\(\int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx\) [1013]

Optimal result

Integrand size = 29, antiderivative size = 30 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=x-4 \left (e^{\frac {5 \left (\frac {10}{3}-2 x\right )}{x}}-\frac {4}{3} \left (-e^3+x\right )\right ) \]

[Out]

19/3*x-4*exp(5/x*(10/3-2*x))-16/3*exp(3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2240} \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=\frac {19 x}{3}-4 e^{\frac {50}{3 x}-10} \]

[In]

Int[(200*E^((50 - 30*x)/(3*x)) + 19*x^2)/(3*x^2),x]

[Out]

-4*E^(-10 + 50/(3*x)) + (19*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{x^2} \, dx \\ & = \frac {1}{3} \int \left (19+\frac {200 e^{-10+\frac {50}{3 x}}}{x^2}\right ) \, dx \\ & = \frac {19 x}{3}+\frac {200}{3} \int \frac {e^{-10+\frac {50}{3 x}}}{x^2} \, dx \\ & = -4 e^{-10+\frac {50}{3 x}}+\frac {19 x}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=-4 e^{-10+\frac {50}{3 x}}+\frac {19 x}{3} \]

[In]

Integrate[(200*E^((50 - 30*x)/(3*x)) + 19*x^2)/(3*x^2),x]

[Out]

-4*E^(-10 + 50/(3*x)) + (19*x)/3

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {19 x}{3}-4 \,{\mathrm e}^{-10+\frac {50}{3 x}}\)	\(15\)
default	\(\frac {19 x}{3}-4 \,{\mathrm e}^{-10+\frac {50}{3 x}}\)	\(15\)
risch	\(\frac {19 x}{3}-4 \,{\mathrm e}^{-\frac {10 \left (3 x -5\right )}{3 x}}\)	\(18\)
parallelrisch	\(\frac {19 x}{3}-4 \,{\mathrm e}^{-\frac {10 \left (3 x -5\right )}{3 x}}\)	\(18\)
parts	\(\frac {19 x}{3}-4 \,{\mathrm e}^{\frac {-30 x +50}{3 x}}\)	\(18\)
norman	\(\frac {\frac {19 x^{2}}{3}-4 x \,{\mathrm e}^{\frac {-30 x +50}{3 x}}}{x}\)	\(25\)

[In]

int(1/3*(200*exp(1/3*(-30*x+50)/x)+19*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

19/3*x-4*exp(-10+50/3/x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=\frac {19}{3} \, x - 4 \, e^{\left (-\frac {10 \, {\left (3 \, x - 5\right )}}{3 \, x}\right )} \]

[In]

integrate(1/3*(200*exp(1/3*(-30*x+50)/x)+19*x^2)/x^2,x, algorithm="fricas")

[Out]

19/3*x - 4*e^(-10/3*(3*x - 5)/x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.50 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=\frac {19 x}{3} - 4 e^{\frac {\frac {50}{3} - 10 x}{x}} \]

[In]

integrate(1/3*(200*exp(1/3*(-30*x+50)/x)+19*x**2)/x**2,x)

[Out]

19*x/3 - 4*exp((50/3 - 10*x)/x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.47 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=\frac {19}{3} \, x - 4 \, e^{\left (\frac {50}{3 \, x} - 10\right )} \]

[In]

integrate(1/3*(200*exp(1/3*(-30*x+50)/x)+19*x^2)/x^2,x, algorithm="maxima")

[Out]

19/3*x - 4*e^(50/3/x - 10)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (21) = 42\).

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=-\frac {\frac {12 \, {\left (3 \, x - 5\right )} e^{\left (-\frac {10 \, {\left (3 \, x - 5\right )}}{3 \, x}\right )}}{x} - 36 \, e^{\left (-\frac {10 \, {\left (3 \, x - 5\right )}}{3 \, x}\right )} + 95}{3 \, {\left (\frac {3 \, x - 5}{x} - 3\right )}} \]

[In]

integrate(1/3*(200*exp(1/3*(-30*x+50)/x)+19*x^2)/x^2,x, algorithm="giac")

[Out]

-1/3*(12*(3*x - 5)*e^(-10/3*(3*x - 5)/x)/x - 36*e^(-10/3*(3*x - 5)/x) + 95)/((3*x - 5)/x - 3)

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.47 \[ \int \frac {200 e^{\frac {50-30 x}{3 x}}+19 x^2}{3 x^2} \, dx=\frac {19\,x}{3}-4\,{\mathrm {e}}^{\frac {50}{3\,x}-10} \]

[In]

int(((200*exp(-(10*x - 50/3)/x))/3 + (19*x^2)/3)/x^2,x)

[Out]

(19*x)/3 - 4*exp(50/(3*x) - 10)