3.53.54 \(\int \frac {6 e^2+e^{\frac {1}{3} (1+4 x)} (100-80 x+16 x^2)}{75-60 x+12 x^2} \, dx\)

Optimal. Leaf size=23 \[ e^{x+\frac {1+x}{3}}+\frac {e^2}{5-2 x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {27, 12, 6742, 2194} \begin {gather*} e^{\frac {4 x}{3}+\frac {1}{3}}+\frac {e^2}{5-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^2 + E^((1 + 4*x)/3)*(100 - 80*x + 16*x^2))/(75 - 60*x + 12*x^2),x]

[Out]

E^(1/3 + (4*x)/3) + E^2/(5 - 2*x)

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 e^2+e^{\frac {1}{3} (1+4 x)} \left (100-80 x+16 x^2\right )}{3 (-5+2 x)^2} \, dx\\ &=\frac {1}{3} \int \frac {6 e^2+e^{\frac {1}{3} (1+4 x)} \left (100-80 x+16 x^2\right )}{(-5+2 x)^2} \, dx\\ &=\frac {1}{3} \int \left (4 e^{\frac {1}{3}+\frac {4 x}{3}}+\frac {6 e^2}{(-5+2 x)^2}\right ) \, dx\\ &=\frac {e^2}{5-2 x}+\frac {4}{3} \int e^{\frac {1}{3}+\frac {4 x}{3}} \, dx\\ &=e^{\frac {1}{3}+\frac {4 x}{3}}+\frac {e^2}{5-2 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 23, normalized size = 1.00 \begin {gather*} e^{\frac {1}{3}+\frac {4 x}{3}}+\frac {e^2}{5-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^2 + E^((1 + 4*x)/3)*(100 - 80*x + 16*x^2))/(75 - 60*x + 12*x^2),x]

[Out]

E^(1/3 + (4*x)/3) + E^2/(5 - 2*x)

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 25, normalized size = 1.09 \begin {gather*} \frac {{\left (2 \, x - 5\right )} e^{\left (\frac {4}{3} \, x + \frac {1}{3}\right )} - e^{2}}{2 \, x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x+100)*exp(4/3*x+1/3)+6*exp(2))/(12*x^2-60*x+75),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.15, size = 30, normalized size = 1.30 \begin {gather*} \frac {2 \, x e^{\left (\frac {4}{3} \, x + \frac {1}{3}\right )} - e^{2} - 5 \, e^{\left (\frac {4}{3} \, x + \frac {1}{3}\right )}}{2 \, x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x+100)*exp(4/3*x+1/3)+6*exp(2))/(12*x^2-60*x+75),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.25, size = 17, normalized size = 0.74

method	result	size
risch	\(-\frac {{\mathrm e}^{2}}{2 \left (x -\frac {5}{2}\right )}+{\mathrm e}^{\frac {4 x}{3}+\frac {1}{3}}\)	\(17\)
derivativedivides	\(-\frac {2 \,{\mathrm e}^{2}}{4 x -10}+{\mathrm e}^{\frac {4 x}{3}+\frac {1}{3}}\)	\(19\)
default	\(-\frac {2 \,{\mathrm e}^{2}}{4 x -10}+{\mathrm e}^{\frac {4 x}{3}+\frac {1}{3}}\)	\(19\)
norman	\(\frac {2 \,{\mathrm e}^{\frac {4 x}{3}+\frac {1}{3}} x -5 \,{\mathrm e}^{\frac {4 x}{3}+\frac {1}{3}}-{\mathrm e}^{2}}{2 x -5}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^2-80*x+100)*exp(4/3*x+1/3)+6*exp(2))/(12*x^2-60*x+75),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, {\left (x^{2} e^{\frac {1}{3}} - 5 \, x e^{\frac {1}{3}}\right )} e^{\left (\frac {4}{3} \, x\right )}}{4 \, x^{2} - 20 \, x + 25} - \frac {50 \, e^{\frac {11}{3}} E_{2}\left (-\frac {4}{3} \, x + \frac {10}{3}\right )}{3 \, {\left (2 \, x - 5\right )}} - \frac {e^{2}}{2 \, x - 5} - 100 \, \int \frac {e^{\left (\frac {4}{3} \, x + \frac {1}{3}\right )}}{8 \, x^{3} - 60 \, x^{2} + 150 \, x - 125}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2-80*x+100)*exp(4/3*x+1/3)+6*exp(2))/(12*x^2-60*x+75),x, algorithm="maxima")

[Out]

4*(x^2*e^(1/3) - 5*x*e^(1/3))*e^(4/3*x)/(4*x^2 - 20*x + 25) - 50/3*e^(11/3)*exp_integral_e(2, -4/3*x + 10/3)/(
2*x - 5) - e^2/(2*x - 5) - 100*integrate(e^(4/3*x + 1/3)/(8*x^3 - 60*x^2 + 150*x - 125), x)

________________________________________________________________________________________

mupad [B]  time = 3.40, size = 18, normalized size = 0.78 \begin {gather*} {\mathrm {e}}^{\frac {4\,x}{3}+\frac {1}{3}}-\frac {{\mathrm {e}}^2}{2\,x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*exp(2) + exp((4*x)/3 + 1/3)*(16*x^2 - 80*x + 100))/(12*x^2 - 60*x + 75),x)

[Out]

exp((4*x)/3 + 1/3) - exp(2)/(2*x - 5)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 19, normalized size = 0.83 \begin {gather*} e^{\frac {4 x}{3} + \frac {1}{3}} - \frac {2 e^{2}}{4 x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**2-80*x+100)*exp(4/3*x+1/3)+6*exp(2))/(12*x**2-60*x+75),x)

[Out]

exp(4*x/3 + 1/3) - 2*exp(2)/(4*x - 10)

________________________________________________________________________________________