\(\int \frac {150-50 x+(150-21 x) \log (\frac {50-7 x}{2 x})}{-50 x^2+7 x^3+(300 x-142 x^2+14 x^3) \log (\frac {50-7 x}{2 x})+(-450+363 x-92 x^2+7 x^3) \log ^2(\frac {50-7 x}{2 x})} \, dx\) [5050]

Optimal result

Integrand size = 97, antiderivative size = 30 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=-2+\frac {x}{x-(3-x) \log \left (-3+\frac {50-x}{2 x}\right )} \]

[Out]

x/(x-ln(1/2*(-x+50)/x-3)*(-x+3))-2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6820, 6843, 32} \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=-\frac {1}{\frac {x}{(x-3) \log \left (\frac {25}{x}-\frac {7}{2}\right )}+1} \]

[In]

Int[(150 - 50*x + (150 - 21*x)*Log[(50 - 7*x)/(2*x)])/(-50*x^2 + 7*x^3 + (300*x - 142*x^2 + 14*x^3)*Log[(50 -
7*x)/(2*x)] + (-450 + 363*x - 92*x^2 + 7*x^3)*Log[(50 - 7*x)/(2*x)]^2),x]

[Out]

-(1 + x/((-3 + x)*Log[-7/2 + 25/x]))^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {50 (-3+x)+3 (-50+7 x) \log \left (-\frac {7}{2}+\frac {25}{x}\right )}{(50-7 x) \left (x+(-3+x) \log \left (-\frac {7}{2}+\frac {25}{x}\right )\right )^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{(-3+x) \log \left (-\frac {7}{2}+\frac {25}{x}\right )}\right ) \\ & = -\frac {1}{1+\frac {x}{(-3+x) \log \left (-\frac {7}{2}+\frac {25}{x}\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=\frac {x}{x+(-3+x) \log \left (-\frac {7}{2}+\frac {25}{x}\right )} \]

[In]

Integrate[(150 - 50*x + (150 - 21*x)*Log[(50 - 7*x)/(2*x)])/(-50*x^2 + 7*x^3 + (300*x - 142*x^2 + 14*x^3)*Log[
(50 - 7*x)/(2*x)] + (-450 + 363*x - 92*x^2 + 7*x^3)*Log[(50 - 7*x)/(2*x)]^2),x]

[Out]

x/(x + (-3 + x)*Log[-7/2 + 25/x])

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
norman	\(\frac {x}{\ln \left (\frac {-7 x +50}{2 x}\right ) x +x -3 \ln \left (\frac {-7 x +50}{2 x}\right )}\)	\(33\)
risch	\(\frac {x}{\ln \left (\frac {-7 x +50}{2 x}\right ) x +x -3 \ln \left (\frac {-7 x +50}{2 x}\right )}\)	\(33\)
parallelrisch	\(\frac {x}{\ln \left (-\frac {7 x -50}{2 x}\right ) x -3 \ln \left (-\frac {7 x -50}{2 x}\right )+x}\)	\(33\)
derivativedivides	\(-\frac {50}{6 \left (-\frac {7}{2}+\frac {25}{x}\right ) \ln \left (-\frac {7}{2}+\frac {25}{x}\right )-29 \ln \left (-\frac {7}{2}+\frac {25}{x}\right )-50}\)	\(34\)
default	\(-\frac {50}{6 \left (-\frac {7}{2}+\frac {25}{x}\right ) \ln \left (-\frac {7}{2}+\frac {25}{x}\right )-29 \ln \left (-\frac {7}{2}+\frac {25}{x}\right )-50}\)	\(34\)

[In]

int(((-21*x+150)*ln(1/2*(-7*x+50)/x)-50*x+150)/((7*x^3-92*x^2+363*x-450)*ln(1/2*(-7*x+50)/x)^2+(14*x^3-142*x^2
+300*x)*ln(1/2*(-7*x+50)/x)+7*x^3-50*x^2),x,method=_RETURNVERBOSE)

[Out]

x/(ln(1/2*(-7*x+50)/x)*x+x-3*ln(1/2*(-7*x+50)/x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=\frac {x}{{\left (x - 3\right )} \log \left (-\frac {7 \, x - 50}{2 \, x}\right ) + x} \]

[In]

integrate(((-21*x+150)*log(1/2*(-7*x+50)/x)-50*x+150)/((7*x^3-92*x^2+363*x-450)*log(1/2*(-7*x+50)/x)^2+(14*x^3
-142*x^2+300*x)*log(1/2*(-7*x+50)/x)+7*x^3-50*x^2),x, algorithm="fricas")

[Out]

x/((x - 3)*log(-1/2*(7*x - 50)/x) + x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=\frac {x}{x + \left (x - 3\right ) \log {\left (\frac {25 - \frac {7 x}{2}}{x} \right )}} \]

[In]

integrate(((-21*x+150)*ln(1/2*(-7*x+50)/x)-50*x+150)/((7*x**3-92*x**2+363*x-450)*ln(1/2*(-7*x+50)/x)**2+(14*x*
*3-142*x**2+300*x)*ln(1/2*(-7*x+50)/x)+7*x**3-50*x**2),x)

[Out]

x/(x + (x - 3)*log((25 - 7*x/2)/x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=-\frac {x}{x {\left (\log \left (2\right ) - 1\right )} + {\left (x - 3\right )} \log \left (x\right ) - {\left (x - 3\right )} \log \left (-7 \, x + 50\right ) - 3 \, \log \left (2\right )} \]

[In]

integrate(((-21*x+150)*log(1/2*(-7*x+50)/x)-50*x+150)/((7*x^3-92*x^2+363*x-450)*log(1/2*(-7*x+50)/x)^2+(14*x^3
-142*x^2+300*x)*log(1/2*(-7*x+50)/x)+7*x^3-50*x^2),x, algorithm="maxima")

[Out]

-x/(x*(log(2) - 1) + (x - 3)*log(x) - (x - 3)*log(-7*x + 50) - 3*log(2))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=\frac {50}{\frac {3 \, {\left (7 \, x - 50\right )} \log \left (-\frac {7 \, x - 50}{2 \, x}\right )}{x} + 29 \, \log \left (-\frac {7 \, x - 50}{2 \, x}\right ) + 50} \]

[In]

integrate(((-21*x+150)*log(1/2*(-7*x+50)/x)-50*x+150)/((7*x^3-92*x^2+363*x-450)*log(1/2*(-7*x+50)/x)^2+(14*x^3
-142*x^2+300*x)*log(1/2*(-7*x+50)/x)+7*x^3-50*x^2),x, algorithm="giac")

[Out]

50/(3*(7*x - 50)*log(-1/2*(7*x - 50)/x)/x + 29*log(-1/2*(7*x - 50)/x) + 50)

Mupad [B] (verification not implemented)

Time = 12.72 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {150-50 x+(150-21 x) \log \left (\frac {50-7 x}{2 x}\right )}{-50 x^2+7 x^3+\left (300 x-142 x^2+14 x^3\right ) \log \left (\frac {50-7 x}{2 x}\right )+\left (-450+363 x-92 x^2+7 x^3\right ) \log ^2\left (\frac {50-7 x}{2 x}\right )} \, dx=\frac {x}{x+\ln \left (-\frac {7\,x-50}{2\,x}\right )\,\left (x-3\right )} \]

[In]

int(-(50*x + log(-((7*x)/2 - 25)/x)*(21*x - 150) - 150)/(log(-((7*x)/2 - 25)/x)^2*(363*x - 92*x^2 + 7*x^3 - 45
0) - 50*x^2 + 7*x^3 + log(-((7*x)/2 - 25)/x)*(300*x - 142*x^2 + 14*x^3)),x)

[Out]

x/(x + log(-(7*x - 50)/(2*x))*(x - 3))