\(\int \frac {1}{2} (6 e^{6 x^2} x+(i \pi +\log (5))^2) \, dx\) [4854]

Optimal result

Integrand size = 25, antiderivative size = 25 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {1}{4} \left (e^{6 x^2}+2 x (i \pi +\log (5))^2\right ) \]

[Out]

1/4*exp(6*x^2)+1/2*x*(ln(5)+I*Pi)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 2240} \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {e^{6 x^2}}{4}+\frac {1}{2} x (\log (5)+i \pi )^2 \]

[In]

Int[(6*E^(6*x^2)*x + (I*Pi + Log[5])^2)/2,x]

[Out]

E^(6*x^2)/4 + (x*(I*Pi + Log[5])^2)/2

Rule 12

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

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx \\ & = \frac {1}{2} x (i \pi +\log (5))^2+3 \int e^{6 x^2} x \, dx \\ & = \frac {e^{6 x^2}}{4}+\frac {1}{2} x (i \pi +\log (5))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {1}{2} \left (\frac {e^{6 x^2}}{2}+x (i \pi +\log (5))^2\right ) \]

[In]

Integrate[(6*E^(6*x^2)*x + (I*Pi + Log[5])^2)/2,x]

[Out]

(E^(6*x^2)/2 + x*(I*Pi + Log[5])^2)/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {{\mathrm e}^{6 x^{2}}}{4}+\frac {x \left (\ln \left (5\right )+i \pi \right )^{2}}{2}\)	\(22\)
parallelrisch	\(\frac {{\mathrm e}^{6 x^{2}}}{4}+\frac {x \left (\ln \left (5\right )+i \pi \right )^{2}}{2}\)	\(22\)
parts	\(\frac {{\mathrm e}^{6 x^{2}}}{4}+\frac {x \left (\ln \left (5\right )+i \pi \right )^{2}}{2}\)	\(22\)
norman	\(\left (i \pi \ln \left (5\right )-\frac {\pi ^{2}}{2}+\frac {\ln \left (5\right )^{2}}{2}\right ) x +\frac {{\mathrm e}^{6 x^{2}}}{4}\)	\(30\)
risch	\(\frac {{\mathrm e}^{6 x^{2}}}{4}-\frac {x \,\pi ^{2}}{2}+i \pi \ln \left (5\right ) x +\frac {x \ln \left (5\right )^{2}}{2}\)	\(30\)

[In]

int(3*x*exp(6*x^2)+1/2*(ln(5)+I*Pi)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*exp(6*x^2)+1/2*x*(ln(5)+I*Pi)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=-\frac {1}{2} \, \pi ^{2} x + i \, \pi x \log \left (5\right ) + \frac {1}{2} \, x \log \left (5\right )^{2} + \frac {1}{4} \, e^{\left (6 \, x^{2}\right )} \]

[In]

integrate(3*x*exp(6*x^2)+1/2*(log(5)+I*pi)^2,x, algorithm="fricas")

[Out]

-1/2*pi^2*x + I*pi*x*log(5) + 1/2*x*log(5)^2 + 1/4*e^(6*x^2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=x \left (- \frac {\pi ^{2}}{2} + \frac {\log {\left (5 \right )}^{2}}{2} + i \pi \log {\left (5 \right )}\right ) + \frac {e^{6 x^{2}}}{4} \]

[In]

integrate(3*x*exp(6*x**2)+1/2*(ln(5)+I*pi)**2,x)

[Out]

x*(-pi**2/2 + log(5)**2/2 + I*pi*log(5)) + exp(6*x**2)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {1}{2} \, {\left (i \, \pi + \log \left (5\right )\right )}^{2} x + \frac {1}{4} \, e^{\left (6 \, x^{2}\right )} \]

[In]

integrate(3*x*exp(6*x^2)+1/2*(log(5)+I*pi)^2,x, algorithm="maxima")

[Out]

1/2*(I*pi + log(5))^2*x + 1/4*e^(6*x^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {1}{2} \, {\left (i \, \pi + \log \left (5\right )\right )}^{2} x + \frac {1}{4} \, e^{\left (6 \, x^{2}\right )} \]

[In]

integrate(3*x*exp(6*x^2)+1/2*(log(5)+I*pi)^2,x, algorithm="giac")

[Out]

1/2*(I*pi + log(5))^2*x + 1/4*e^(6*x^2)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1}{2} \left (6 e^{6 x^2} x+(i \pi +\log (5))^2\right ) \, dx=\frac {{\mathrm {e}}^{6\,x^2}}{4}+\frac {x\,{\left (\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )}^2}{2} \]

[In]

int(3*x*exp(6*x^2) + (Pi*1i + log(5))^2/2,x)

[Out]

exp(6*x^2)/4 + (x*(Pi*1i + log(5))^2)/2