\(\int \frac {e^x (-15-50 x^2+10 e^{2 x} x^2+(15 x+5 e^{2 x} x^2-50 x^3) \log (\frac {-3-e^{2 x} x+10 x^2}{x}))}{3 x+e^{2 x} x^2-10 x^3} \, dx\) [7775]

Optimal result

Integrand size = 81, antiderivative size = 22 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 e^x \log \left (-e^{2 x}-\frac {3}{x}+10 x\right ) \]

[Out]

5*exp(x+ln(ln(-exp(x)^2+10*x-3/x)))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2326} \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 e^x \log \left (-\frac {-10 x^2+e^{2 x} x+3}{x}\right ) \]

[In]

Int[(E^x*(-15 - 50*x^2 + 10*E^(2*x)*x^2 + (15*x + 5*E^(2*x)*x^2 - 50*x^3)*Log[(-3 - E^(2*x)*x + 10*x^2)/x]))/(
3*x + E^(2*x)*x^2 - 10*x^3),x]

[Out]

5*E^x*Log[-((3 + E^(2*x)*x - 10*x^2)/x)]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 5 e^x \log \left (-\frac {3+e^{2 x} x-10 x^2}{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 e^x \log \left (-e^{2 x}-\frac {3}{x}+10 x\right ) \]

[In]

Integrate[(E^x*(-15 - 50*x^2 + 10*E^(2*x)*x^2 + (15*x + 5*E^(2*x)*x^2 - 50*x^3)*Log[(-3 - E^(2*x)*x + 10*x^2)/
x]))/(3*x + E^(2*x)*x^2 - 10*x^3),x]

[Out]

5*E^x*Log[-E^(2*x) - 3/x + 10*x]

Maple [A] (verified)

Time = 183.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(5 \,{\mathrm e}^{\ln \left (\ln \left (-\frac {x \,{\mathrm e}^{2 x}-10 x^{2}+3}{x}\right )\right )+x}\)	\(26\)
risch	\(5 \,{\mathrm e}^{x} \ln \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )-5 \,{\mathrm e}^{x} \ln \left (x \right )+\frac {5 i \pi \,\operatorname {csgn}\left (i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )}^{2} {\mathrm e}^{x}}{2}-\frac {5 i \pi \,\operatorname {csgn}\left (i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}+\frac {5 i \pi {\operatorname {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}-\frac {5 i \pi {\operatorname {csgn}\left (\frac {i \left (-\frac {x \,{\mathrm e}^{2 x}}{10}+x^{2}-\frac {3}{10}\right )}{x}\right )}^{3} {\mathrm e}^{x}}{2}+5 \,{\mathrm e}^{x} \ln \left (5\right )+5 \,{\mathrm e}^{x} \ln \left (2\right )\)	\(189\)

[In]

int(((5*exp(x)^2*x^2-50*x^3+15*x)*ln((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(ln(ln((-x*exp(x)
^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/ln((-x*exp(x)^2+10*x^2-3)/x),x,method=_RETURNVERBOSE)

[Out]

5*exp(ln(ln(-(x*exp(x)^2-10*x^2+3)/x))+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 \, e^{x} \log \left (\frac {10 \, x^{2} - x e^{\left (2 \, x\right )} - 3}{x}\right ) \]

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="fricas")

[Out]

5*e^x*log((10*x^2 - x*e^(2*x) - 3)/x)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=\text {Timed out} \]

[In]

integrate(((5*exp(x)**2*x**2-50*x**3+15*x)*ln((-x*exp(x)**2+10*x**2-3)/x)+10*exp(x)**2*x**2-50*x**2-15)*exp(ln
(ln((-x*exp(x)**2+10*x**2-3)/x))+x)/(exp(x)**2*x**2-10*x**3+3*x)/ln((-x*exp(x)**2+10*x**2-3)/x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 \, e^{x} \log \left (10 \, x^{2} - x e^{\left (2 \, x\right )} - 3\right ) - 5 \, e^{x} \log \left (x\right ) \]

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="maxima")

[Out]

5*e^x*log(10*x^2 - x*e^(2*x) - 3) - 5*e^x*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5 \, e^{x} \log \left (10 \, x^{2} - x e^{\left (2 \, x\right )} - 3\right ) - 5 \, e^{x} \log \left (x\right ) \]

[In]

integrate(((5*exp(x)^2*x^2-50*x^3+15*x)*log((-x*exp(x)^2+10*x^2-3)/x)+10*exp(x)^2*x^2-50*x^2-15)*exp(log(log((
-x*exp(x)^2+10*x^2-3)/x))+x)/(exp(x)^2*x^2-10*x^3+3*x)/log((-x*exp(x)^2+10*x^2-3)/x),x, algorithm="giac")

[Out]

5*e^x*log(10*x^2 - x*e^(2*x) - 3) - 5*e^x*log(x)

Mupad [B] (verification not implemented)

Time = 13.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^x \left (-15-50 x^2+10 e^{2 x} x^2+\left (15 x+5 e^{2 x} x^2-50 x^3\right ) \log \left (\frac {-3-e^{2 x} x+10 x^2}{x}\right )\right )}{3 x+e^{2 x} x^2-10 x^3} \, dx=5\,{\mathrm {e}}^x\,\ln \left (10\,x-{\mathrm {e}}^{2\,x}-\frac {3}{x}\right ) \]

[In]

int((exp(x + log(log(-(x*exp(2*x) - 10*x^2 + 3)/x)))*(10*x^2*exp(2*x) + log(-(x*exp(2*x) - 10*x^2 + 3)/x)*(15*
x + 5*x^2*exp(2*x) - 50*x^3) - 50*x^2 - 15))/(log(-(x*exp(2*x) - 10*x^2 + 3)/x)*(3*x + x^2*exp(2*x) - 10*x^3))
,x)

[Out]

5*exp(x)*log(10*x - exp(2*x) - 3/x)