\(\int \frac {-20 x+(50+20 x) \log (\frac {1}{2} (5+2 x))}{605 x^2+242 x^3+(1100 x+440 x^2) \log (\frac {1}{2} (5+2 x))+(500+200 x) \log ^2(\frac {1}{2} (5+2 x))} \, dx\) [8992]

Optimal result

Integrand size = 72, antiderivative size = 16 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{x+10 \left (x+\log \left (\frac {5}{2}+x\right )\right )} \]

[Out]

x/(11*x+10*ln(5/2+x))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6820, 12, 6843, 32} \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=-\frac {10}{11 \left (\frac {11 x}{\log \left (x+\frac {5}{2}\right )}+10\right )} \]

[In]

Int[(-20*x + (50 + 20*x)*Log[(5 + 2*x)/2])/(605*x^2 + 242*x^3 + (1100*x + 440*x^2)*Log[(5 + 2*x)/2] + (500 + 2
00*x)*Log[(5 + 2*x)/2]^2),x]

[Out]

-10/(11*(10 + (11*x)/Log[5/2 + x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {10 \left (-2 x+(5+2 x) \log \left (\frac {5}{2}+x\right )\right )}{(5+2 x) \left (11 x+10 \log \left (\frac {5}{2}+x\right )\right )^2} \, dx \\ & = 10 \int \frac {-2 x+(5+2 x) \log \left (\frac {5}{2}+x\right )}{(5+2 x) \left (11 x+10 \log \left (\frac {5}{2}+x\right )\right )^2} \, dx \\ & = 10 \text {Subst}\left (\int \frac {1}{(10+11 x)^2} \, dx,x,\frac {x}{\log \left (\frac {5}{2}+x\right )}\right ) \\ & = -\frac {10}{11 \left (10+\frac {11 x}{\log \left (\frac {5}{2}+x\right )}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {10 x}{110 x+100 \log \left (\frac {5}{2}+x\right )} \]

[In]

Integrate[(-20*x + (50 + 20*x)*Log[(5 + 2*x)/2])/(605*x^2 + 242*x^3 + (1100*x + 440*x^2)*Log[(5 + 2*x)/2] + (5
00 + 200*x)*Log[(5 + 2*x)/2]^2),x]

[Out]

(10*x)/(110*x + 100*Log[5/2 + x])

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x}{11 x +10 \ln \left (\frac {5}{2}+x \right )}\)	\(15\)
parallelrisch	\(\frac {x}{11 x +10 \ln \left (\frac {5}{2}+x \right )}\)	\(15\)
derivativedivides	\(\frac {2 x}{22 x +20 \ln \left (\frac {5}{2}+x \right )}\)	\(16\)
default	\(\frac {2 x}{22 x +20 \ln \left (\frac {5}{2}+x \right )}\)	\(16\)
norman	\(-\frac {10 \ln \left (\frac {5}{2}+x \right )}{11 \left (11 x +10 \ln \left (\frac {5}{2}+x \right )\right )}\)	\(19\)

[In]

int(((20*x+50)*ln(5/2+x)-20*x)/((200*x+500)*ln(5/2+x)^2+(440*x^2+1100*x)*ln(5/2+x)+242*x^3+605*x^2),x,method=_
RETURNVERBOSE)

[Out]

x/(11*x+10*ln(5/2+x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{11 \, x + 10 \, \log \left (x + \frac {5}{2}\right )} \]

[In]

integrate(((20*x+50)*log(5/2+x)-20*x)/((200*x+500)*log(5/2+x)^2+(440*x^2+1100*x)*log(5/2+x)+242*x^3+605*x^2),x
, algorithm="fricas")

[Out]

x/(11*x + 10*log(x + 5/2))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{11 x + 10 \log {\left (x + \frac {5}{2} \right )}} \]

[In]

integrate(((20*x+50)*ln(5/2+x)-20*x)/((200*x+500)*ln(5/2+x)**2+(440*x**2+1100*x)*ln(5/2+x)+242*x**3+605*x**2),
x)

[Out]

x/(11*x + 10*log(x + 5/2))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{11 \, x - 10 \, \log \left (2\right ) + 10 \, \log \left (2 \, x + 5\right )} \]

[In]

integrate(((20*x+50)*log(5/2+x)-20*x)/((200*x+500)*log(5/2+x)^2+(440*x^2+1100*x)*log(5/2+x)+242*x^3+605*x^2),x
, algorithm="maxima")

[Out]

x/(11*x - 10*log(2) + 10*log(2*x + 5))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{11 \, x + 10 \, \log \left (x + \frac {5}{2}\right )} \]

[In]

integrate(((20*x+50)*log(5/2+x)-20*x)/((200*x+500)*log(5/2+x)^2+(440*x^2+1100*x)*log(5/2+x)+242*x^3+605*x^2),x
, algorithm="giac")

[Out]

x/(11*x + 10*log(x + 5/2))

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-20 x+(50+20 x) \log \left (\frac {1}{2} (5+2 x)\right )}{605 x^2+242 x^3+\left (1100 x+440 x^2\right ) \log \left (\frac {1}{2} (5+2 x)\right )+(500+200 x) \log ^2\left (\frac {1}{2} (5+2 x)\right )} \, dx=\frac {x}{11\,x+10\,\ln \left (x+\frac {5}{2}\right )} \]

[In]

int(-(20*x - log(x + 5/2)*(20*x + 50))/(log(x + 5/2)*(1100*x + 440*x^2) + log(x + 5/2)^2*(200*x + 500) + 605*x
^2 + 242*x^3),x)

[Out]

x/(11*x + 10*log(x + 5/2))