\(\int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx\) [6716]

Optimal result

Integrand size = 20, antiderivative size = 25 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=-e^5+16 \left (-5+x^2\right )+\frac {-4+e^4+x+\log (4)}{x} \]

[Out]

16*x^2-80-exp(5)+(exp(4)-4+x+2*ln(2))/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {14} \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=16 x^2-\frac {4-e^4-\log (4)}{x} \]

[In]

Int[(4 - E^4 + 32*x^3 - Log[4])/x^2,x]

[Out]

16*x^2 - (4 - E^4 - Log[4])/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}& = \int \left (32 x+\frac {4-e^4-\log (4)}{x^2}\right ) \, dx \\ & = 16 x^2-\frac {4-e^4-\log (4)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=\frac {-4+e^4+16 x^3+\log (4)}{x} \]

[In]

Integrate[(4 - E^4 + 32*x^3 - Log[4])/x^2,x]

[Out]

(-4 + E^4 + 16*x^3 + Log[4])/x

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72

method	result	size
gosper	\(\frac {16 x^{3}+2 \ln \left (2\right )+{\mathrm e}^{4}-4}{x}\)	\(18\)
norman	\(\frac {16 x^{3}+2 \ln \left (2\right )+{\mathrm e}^{4}-4}{x}\)	\(18\)
parallelrisch	\(\frac {16 x^{3}+2 \ln \left (2\right )+{\mathrm e}^{4}-4}{x}\)	\(18\)
default	\(16 x^{2}-\frac {-2 \ln \left (2\right )-{\mathrm e}^{4}+4}{x}\)	\(22\)
risch	\(16 x^{2}+\frac {2 \ln \left (2\right )}{x}+\frac {{\mathrm e}^{4}}{x}-\frac {4}{x}\)	\(25\)

[In]

int((32*x^3-2*ln(2)+4-exp(4))/x^2,x,method=_RETURNVERBOSE)

[Out]

(16*x^3+2*ln(2)+exp(4)-4)/x

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=\frac {16 \, x^{3} + e^{4} + 2 \, \log \left (2\right ) - 4}{x} \]

[In]

integrate((32*x^3-2*log(2)+4-exp(4))/x^2,x, algorithm="fricas")

[Out]

(16*x^3 + e^4 + 2*log(2) - 4)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=16 x^{2} + \frac {-4 + 2 \log {\left (2 \right )} + e^{4}}{x} \]

[In]

integrate((32*x**3-2*ln(2)+4-exp(4))/x**2,x)

[Out]

16*x**2 + (-4 + 2*log(2) + exp(4))/x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=16 \, x^{2} + \frac {e^{4} + 2 \, \log \left (2\right ) - 4}{x} \]

[In]

integrate((32*x^3-2*log(2)+4-exp(4))/x^2,x, algorithm="maxima")

[Out]

16*x^2 + (e^4 + 2*log(2) - 4)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=16 \, x^{2} + \frac {e^{4} + 2 \, \log \left (2\right ) - 4}{x} \]

[In]

integrate((32*x^3-2*log(2)+4-exp(4))/x^2,x, algorithm="giac")

[Out]

16*x^2 + (e^4 + 2*log(2) - 4)/x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {4-e^4+32 x^3-\log (4)}{x^2} \, dx=\frac {{\mathrm {e}}^4+\ln \left (4\right )-4}{x}+16\,x^2 \]

[In]

int(-(exp(4) + 2*log(2) - 32*x^3 - 4)/x^2,x)

[Out]

(exp(4) + log(4) - 4)/x + 16*x^2