\(\int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx\) [6502]

Optimal result

Integrand size = 38, antiderivative size = 25 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=\log \left (-x+\frac {1+e^{\frac {6}{e^{20}}}-\frac {x^2}{25}}{x}\right ) \]

[Out]

ln((exp(6/exp(5)^4)-1/25*x^2+1)/x-x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6, 1607, 457, 78} \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=\log \left (25 \left (1+e^{\frac {6}{e^{20}}}\right )-26 x^2\right )-\log (x) \]

[In]

Int[(-25 - 25*E^(6/E^20) - 26*x^2)/(25*x + 25*E^(6/E^20)*x - 26*x^3),x]

[Out]

-Log[x] + Log[25*(1 + E^(6/E^20)) - 26*x^2]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{\left (25+25 e^{\frac {6}{e^{20}}}\right ) x-26 x^3} \, dx \\ & = \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{x \left (25+25 e^{\frac {6}{e^{20}}}-26 x^2\right )} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x}{\left (25+25 e^{\frac {6}{e^{20}}}-26 x\right ) x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {52}{25+25 e^{\frac {6}{e^{20}}}-26 x}-\frac {1}{x}\right ) \, dx,x,x^2\right ) \\ & = -\log (x)+\log \left (25 \left (1+e^{\frac {6}{e^{20}}}\right )-26 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=-\log (x)+\log \left (25+25 e^{\frac {6}{e^{20}}}-26 x^2\right ) \]

[In]

Integrate[(-25 - 25*E^(6/E^20) - 26*x^2)/(25*x + 25*E^(6/E^20)*x - 26*x^3),x]

[Out]

-Log[x] + Log[25 + 25*E^(6/E^20) - 26*x^2]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

method	result	size
default	\(-\ln \left (x \right )+\ln \left (26 x^{2}-25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}-25\right )\)	\(21\)
risch	\(-\ln \left (x \right )+\ln \left (26 x^{2}-25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}-25\right )\)	\(21\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (x^{2}-\frac {25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}}{26}-\frac {25}{26}\right )\)	\(21\)
norman	\(-\ln \left (x \right )+\ln \left (-26 x^{2}+25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}+25\right )\)	\(23\)
meijerg	\(-\frac {25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}} \left (2 \ln \left (x \right )+\ln \left (2\right )+\ln \left (13\right )-2 \ln \left (5\right )-\ln \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right )+i \pi -\ln \left (1-\frac {26 x^{2}}{25 \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right )}\right )\right )}{2 \left (25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}+25\right )}+\frac {25 \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right ) \ln \left (1-\frac {26 x^{2}}{25 \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right )}\right )}{2 \left (25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}+25\right )}-\frac {25 \left (2 \ln \left (x \right )+\ln \left (2\right )+\ln \left (13\right )-2 \ln \left (5\right )-\ln \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right )+i \pi -\ln \left (1-\frac {26 x^{2}}{25 \left ({\mathrm e}^{6 \,{\mathrm e}^{-20}}+1\right )}\right )\right )}{2 \left (25 \,{\mathrm e}^{6 \,{\mathrm e}^{-20}}+25\right )}\)	\(162\)

[In]

int((-25*exp(6/exp(5)^4)-26*x^2-25)/(25*x*exp(6/exp(5)^4)-26*x^3+25*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(26*x^2-25*exp(6*exp(-20))-25)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=\log \left (26 \, x^{2} - 25 \, e^{\left (6 \, e^{\left (-20\right )}\right )} - 25\right ) - \log \left (x\right ) \]

[In]

integrate((-25*exp(6/exp(5)^4)-26*x^2-25)/(25*x*exp(6/exp(5)^4)-26*x^3+25*x),x, algorithm="fricas")

[Out]

log(26*x^2 - 25*e^(6*e^(-20)) - 25) - log(x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} - \frac {25 e^{\frac {6}{e^{20}}}}{26} - \frac {25}{26} \right )} \]

[In]

integrate((-25*exp(6/exp(5)**4)-26*x**2-25)/(25*x*exp(6/exp(5)**4)-26*x**3+25*x),x)

[Out]

-log(x) + log(x**2 - 25*exp(6*exp(-20))/26 - 25/26)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=\log \left (26 \, x^{2} - 25 \, e^{\left (6 \, e^{\left (-20\right )}\right )} - 25\right ) - \log \left (x\right ) \]

[In]

integrate((-25*exp(6/exp(5)^4)-26*x^2-25)/(25*x*exp(6/exp(5)^4)-26*x^3+25*x),x, algorithm="maxima")

[Out]

log(26*x^2 - 25*e^(6*e^(-20)) - 25) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=-\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left ({\left | 26 \, x^{2} - 25 \, e^{\left (6 \, e^{\left (-20\right )}\right )} - 25 \right |}\right ) \]

[In]

integrate((-25*exp(6/exp(5)^4)-26*x^2-25)/(25*x*exp(6/exp(5)^4)-26*x^3+25*x),x, algorithm="giac")

[Out]

-1/2*log(x^2) + log(abs(26*x^2 - 25*e^(6*e^(-20)) - 25))

Mupad [B] (verification not implemented)

Time = 12.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25-25 e^{\frac {6}{e^{20}}}-26 x^2}{25 x+25 e^{\frac {6}{e^{20}}} x-26 x^3} \, dx=\ln \left (-52\,x^2+50\,{\mathrm {e}}^{6\,{\mathrm {e}}^{-20}}+50\right )-\ln \left (x\right ) \]

[In]

int(-(25*exp(6*exp(-20)) + 26*x^2 + 25)/(25*x + 25*x*exp(6*exp(-20)) - 26*x^3),x)

[Out]

log(50*exp(6*exp(-20)) - 52*x^2 + 50) - log(x)