\(\int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx\) [3577]

Optimal result

Integrand size = 29, antiderivative size = 18 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=5 \left (x+\frac {2 x^2}{3}+\log (x-\log (5))\right ) \]

[Out]

10/3*x^2+5*x+5*ln(-ln(5)+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1864} \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {10 x^2}{3}+5 x+5 \log (x-\log (5)) \]

[In]

Int[(-15 - 15*x - 20*x^2 + (15 + 20*x)*Log[5])/(-3*x + 3*Log[5]),x]

[Out]

5*x + (10*x^2)/3 + 5*Log[x - Log[5]]

Rule 1864

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{align*} \text {integral}& = \int \left (5+\frac {20 x}{3}+\frac {5}{x-\log (5)}\right ) \, dx \\ & = 5 x+\frac {10 x^2}{3}+5 \log (x-\log (5)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {5}{3} \left (3 x+2 x^2-\log (5) (3+\log (25))+3 \log (x-\log (5))\right ) \]

[In]

Integrate[(-15 - 15*x - 20*x^2 + (15 + 20*x)*Log[5])/(-3*x + 3*Log[5]),x]

[Out]

(5*(3*x + 2*x^2 - Log[5]*(3 + Log[25]) + 3*Log[x - Log[5]]))/3

Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {10 x^{2}}{3}+5 x +5 \ln \left (-\ln \left (5\right )+x \right )\)	\(19\)
risch	\(\frac {10 x^{2}}{3}+5 x +5 \ln \left (-\ln \left (5\right )+x \right )\)	\(19\)
parallelrisch	\(\frac {10 x^{2}}{3}+5 x +5 \ln \left (-\ln \left (5\right )+x \right )\)	\(19\)
norman	\(5 x +\frac {10 x^{2}}{3}+5 \ln \left (3 \ln \left (5\right )-3 x \right )\)	\(21\)
meijerg	\(-5 \ln \left (5\right ) \ln \left (1-\frac {x}{\ln \left (5\right )}\right )+5 \ln \left (1-\frac {x}{\ln \left (5\right )}\right )-\left (-\frac {20 \ln \left (5\right )}{3}+5\right ) \ln \left (5\right ) \left (-\frac {x}{\ln \left (5\right )}-\ln \left (1-\frac {x}{\ln \left (5\right )}\right )\right )+\frac {20 \ln \left (5\right )^{2} \left (\frac {x \left (\frac {3 x}{\ln \left (5\right )}+6\right )}{6 \ln \left (5\right )}+\ln \left (1-\frac {x}{\ln \left (5\right )}\right )\right )}{3}\)	\(91\)

[In]

int(((20*x+15)*ln(5)-20*x^2-15*x-15)/(3*ln(5)-3*x),x,method=_RETURNVERBOSE)

[Out]

10/3*x^2+5*x+5*ln(-ln(5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {10}{3} \, x^{2} + 5 \, x + 5 \, \log \left (x - \log \left (5\right )\right ) \]

[In]

integrate(((20*x+15)*log(5)-20*x^2-15*x-15)/(3*log(5)-3*x),x, algorithm="fricas")

[Out]

10/3*x^2 + 5*x + 5*log(x - log(5))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {10 x^{2}}{3} + 5 x + 5 \log {\left (x - \log {\left (5 \right )} \right )} \]

[In]

integrate(((20*x+15)*ln(5)-20*x**2-15*x-15)/(3*ln(5)-3*x),x)

[Out]

10*x**2/3 + 5*x + 5*log(x - log(5))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {10}{3} \, x^{2} + 5 \, x + 5 \, \log \left (x - \log \left (5\right )\right ) \]

[In]

integrate(((20*x+15)*log(5)-20*x^2-15*x-15)/(3*log(5)-3*x),x, algorithm="maxima")

[Out]

10/3*x^2 + 5*x + 5*log(x - log(5))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=\frac {10}{3} \, x^{2} + 5 \, x + 5 \, \log \left ({\left | x - \log \left (5\right ) \right |}\right ) \]

[In]

integrate(((20*x+15)*log(5)-20*x^2-15*x-15)/(3*log(5)-3*x),x, algorithm="giac")

[Out]

10/3*x^2 + 5*x + 5*log(abs(x - log(5)))

Mupad [B] (verification not implemented)

Time = 8.94 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-15-15 x-20 x^2+(15+20 x) \log (5)}{-3 x+3 \log (5)} \, dx=5\,x+5\,\ln \left (x-\ln \left (5\right )\right )+\frac {10\,x^2}{3} \]

[In]

int((15*x - log(5)*(20*x + 15) + 20*x^2 + 15)/(3*x - 3*log(5)),x)

[Out]

5*x + 5*log(x - log(5)) + (10*x^2)/3