\(\int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x (x-30 x^2)+e^{2 x} (4 x-30 x^2)} \, dx\) [997]

Optimal result

Integrand size = 96, antiderivative size = 28 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\log \left (-x+\frac {x}{3 \left (-e^x-e^{2 x}+5 x\right )}\right ) \]

[Out]

ln(x/(15*x-3*exp(x)^2-3*exp(x))-x)

Rubi [F]

\[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx \]

[In]

Int[(6*E^(3*x) + 3*E^(4*x) + E^(2*x)*(4 - 32*x) + E^x*(1 - 31*x) + 75*x^2)/(6*E^(3*x)*x + 3*E^(4*x)*x - 5*x^2
+ 75*x^3 + E^x*(x - 30*x^2) + E^(2*x)*(4*x - 30*x^2)),x]

[Out]

Log[x] - 17*Defer[Int][(1 + 3*E^x + 3*E^(2*x) - 15*x)^(-1), x] - 3*Defer[Int][E^x/(1 + 3*E^x + 3*E^(2*x) - 15*
x), x] + 5*Defer[Int][(E^x + E^(2*x) - 5*x)^(-1), x] + Defer[Int][E^x/(E^x + E^(2*x) - 5*x), x] + 30*Defer[Int
][x/(1 + 3*E^x + 3*E^(2*x) - 15*x), x] - 10*Defer[Int][x/(E^x + E^(2*x) - 5*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {17+3 e^x-30 x}{1+3 e^x+3 e^{2 x}-15 x}+\frac {5+e^x-10 x}{e^x+e^{2 x}-5 x}+\frac {1}{x}\right ) \, dx \\ & = \log (x)-\int \frac {17+3 e^x-30 x}{1+3 e^x+3 e^{2 x}-15 x} \, dx+\int \frac {5+e^x-10 x}{e^x+e^{2 x}-5 x} \, dx \\ & = \log (x)-\int \left (\frac {17}{1+3 e^x+3 e^{2 x}-15 x}+\frac {3 e^x}{1+3 e^x+3 e^{2 x}-15 x}-\frac {30 x}{1+3 e^x+3 e^{2 x}-15 x}\right ) \, dx+\int \left (\frac {5}{e^x+e^{2 x}-5 x}+\frac {e^x}{e^x+e^{2 x}-5 x}-\frac {10 x}{e^x+e^{2 x}-5 x}\right ) \, dx \\ & = \log (x)-3 \int \frac {e^x}{1+3 e^x+3 e^{2 x}-15 x} \, dx+5 \int \frac {1}{e^x+e^{2 x}-5 x} \, dx-10 \int \frac {x}{e^x+e^{2 x}-5 x} \, dx-17 \int \frac {1}{1+3 e^x+3 e^{2 x}-15 x} \, dx+30 \int \frac {x}{1+3 e^x+3 e^{2 x}-15 x} \, dx+\int \frac {e^x}{e^x+e^{2 x}-5 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=2 \text {arctanh}\left (1+6 e^x+6 e^{2 x}-30 x\right )+\log (x) \]

[In]

Integrate[(6*E^(3*x) + 3*E^(4*x) + E^(2*x)*(4 - 32*x) + E^x*(1 - 31*x) + 75*x^2)/(6*E^(3*x)*x + 3*E^(4*x)*x -
5*x^2 + 75*x^3 + E^x*(x - 30*x^2) + E^(2*x)*(4*x - 30*x^2)),x]

[Out]

2*ArcTanh[1 + 6*E^x + 6*E^(2*x) - 30*x] + Log[x]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(\ln \left (x \right )+\ln \left ({\mathrm e}^{2 x}-5 x +{\mathrm e}^{x}+\frac {1}{3}\right )-\ln \left ({\mathrm e}^{2 x}-5 x +{\mathrm e}^{x}\right )\)	\(29\)
parallelrisch	\(\ln \left (x \right )-\ln \left (-\frac {{\mathrm e}^{2 x}}{5}+x -\frac {{\mathrm e}^{x}}{5}\right )+\ln \left (-\frac {{\mathrm e}^{2 x}}{5}+x -\frac {{\mathrm e}^{x}}{5}-\frac {1}{15}\right )\)	\(33\)
norman	\(-\ln \left (-{\mathrm e}^{2 x}+5 x -{\mathrm e}^{x}\right )+\ln \left (x \right )+\ln \left (-3 \,{\mathrm e}^{2 x}+15 x -3 \,{\mathrm e}^{x}-1\right )\)	\(37\)

[In]

int((3*exp(x)^4+6*exp(x)^3+(-32*x+4)*exp(x)^2+(-31*x+1)*exp(x)+75*x^2)/(3*x*exp(x)^4+6*x*exp(x)^3+(-30*x^2+4*x
)*exp(x)^2+(-30*x^2+x)*exp(x)+75*x^3-5*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(exp(2*x)-5*x+exp(x)+1/3)-ln(exp(2*x)-5*x+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\log \left (x\right ) - \log \left (-5 \, x + e^{\left (2 \, x\right )} + e^{x}\right ) + \log \left (-15 \, x + 3 \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 1\right ) \]

[In]

integrate((3*exp(x)^4+6*exp(x)^3+(-32*x+4)*exp(x)^2+(-31*x+1)*exp(x)+75*x^2)/(3*x*exp(x)^4+6*x*exp(x)^3+(-30*x
^2+4*x)*exp(x)^2+(-30*x^2+x)*exp(x)+75*x^3-5*x^2),x, algorithm="fricas")

[Out]

log(x) - log(-5*x + e^(2*x) + e^x) + log(-15*x + 3*e^(2*x) + 3*e^x + 1)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\log {\left (x \right )} - \log {\left (- 5 x + e^{2 x} + e^{x} \right )} + \log {\left (- 5 x + e^{2 x} + e^{x} + \frac {1}{3} \right )} \]

[In]

integrate((3*exp(x)**4+6*exp(x)**3+(-32*x+4)*exp(x)**2+(-31*x+1)*exp(x)+75*x**2)/(3*x*exp(x)**4+6*x*exp(x)**3+
(-30*x**2+4*x)*exp(x)**2+(-30*x**2+x)*exp(x)+75*x**3-5*x**2),x)

[Out]

log(x) - log(-5*x + exp(2*x) + exp(x)) + log(-5*x + exp(2*x) + exp(x) + 1/3)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\log \left (x\right ) + \log \left (-5 \, x + e^{\left (2 \, x\right )} + e^{x} + \frac {1}{3}\right ) - \log \left (-5 \, x + e^{\left (2 \, x\right )} + e^{x}\right ) \]

[In]

integrate((3*exp(x)^4+6*exp(x)^3+(-32*x+4)*exp(x)^2+(-31*x+1)*exp(x)+75*x^2)/(3*x*exp(x)^4+6*x*exp(x)^3+(-30*x
^2+4*x)*exp(x)^2+(-30*x^2+x)*exp(x)+75*x^3-5*x^2),x, algorithm="maxima")

[Out]

log(x) + log(-5*x + e^(2*x) + e^x + 1/3) - log(-5*x + e^(2*x) + e^x)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=-\log \left (5 \, x - e^{\left (2 \, x\right )} - e^{x}\right ) + \log \left (x\right ) + \log \left (-15 \, x + 3 \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 1\right ) \]

[In]

integrate((3*exp(x)^4+6*exp(x)^3+(-32*x+4)*exp(x)^2+(-31*x+1)*exp(x)+75*x^2)/(3*x*exp(x)^4+6*x*exp(x)^3+(-30*x
^2+4*x)*exp(x)^2+(-30*x^2+x)*exp(x)+75*x^3-5*x^2),x, algorithm="giac")

[Out]

-log(5*x - e^(2*x) - e^x) + log(x) + log(-15*x + 3*e^(2*x) + 3*e^x + 1)

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {6 e^{3 x}+3 e^{4 x}+e^{2 x} (4-32 x)+e^x (1-31 x)+75 x^2}{6 e^{3 x} x+3 e^{4 x} x-5 x^2+75 x^3+e^x \left (x-30 x^2\right )+e^{2 x} \left (4 x-30 x^2\right )} \, dx=\ln \left (15\,x-3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^x-1\right )-\ln \left (5\,x-{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\right )+\ln \left (x\right ) \]

[In]

int((6*exp(3*x) + 3*exp(4*x) - exp(x)*(31*x - 1) - exp(2*x)*(32*x - 4) + 75*x^2)/(exp(2*x)*(4*x - 30*x^2) + 6*
x*exp(3*x) + 3*x*exp(4*x) + exp(x)*(x - 30*x^2) - 5*x^2 + 75*x^3),x)

[Out]

log(15*x - 3*exp(2*x) - 3*exp(x) - 1) - log(5*x - exp(2*x) - exp(x)) + log(x)