3.23.48 \(\int \frac {x (i \pi +\log (5))}{18 e^3} \, dx\)

Optimal. Leaf size=18 \[ \frac {x^2 (i \pi +\log (5))}{36 e^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 30} \begin {gather*} \frac {x^2 (\log (5)+i \pi )}{36 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(I*Pi + Log[5]))/(18*E^3),x]

[Out]

(x^2*(I*Pi + Log[5]))/(36*E^3)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {(i \pi +\log (5)) \int x \, dx}{18 e^3}\\ &=\frac {x^2 (i \pi +\log (5))}{36 e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 26, normalized size = 1.44 \begin {gather*} \frac {i \pi x^2}{36 e^3}+\frac {x^2 \log (5)}{36 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(I*Pi + Log[5]))/(18*E^3),x]

[Out]

((I/36)*Pi*x^2)/E^3 + (x^2*Log[5])/(36*E^3)

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fricas [A]  time = 1.12, size = 17, normalized size = 0.94 \begin {gather*} \frac {1}{36} \, {\left (i \, \pi x^{2} + x^{2} \log \relax (5)\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*(log(5)+I*pi)/exp(3),x, algorithm="fricas")

[Out]

1/36*(I*pi*x^2 + x^2*log(5))*e^(-3)

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giac [A]  time = 0.28, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{36} \, {\left (i \, \pi + \log \relax (5)\right )} x^{2} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*(log(5)+I*pi)/exp(3),x, algorithm="giac")

[Out]

1/36*(I*pi + log(5))*x^2*e^(-3)

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maple [A]  time = 0.03, size = 17, normalized size = 0.94

method	result	size
gosper	\(\frac {x^{2} {\mathrm e}^{-3} \left (\ln \relax (5)+i \pi \right )}{36}\)	\(17\)
default	\(\frac {x^{2} {\mathrm e}^{-3} \left (\ln \relax (5)+i \pi \right )}{36}\)	\(17\)
norman	\(\frac {x^{2} {\mathrm e}^{-3} \left (\ln \relax (5)+i \pi \right )}{36}\)	\(17\)
risch	\(\frac {i \pi \,{\mathrm e}^{-3} x^{2}}{36}+\frac {\ln \relax (5) {\mathrm e}^{-3} x^{2}}{36}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*x*(ln(5)+I*Pi)/exp(3),x,method=_RETURNVERBOSE)

[Out]

1/36*x^2/exp(3)*(ln(5)+I*Pi)

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maxima [A]  time = 0.40, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{36} \, {\left (i \, \pi + \log \relax (5)\right )} x^{2} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*(log(5)+I*pi)/exp(3),x, algorithm="maxima")

[Out]

1/36*(I*pi + log(5))*x^2*e^(-3)

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mupad [B]  time = 1.23, size = 14, normalized size = 0.78 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-3}\,\left (\ln \relax (5)+\Pi \,1{}\mathrm {i}\right )}{36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(-3)*(Pi*1i + log(5)))/18,x)

[Out]

(x^2*exp(-3)*(Pi*1i + log(5)))/36

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sympy [A]  time = 0.05, size = 14, normalized size = 0.78 \begin {gather*} \frac {x^{2} \left (\log {\relax (5 )} + i \pi \right )}{36 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*(ln(5)+I*pi)/exp(3),x)

[Out]

x**2*(log(5) + I*pi)*exp(-3)/36

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