\(\int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx\) [757]

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=x-\log (3)+\left (\left (2+e^3\right )^2-(5-x)^2+\frac {x}{5}\right ) \log (4) \]

[Out]

x-ln(3)+2*ln(2)*(1/5*x+(2+exp(3))^2-(5-x)^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12} \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=x-\frac {1}{100} (51-10 x)^2 \log (4) \]

[In]

Int[(5 + (51 - 10*x)*Log[4])/5,x]

[Out]

x - ((51 - 10*x)^2*Log[4])/100

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int (5+(51-10 x) \log (4)) \, dx \\ & = x-\frac {1}{100} (51-10 x)^2 \log (4) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.52 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=x+\frac {51}{5} x \log (4)-x^2 \log (4) \]

[In]

Integrate[(5 + (51 - 10*x)*Log[4])/5,x]

[Out]

x + (51*x*Log[4])/5 - x^2*Log[4]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {x \left (10 x \ln \left (2\right )-102 \ln \left (2\right )-5\right )}{5}\)	\(15\)
default	\(-2 x^{2} \ln \left (2\right )+\frac {102 x \ln \left (2\right )}{5}+x\)	\(15\)
risch	\(-2 x^{2} \ln \left (2\right )+\frac {102 x \ln \left (2\right )}{5}+x\)	\(15\)
parts	\(-2 x^{2} \ln \left (2\right )+\frac {102 x \ln \left (2\right )}{5}+x\)	\(15\)
parallelrisch	\(\frac {2 \ln \left (2\right ) \left (-5 x^{2}+51 x \right )}{5}+x\)	\(16\)
norman	\(\left (\frac {102 \ln \left (2\right )}{5}+1\right ) x -2 x^{2} \ln \left (2\right )\)	\(17\)

[In]

int(2/5*(-10*x+51)*ln(2)+1,x,method=_RETURNVERBOSE)

[Out]

-1/5*x*(10*x*ln(2)-102*ln(2)-5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=-\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \left (2\right ) + x \]

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="fricas")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=- 2 x^{2} \log {\left (2 \right )} + x \left (1 + \frac {102 \log {\left (2 \right )}}{5}\right ) \]

[In]

integrate(2/5*(-10*x+51)*ln(2)+1,x)

[Out]

-2*x**2*log(2) + x*(1 + 102*log(2)/5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=-\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \left (2\right ) + x \]

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="maxima")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=-\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \left (2\right ) + x \]

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="giac")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

Mupad [B] (verification not implemented)

Time = 8.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.42 \[ \int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx=x-\frac {\ln \left (2\right )\,{\left (10\,x-51\right )}^2}{50} \]

[In]

int(1 - (2*log(2)*(10*x - 51))/5,x)

[Out]

x - (log(2)*(10*x - 51)^2)/50