {\left (20 \right )} + 4}{4 \log {\left (20 \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x+1)*exp(3)-2*x**2)*ln(20)-4*x-4)/(2*(x*exp(3)-2*x**2)*ln(20)-4*x),x)

[Out]

x + log(x) - log(x + (-2*exp(3)*log(20) + 4)/(4*log(20)))

________________________________________________________________________________________