\(\int \frac {1}{3} (14 x-6 (i \pi +\log (\frac {1}{20} \log (\frac {25}{16})))) \, dx\) [4267]

Optimal result

Integrand size = 25, antiderivative size = 29 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {x^2}{3}-2 x \left (i \pi -x+\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right ) \]

[Out]

1/3*x^2-2*(ln(1/20*ln(16/25))-x)*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {9} \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {1}{21} \left (7 x-3 \left (\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )+i \pi \right )\right )^2 \]

[In]

Int[(14*x - 6*(I*Pi + Log[Log[25/16]/20]))/3,x]

[Out]

(7*x - 3*(I*Pi + Log[Log[25/16]/20]))^2/21

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{21} \left (7 x-3 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=-2 i \pi x+\frac {7 x^2}{3}-2 x \log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right ) \]

[In]

Integrate[(14*x - 6*(I*Pi + Log[Log[25/16]/20]))/3,x]

[Out]

(-2*I)*Pi*x + (7*x^2)/3 - 2*x*Log[Log[25/16]/20]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.52

method	result	size
gosper	\(-\frac {x \left (-7 x +6 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right )\right )}{3}\)	\(15\)
default	\(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\)	\(15\)
parallelrisch	\(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\)	\(15\)
parts	\(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\)	\(15\)
risch	\(-2 \ln \left (-\frac {\ln \left (5\right )}{10}+\frac {\ln \left (2\right )}{5}\right ) x +\frac {7 x^{2}}{3}\)	\(20\)
norman	\(\left (-2 i \pi +2 \ln \left (5\right )+2 \ln \left (10\right )-2 \ln \left (\ln \left (5\right )\right )-2 \ln \left (\ln \left (2\right )\right )\right ) x +\frac {7 x^{2}}{3}\)	\(32\)

[In]

int(-2*ln(1/20*ln(16/25))+14/3*x,x,method=_RETURNVERBOSE)

[Out]

-1/3*x*(-7*x+6*ln(1/20*ln(16/25)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]

[In]

integrate(-2*log(1/20*log(16/25))+14/3*x,x, algorithm="fricas")

[Out]

7/3*x^2 - 2*x*log(1/20*log(16/25))

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7 x^{2}}{3} + x \left (- 2 \log {\left (- \frac {\log {\left (2 \right )}}{5} + \frac {\log {\left (5 \right )}}{10} \right )} - 2 i \pi \right ) \]

[In]

integrate(-2*ln(1/20*ln(16/25))+14/3*x,x)

[Out]

7*x**2/3 + x*(-2*log(-log(2)/5 + log(5)/10) - 2*I*pi)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]

[In]

integrate(-2*log(1/20*log(16/25))+14/3*x,x, algorithm="maxima")

[Out]

7/3*x^2 - 2*x*log(1/20*log(16/25))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]

[In]

integrate(-2*log(1/20*log(16/25))+14/3*x,x, algorithm="giac")

[Out]

7/3*x^2 - 2*x*log(1/20*log(16/25))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7\,x^2}{3}-2\,x\,\ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) \]

[In]

int((14*x)/3 - 2*log(log(16/25)/20),x)

[Out]

(7*x^2)/3 - 2*x*log(log(16/25)/20)