\(\int \frac {4+100 x^2-\log (\frac {1}{5} e^{50 x^2} x^4)}{4 x^2} \, dx\) [8192]

Optimal result

Integrand size = 31, antiderivative size = 22 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x} \]

[Out]

1/4/x*ln(1/5*x^4*exp(50*x^2))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2631} \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x} \]

[In]

Int[(4 + 100*x^2 - Log[(E^(50*x^2)*x^4)/5])/(4*x^2),x]

[Out]

Log[(E^(50*x^2)*x^4)/5]/(4*x)

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]]

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{x^2} \, dx \\ & = \frac {1}{4} \int \left (\frac {4 \left (1+25 x^2\right )}{x^2}-\frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int \frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{x^2} \, dx\right )+\int \frac {1+25 x^2}{x^2} \, dx \\ & = \frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x}-\frac {1}{4} \int 4 \left (25+\frac {1}{x^2}\right ) \, dx+\int \left (25+\frac {1}{x^2}\right ) \, dx \\ & = -\frac {1}{x}+25 x+\frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x}-\int \left (25+\frac {1}{x^2}\right ) \, dx \\ & = \frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x} \]

[In]

Integrate[(4 + 100*x^2 - Log[(E^(50*x^2)*x^4)/5])/(4*x^2),x]

[Out]

Log[(E^(50*x^2)*x^4)/5]/(4*x)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\ln \left (\frac {x^{4} {\mathrm e}^{50 x^{2}}}{5}\right )}{4 x}\)	\(18\)
norman	\(\frac {\ln \left (\frac {x^{4} {\mathrm e}^{50 x^{2}}}{5}\right )}{4 x}\)	\(18\)
parallelrisch	\(\frac {\ln \left (\frac {x^{4} {\mathrm e}^{50 x^{2}}}{5}\right )}{4 x}\)	\(18\)
parts	\(\frac {\ln \left (\frac {x^{4} {\mathrm e}^{50 x^{2}}}{5}\right )}{4 x}\)	\(18\)
risch	\(\frac {\ln \left ({\mathrm e}^{50 x^{2}}\right )}{4 x}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{4} {\mathrm e}^{50 x^{2}}\right )^{3}-i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{50 x^{2}}\right )^{2}+i \pi \operatorname {csgn}\left (i x^{4}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i {\mathrm e}^{50 x^{2}}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{50 x^{2}}\right )-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{50 x^{2}}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{50 x^{2}}\right )^{2}+i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (5\right )-8 \ln \left (x \right )}{8 x}\)	\(333\)

[In]

int(1/4*(-ln(1/5*x^4*exp(50*x^2))+100*x^2+4)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4/x*ln(1/5*x^4*exp(50*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {\log \left (\frac {1}{5} \, x^{4} e^{\left (50 \, x^{2}\right )}\right )}{4 \, x} \]

[In]

integrate(1/4*(-log(1/5*x^4*exp(50*x^2))+100*x^2+4)/x^2,x, algorithm="fricas")

[Out]

1/4*log(1/5*x^4*e^(50*x^2))/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {\log {\left (\frac {x^{4} e^{50 x^{2}}}{5} \right )}}{4 x} \]

[In]

integrate(1/4*(-ln(1/5*x**4*exp(50*x**2))+100*x**2+4)/x**2,x)

[Out]

log(x**4*exp(50*x**2)/5)/(4*x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (17) = 34\).

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=25 \, x - \frac {25 \, x^{2} - 1}{x} + \frac {\log \left (\frac {1}{5} \, x^{4} e^{\left (50 \, x^{2}\right )}\right )}{4 \, x} - \frac {1}{x} \]

[In]

integrate(1/4*(-log(1/5*x^4*exp(50*x^2))+100*x^2+4)/x^2,x, algorithm="maxima")

[Out]

25*x - (25*x^2 - 1)/x + 1/4*log(1/5*x^4*e^(50*x^2))/x - 1/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {25}{2} \, x + \frac {\log \left (\frac {1}{5} \, x^{4}\right )}{4 \, x} \]

[In]

integrate(1/4*(-log(1/5*x^4*exp(50*x^2))+100*x^2+4)/x^2,x, algorithm="giac")

[Out]

25/2*x + 1/4*log(1/5*x^4)/x

Mupad [B] (verification not implemented)

Time = 13.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {4+100 x^2-\log \left (\frac {1}{5} e^{50 x^2} x^4\right )}{4 x^2} \, dx=\frac {25\,x}{2}+\frac {\ln \left (\frac {x^4}{5}\right )}{4\,x} \]

[In]

int((25*x^2 - log((x^4*exp(50*x^2))/5)/4 + 1)/x^2,x)

[Out]

(25*x)/2 + log(x^4/5)/(4*x)