\(\int \frac {4+2 e^8 x^3 (i \pi +\log (\frac {59}{16}))}{e^8 x^2} \, dx\) [7888]

Optimal result

Integrand size = 27, antiderivative size = 27 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-\frac {\frac {4}{e^8}+x}{x}+x^2 \left (i \pi +\log \left (\frac {59}{16}\right )\right ) \]

[Out]

x^2*(ln(59/16)+I*Pi)-(4/exp(4)^2+x)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 14} \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-\frac {4}{e^8 x}+x^2 \left (\log \left (\frac {59}{16}\right )+i \pi \right ) \]

[In]

Int[(4 + 2*E^8*x^3*(I*Pi + Log[59/16]))/(E^8*x^2),x]

[Out]

-4/(E^8*x) + x^2*(I*Pi + Log[59/16])

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{x^2} \, dx}{e^8} \\ & = \frac {\int \left (\frac {4}{x^2}+2 e^8 x \left (i \pi +\log \left (\frac {59}{16}\right )\right )\right ) \, dx}{e^8} \\ & = -\frac {4}{e^8 x}+x^2 \left (i \pi +\log \left (\frac {59}{16}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-\frac {4}{e^8 x}+x^2 \left (i \pi +\log \left (\frac {59}{16}\right )\right ) \]

[In]

Integrate[(4 + 2*E^8*x^3*(I*Pi + Log[59/16]))/(E^8*x^2),x]

[Out]

-4/(E^8*x) + x^2*(I*Pi + Log[59/16])

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
default	\(2 \,{\mathrm e}^{-8} \left (\frac {{\mathrm e}^{8} \left (\ln \left (\frac {59}{16}\right )+i \pi \right ) x^{2}}{2}-\frac {2}{x}\right )\)	\(27\)
risch	\(-4 x^{2} \ln \left (2\right )+\ln \left (59\right ) x^{2}+i \pi \,x^{2}-\frac {4 \,{\mathrm e}^{-8}}{x}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{-8} \left (i \pi \,x^{3} {\mathrm e}^{8}+\ln \left (\frac {59}{16}\right ) x^{3} {\mathrm e}^{8}-4\right )}{x}\)	\(32\)
norman	\(\frac {\left ({\mathrm e}^{4} \left (\ln \left (59\right )-4 \ln \left (2\right )+i \pi \right ) x^{3}-4 \,{\mathrm e}^{-4}\right ) {\mathrm e}^{-4}}{x}\)	\(33\)
gosper	\(\frac {\left (-i \ln \left (\frac {59}{16}\right ) x^{3} {\mathrm e}^{8}+\pi \,x^{3} {\mathrm e}^{8}+4 i\right ) \left (i \pi \,x^{3} {\mathrm e}^{8}+\ln \left (\frac {59}{16}\right ) x^{3} {\mathrm e}^{8}+2\right ) {\mathrm e}^{-8}}{x \left (-i \ln \left (\frac {59}{16}\right ) x^{3} {\mathrm e}^{8}+\pi \,x^{3} {\mathrm e}^{8}-2 i\right )}\)	\(82\)

[In]

int((2*x^3*exp(4)^2*(ln(59/16)+I*Pi)+4)/x^2/exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

2/exp(4)^2*(1/2*exp(8)*(ln(59/16)+I*Pi)*x^2-2/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=\frac {{\left (i \, \pi x^{3} e^{8} + x^{3} e^{8} \log \left (\frac {59}{16}\right ) - 4\right )} e^{\left (-8\right )}}{x} \]

[In]

integrate((2*x^3*exp(4)^2*(log(59/16)+I*pi)+4)/x^2/exp(4)^2,x, algorithm="fricas")

[Out]

(I*pi*x^3*e^8 + x^3*e^8*log(59/16) - 4)*e^(-8)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=\frac {x^{2} \left (- 4 e^{8} \log {\left (2 \right )} + e^{8} \log {\left (59 \right )} + i \pi e^{8}\right ) - \frac {4}{x}}{e^{8}} \]

[In]

integrate((2*x**3*exp(4)**2*(ln(59/16)+I*pi)+4)/x**2/exp(4)**2,x)

[Out]

(x**2*(-4*exp(8)*log(2) + exp(8)*log(59) + I*pi*exp(8)) - 4/x)*exp(-8)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-{\left ({\left (-i \, \pi e^{8} - e^{8} \log \left (\frac {59}{16}\right )\right )} x^{2} + \frac {4}{x}\right )} e^{\left (-8\right )} \]

[In]

integrate((2*x^3*exp(4)^2*(log(59/16)+I*pi)+4)/x^2/exp(4)^2,x, algorithm="maxima")

[Out]

-((-I*pi*e^8 - e^8*log(59/16))*x^2 + 4/x)*e^(-8)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-{\left (-i \, \pi x^{2} e^{8} - x^{2} e^{8} \log \left (\frac {59}{16}\right ) + \frac {4}{x}\right )} e^{\left (-8\right )} \]

[In]

integrate((2*x^3*exp(4)^2*(log(59/16)+I*pi)+4)/x^2/exp(4)^2,x, algorithm="giac")

[Out]

-(-I*pi*x^2*e^8 - x^2*e^8*log(59/16) + 4/x)*e^(-8)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {4+2 e^8 x^3 \left (i \pi +\log \left (\frac {59}{16}\right )\right )}{e^8 x^2} \, dx=-\frac {4\,{\mathrm {e}}^{-8}}{x}+x^2\,\left (\ln \left (\frac {59}{16}\right )+\Pi \,1{}\mathrm {i}\right ) \]

[In]

int((exp(-8)*(2*x^3*exp(8)*(Pi*1i + log(59/16)) + 4))/x^2,x)

[Out]

x^2*(Pi*1i + log(59/16)) - (4*exp(-8))/x