3.93.52 \(\int (21+i \pi +18 x+\log (\frac {1}{4} (-5+e^2))) \, dx\)

Optimal. Leaf size=28 \[ x \left (i \pi +4 (-1+x)+5 (5+x)+\log \left (\frac {1}{4} \left (-5+e^2\right )\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 1, number of rules used = 0, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} 9 x^2+x \left (21+i \pi +\log \left (\frac {1}{4} \left (e^2-5\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[21 + I*Pi + 18*x + Log[(-5 + E^2)/4],x]

[Out]

9*x^2 + x*(21 + I*Pi + Log[(-5 + E^2)/4])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=9 x^2+x \left (21+i \pi +\log \left (\frac {1}{4} \left (-5+e^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 0.96 \begin {gather*} 21 x+i \pi x+9 x^2+x \log \left (\frac {1}{4} \left (-5+e^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[21 + I*Pi + 18*x + Log[(-5 + E^2)/4],x]

[Out]

21*x + I*Pi*x + 9*x^2 + x*Log[(-5 + E^2)/4]

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fricas [A]  time = 0.50, size = 18, normalized size = 0.64 \begin {gather*} 9 \, x^{2} + x \log \left (-\frac {1}{4} \, e^{2} + \frac {5}{4}\right ) + 21 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/4-1/4*exp(2))+18*x+21,x, algorithm="fricas")

[Out]

9*x^2 + x*log(-1/4*e^2 + 5/4) + 21*x

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giac [A]  time = 0.14, size = 18, normalized size = 0.64 \begin {gather*} 9 \, x^{2} + x \log \left (-\frac {1}{4} \, e^{2} + \frac {5}{4}\right ) + 21 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/4-1/4*exp(2))+18*x+21,x, algorithm="giac")

[Out]

9*x^2 + x*log(-1/4*e^2 + 5/4) + 21*x

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maple [A]  time = 0.03, size = 19, normalized size = 0.68

method	result	size
gosper	\(x \ln \left (\frac {5}{4}-\frac {{\mathrm e}^{2}}{4}\right )+9 x^{2}+21 x\)	\(19\)
default	\(x \ln \left (\frac {5}{4}-\frac {{\mathrm e}^{2}}{4}\right )+9 x^{2}+21 x\)	\(19\)
norman	\(\left (-2 \ln \relax (2)+\ln \left (5-{\mathrm e}^{2}\right )+21\right ) x +9 x^{2}\)	\(22\)
risch	\(-2 x \ln \relax (2)+x \ln \left (5-{\mathrm e}^{2}\right )+9 x^{2}+21 x\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(5/4-1/4*exp(2))+18*x+21,x,method=_RETURNVERBOSE)

[Out]

x*ln(5/4-1/4*exp(2))+9*x^2+21*x

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maxima [A]  time = 0.36, size = 18, normalized size = 0.64 \begin {gather*} 9 \, x^{2} + x \log \left (-\frac {1}{4} \, e^{2} + \frac {5}{4}\right ) + 21 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/4-1/4*exp(2))+18*x+21,x, algorithm="maxima")

[Out]

9*x^2 + x*log(-1/4*e^2 + 5/4) + 21*x

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mupad [B]  time = 5.57, size = 17, normalized size = 0.61 \begin {gather*} 9\,x^2+\left (\ln \left (\frac {5}{4}-\frac {{\mathrm {e}}^2}{4}\right )+21\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(18*x + log(5/4 - exp(2)/4) + 21,x)

[Out]

x*(log(5/4 - exp(2)/4) + 21) + 9*x^2

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sympy [A]  time = 0.06, size = 20, normalized size = 0.71 \begin {gather*} 9 x^{2} + x \left (\log {\left (- \frac {5}{4} + \frac {e^{2}}{4} \right )} + 21 + i \pi \right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(5/4-1/4*exp(2))+18*x+21,x)

[Out]

9*x**2 + x*(log(-5/4 + exp(2)/4) + 21 + I*pi)

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