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

Optimal. Leaf size=25 \[ \frac {x^2}{5}-\frac {5 (-3+x) (i \pi +\log (4))}{\log (4)} \]

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Rubi [A]  time = 0.00, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {9} \begin {gather*} \frac {(2 x \log (4)-25 (\log (4)+i \pi ))^2}{20 \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \text {integral} &=\frac {(2 x \log (4)-25 (i \pi +\log (4)))^2}{20 \log ^2(4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.84 \begin {gather*} -5 x+\frac {x^2}{5}-\frac {5 i \pi x}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.64, size = 22, normalized size = 0.88 \begin {gather*} \frac {-25 i \, \pi x + 2 \, {\left (x^{2} - 25 \, x\right )} \log \relax (2)}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.21, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 \, x^{2} \log \relax (2) - 25 i \, \pi x - 50 \, x \log \relax (2)}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 19, normalized size = 0.76

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

Verification of antiderivative is not currently implemented for this CAS.

[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)*Pi*x

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maxima [A]  time = 0.51, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 \, x^{2} \log \relax (2) - 25 i \, \pi x - 50 \, x \log \relax (2)}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.07, size = 20, normalized size = 0.80 \begin {gather*} \frac {5\,{\left (\ln \left (32\right )-\frac {2\,x\,\ln \relax (2)}{5}+\frac {\Pi \,5{}\mathrm {i}}{2}\right )}^2}{4\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.05, size = 22, normalized size = 0.88 \begin {gather*} \frac {x^{2}}{5} + \frac {x \left (- 10 \log {\relax (2 )} - 5 i \pi \right )}{2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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