\(\int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx\) [2551]

Optimal result

Integrand size = 24, antiderivative size = 25 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {x^2}{5}-\frac {5 (-3+x) (i \pi +\log (4))}{\log (4)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {9} \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {(2 x \log (4)-25 (\log (4)+i \pi ))^2}{20 \log ^2(4)} \]

[In]

Int[(2*x*Log[4] - 25*(I*Pi + Log[4]))/(5*Log[4]),x]

[Out]

(2*x*Log[4] - 25*(I*Pi + Log[4]))^2/(20*Log[4]^2)

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 {(2 x \log (4)-25 (i \pi +\log (4)))^2}{20 \log ^2(4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=-5 x+\frac {x^2}{5}-\frac {5 i \pi x}{\log (4)} \]

[In]

Integrate[(2*x*Log[4] - 25*(I*Pi + Log[4]))/(5*Log[4]),x]

[Out]

-5*x + x^2/5 - ((5*I)*Pi*x)/Log[4]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {x^{2}}{5}-5 x -\frac {5 i x \pi }{2 \ln \left (2\right )}\)	\(19\)
norman	\(\frac {x^{2}}{5}-\frac {5 \left (2 \ln \left (2\right )+i \pi \right ) x}{2 \ln \left (2\right )}\)	\(23\)
parallelrisch	\(\frac {2 x^{2} \ln \left (2\right )+\left (-25 i \pi -50 \ln \left (2\right )\right ) x}{10 \ln \left (2\right )}\)	\(26\)
default	\(-\frac {i \left (2 i \ln \left (2\right ) x^{2}-50 i x \ln \left (2\right )+25 \pi x \right )}{10 \ln \left (2\right )}\)	\(27\)
gosper	\(-\frac {x \left (2 i x \ln \left (2\right )-50 i \ln \left (2\right )+25 \pi \right ) \left (25 i \pi -4 x \ln \left (2\right )+50 \ln \left (2\right )\right )}{10 \left (4 i x \ln \left (2\right )-50 i \ln \left (2\right )+25 \pi \right ) \ln \left (2\right )}\)	\(54\)

[In]

int(1/10*(4*x*ln(2)-50*ln(2)-25*I*Pi)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/5*x^2-5*x-5/2*I/ln(2)*x*Pi

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {-25 i \, \pi x + 2 \, {\left (x^{2} - 25 \, x\right )} \log \left (2\right )}{10 \, \log \left (2\right )} \]

[In]

integrate(1/10*(4*x*log(2)-50*log(2)-25*I*pi)/log(2),x, algorithm="fricas")

[Out]

1/10*(-25*I*pi*x + 2*(x^2 - 25*x)*log(2))/log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {x^{2}}{5} + \frac {x \left (- 10 \log {\left (2 \right )} - 5 i \pi \right )}{2 \log {\left (2 \right )}} \]

[In]

integrate(1/10*(4*x*ln(2)-50*ln(2)-25*I*pi)/ln(2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {2 \, x^{2} \log \left (2\right ) - 25 i \, \pi x - 50 \, x \log \left (2\right )}{10 \, \log \left (2\right )} \]

[In]

integrate(1/10*(4*x*log(2)-50*log(2)-25*I*pi)/log(2),x, algorithm="maxima")

[Out]

1/10*(2*x^2*log(2) - 25*I*pi*x - 50*x*log(2))/log(2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {2 \, x^{2} \log \left (2\right ) - 25 i \, \pi x - 50 \, x \log \left (2\right )}{10 \, \log \left (2\right )} \]

[In]

integrate(1/10*(4*x*log(2)-50*log(2)-25*I*pi)/log(2),x, algorithm="giac")

[Out]

1/10*(2*x^2*log(2) - 25*I*pi*x - 50*x*log(2))/log(2)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2 x \log (4)-25 (i \pi +\log (4))}{5 \log (4)} \, dx=\frac {5\,{\left (\ln \left (32\right )-\frac {2\,x\,\ln \left (2\right )}{5}+\frac {\Pi \,5{}\mathrm {i}}{2}\right )}^2}{4\,{\ln \left (2\right )}^2} \]

[In]

int(-((Pi*5i)/2 + 5*log(2) - (2*x*log(2))/5)/log(2),x)

[Out]

(5*((Pi*5i)/2 + log(32) - (2*x*log(2))/5)^2)/(4*log(2)^2)