\(\int \frac {e^{\frac {15-3 x^2+x \log (\frac {1}{16} (16+8 e^x x+e^{2 x} x^2))}{3 x}} (-60-12 x^2+e^x (-15 x+2 x^2-x^3))}{12 x^2+3 e^x x^3} \, dx\) [6745]

Optimal result

Integrand size = 82, antiderivative size = 28 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=e^{\frac {5}{x}-x} \sqrt [3]{\left (1+\frac {e^x x}{4}\right )^2} \]

[Out]

exp(5/x-x+1/3*ln((1/4*exp(x)*x+1)^2))

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6838} \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=\frac {e^{\frac {5-x^2}{x}} \sqrt [3]{e^{2 x} x^2+8 e^x x+16}}{2 \sqrt [3]{2}} \]

[In]

Int[(E^((15 - 3*x^2 + x*Log[(16 + 8*E^x*x + E^(2*x)*x^2)/16])/(3*x))*(-60 - 12*x^2 + E^x*(-15*x + 2*x^2 - x^3)
))/(12*x^2 + 3*E^x*x^3),x]

[Out]

(E^((5 - x^2)/x)*(16 + 8*E^x*x + E^(2*x)*x^2)^(1/3))/(2*2^(1/3))

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{\frac {5-x^2}{x}} \sqrt [3]{16+8 e^x x+e^{2 x} x^2}}{2 \sqrt [3]{2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.46 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=\frac {e^{\frac {5}{x}-x} \sqrt [3]{\left (4+e^x x\right )^2}}{2 \sqrt [3]{2}} \]

[In]

Integrate[(E^((15 - 3*x^2 + x*Log[(16 + 8*E^x*x + E^(2*x)*x^2)/16])/(3*x))*(-60 - 12*x^2 + E^x*(-15*x + 2*x^2
- x^3)))/(12*x^2 + 3*E^x*x^3),x]

[Out]

(E^(5/x - x)*((4 + E^x*x)^2)^(1/3))/(2*2^(1/3))

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\({\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{2 x} x^{2}}{16}+\frac {{\mathrm e}^{x} x}{2}+1\right )-3 x^{2}+15}{3 x}}\)	\(33\)
risch	\(\frac {\left ({\mathrm e}^{x} x +4\right )^{\frac {2}{3}} 2^{\frac {2}{3}} {\mathrm e}^{-\frac {i x \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +4\right )^{2}\right )}^{3}-2 i x \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +4\right )^{2}\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +4\right )\right )+i x \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +4\right )^{2}\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +4\right )\right )}^{2}+6 x^{2}-30}{6 x}}}{4}\)	\(104\)

[In]

int(((-x^3+2*x^2-15*x)*exp(x)-12*x^2-60)*exp(1/3*(x*ln(1/16*exp(x)^2*x^2+1/2*exp(x)*x+1)-3*x^2+15)/x)/(3*exp(x
)*x^3+12*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*(x*ln(1/16*exp(x)^2*x^2+1/2*exp(x)*x+1)-3*x^2+15)/x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=e^{\left (-\frac {3 \, x^{2} - x \log \left (\frac {1}{16} \, x^{2} e^{\left (2 \, x\right )} + \frac {1}{2} \, x e^{x} + 1\right ) - 15}{3 \, x}\right )} \]

[In]

integrate(((-x^3+2*x^2-15*x)*exp(x)-12*x^2-60)*exp(1/3*(x*log(1/16*exp(x)^2*x^2+1/2*exp(x)*x+1)-3*x^2+15)/x)/(
3*exp(x)*x^3+12*x^2),x, algorithm="fricas")

[Out]

e^(-1/3*(3*x^2 - x*log(1/16*x^2*e^(2*x) + 1/2*x*e^x + 1) - 15)/x)

Sympy [A] (verification not implemented)

Time = 34.44 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=e^{\frac {- x^{2} + \frac {x \log {\left (\frac {x^{2} e^{2 x}}{16} + \frac {x e^{x}}{2} + 1 \right )}}{3} + 5}{x}} \]

[In]

integrate(((-x**3+2*x**2-15*x)*exp(x)-12*x**2-60)*exp(1/3*(x*ln(1/16*exp(x)**2*x**2+1/2*exp(x)*x+1)-3*x**2+15)
/x)/(3*exp(x)*x**3+12*x**2),x)

[Out]

exp((-x**2 + x*log(x**2*exp(2*x)/16 + x*exp(x)/2 + 1)/3 + 5)/x)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=\frac {1}{4} \cdot 2^{\frac {2}{3}} {\left (x e^{x} + 4\right )}^{\frac {2}{3}} e^{\left (-x + \frac {5}{x}\right )} \]

[In]

integrate(((-x^3+2*x^2-15*x)*exp(x)-12*x^2-60)*exp(1/3*(x*log(1/16*exp(x)^2*x^2+1/2*exp(x)*x+1)-3*x^2+15)/x)/(
3*exp(x)*x^3+12*x^2),x, algorithm="maxima")

[Out]

1/4*2^(2/3)*(x*e^x + 4)^(2/3)*e^(-x + 5/x)

Giac [A] (verification not implemented)

none

Time = 0.87 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=e^{\left (-x + \frac {5}{x} + \frac {1}{3} \, \log \left (\frac {1}{16} \, x^{2} e^{\left (2 \, x\right )} + \frac {1}{2} \, x e^{x} + 1\right )\right )} \]

[In]

integrate(((-x^3+2*x^2-15*x)*exp(x)-12*x^2-60)*exp(1/3*(x*log(1/16*exp(x)^2*x^2+1/2*exp(x)*x+1)-3*x^2+15)/x)/(
3*exp(x)*x^3+12*x^2),x, algorithm="giac")

[Out]

e^(-x + 5/x + 1/3*log(1/16*x^2*e^(2*x) + 1/2*x*e^x + 1))

Mupad [B] (verification not implemented)

Time = 18.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {15-3 x^2+x \log \left (\frac {1}{16} \left (16+8 e^x x+e^{2 x} x^2\right )\right )}{3 x}} \left (-60-12 x^2+e^x \left (-15 x+2 x^2-x^3\right )\right )}{12 x^2+3 e^x x^3} \, dx=\frac {{16}^{2/3}\,{\mathrm {e}}^{\frac {5}{x}-x}\,{\left (x^2\,{\mathrm {e}}^{2\,x}+8\,x\,{\mathrm {e}}^x+16\right )}^{1/3}}{16} \]

[In]

int(-(exp(((x*log((x^2*exp(2*x))/16 + (x*exp(x))/2 + 1))/3 - x^2 + 5)/x)*(exp(x)*(15*x - 2*x^2 + x^3) + 12*x^2
+ 60))/(3*x^3*exp(x) + 12*x^2),x)

[Out]

(16^(2/3)*exp(5/x - x)*(x^2*exp(2*x) + 8*x*exp(x) + 16)^(1/3))/16