\(\int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} (94 x+21 x^2+e^{4 x} (-9+12 x))}{3 x^4} \, dx\) [9290]

Optimal result

Integrand size = 57, antiderivative size = 27 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\frac {\frac {1}{3}+\frac {e^{4 x}}{x}+x-(4+x)^2}{x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {12, 6838} \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\frac {-3 x^3-21 x^2-47 x+3 e^{4 x}}{3 x^3}} \]

[In]

Int[(E^((3*E^(4*x) - 47*x - 21*x^2 - 3*x^3)/(3*x^3))*(94*x + 21*x^2 + E^(4*x)*(-9 + 12*x)))/(3*x^4),x]

[Out]

E^((3*E^(4*x) - 47*x - 21*x^2 - 3*x^3)/(3*x^3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {1}{3} \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{x^4} \, dx \\ & = e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{-1+\frac {e^{4 x}}{x^3}-\frac {47}{3 x^2}-\frac {7}{x}} \]

[In]

Integrate[(E^((3*E^(4*x) - 47*x - 21*x^2 - 3*x^3)/(3*x^3))*(94*x + 21*x^2 + E^(4*x)*(-9 + 12*x)))/(3*x^4),x]

[Out]

E^(-1 + E^(4*x)/x^3 - 47/(3*x^2) - 7/x)

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
risch	\({\mathrm e}^{-\frac {-3 \,{\mathrm e}^{4 x}+3 x^{3}+21 x^{2}+47 x}{3 x^{3}}}\)	\(27\)
norman	\({\mathrm e}^{\frac {3 \,{\mathrm e}^{4 x}-3 x^{3}-21 x^{2}-47 x}{3 x^{3}}}\)	\(29\)
parallelrisch	\({\mathrm e}^{\frac {3 \,{\mathrm e}^{4 x}-3 x^{3}-21 x^{2}-47 x}{3 x^{3}}}\)	\(29\)

[In]

int(1/3*((12*x-9)*exp(2*x)^2+21*x^2+94*x)*exp(1/3*(3*exp(2*x)^2-3*x^3-21*x^2-47*x)/x^3)/x^4,x,method=_RETURNVE
RBOSE)

[Out]

exp(-1/3*(-3*exp(4*x)+3*x^3+21*x^2+47*x)/x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\left (-\frac {3 \, x^{3} + 21 \, x^{2} + 47 \, x - 3 \, e^{\left (4 \, x\right )}}{3 \, x^{3}}\right )} \]

[In]

integrate(1/3*((12*x-9)*exp(2*x)^2+21*x^2+94*x)*exp(1/3*(3*exp(2*x)^2-3*x^3-21*x^2-47*x)/x^3)/x^4,x, algorithm
="fricas")

[Out]

e^(-1/3*(3*x^3 + 21*x^2 + 47*x - 3*e^(4*x))/x^3)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\frac {- x^{3} - 7 x^{2} - \frac {47 x}{3} + e^{4 x}}{x^{3}}} \]

[In]

integrate(1/3*((12*x-9)*exp(2*x)**2+21*x**2+94*x)*exp(1/3*(3*exp(2*x)**2-3*x**3-21*x**2-47*x)/x**3)/x**4,x)

[Out]

exp((-x**3 - 7*x**2 - 47*x/3 + exp(4*x))/x**3)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\left (-\frac {7}{x} - \frac {47}{3 \, x^{2}} + \frac {e^{\left (4 \, x\right )}}{x^{3}} - 1\right )} \]

[In]

integrate(1/3*((12*x-9)*exp(2*x)^2+21*x^2+94*x)*exp(1/3*(3*exp(2*x)^2-3*x^3-21*x^2-47*x)/x^3)/x^4,x, algorithm
="maxima")

[Out]

e^(-7/x - 47/3/x^2 + e^(4*x)/x^3 - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx=e^{\left (-\frac {7}{x} - \frac {47}{3 \, x^{2}} + \frac {e^{\left (4 \, x\right )}}{x^{3}} - 1\right )} \]

[In]

integrate(1/3*((12*x-9)*exp(2*x)^2+21*x^2+94*x)*exp(1/3*(3*exp(2*x)^2-3*x^3-21*x^2-47*x)/x^3)/x^4,x, algorithm
="giac")

[Out]

e^(-7/x - 47/3/x^2 + e^(4*x)/x^3 - 1)

Mupad [B] (verification not implemented)

Time = 12.59 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {3 e^{4 x}-47 x-21 x^2-3 x^3}{3 x^3}} \left (94 x+21 x^2+e^{4 x} (-9+12 x)\right )}{3 x^4} \, dx={\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {7}{x}}\,{\mathrm {e}}^{-\frac {47}{3\,x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,x}}{x^3}} \]

[In]

int((exp(-((47*x)/3 - exp(4*x) + 7*x^2 + x^3)/x^3)*(94*x + exp(4*x)*(12*x - 9) + 21*x^2))/(3*x^4),x)

[Out]

exp(-1)*exp(-7/x)*exp(-47/(3*x^2))*exp(exp(4*x)/x^3)