\(\int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx\) [8486]

Optimal result

Integrand size = 28, antiderivative size = 15 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {13+e^{3 x/4}}{-4+x} \]

[Out]

(exp(3/4*x)+13)/(x-4)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 12, 6874, 2228} \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=-\frac {e^{3 x/4}}{4-x}-\frac {13}{4-x} \]

[In]

Int[(-52 + E^((3*x)/4)*(-16 + 3*x))/(64 - 32*x + 4*x^2),x]

[Out]

-13/(4 - x) - E^((3*x)/4)/(4 - 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 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-52+e^{3 x/4} (-16+3 x)}{4 (-4+x)^2} \, dx \\ & = \frac {1}{4} \int \frac {-52+e^{3 x/4} (-16+3 x)}{(-4+x)^2} \, dx \\ & = \frac {1}{4} \int \left (-\frac {52}{(-4+x)^2}+\frac {e^{3 x/4} (-16+3 x)}{(-4+x)^2}\right ) \, dx \\ & = -\frac {13}{4-x}+\frac {1}{4} \int \frac {e^{3 x/4} (-16+3 x)}{(-4+x)^2} \, dx \\ & = -\frac {13}{4-x}-\frac {e^{3 x/4}}{4-x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {13+e^{3 x/4}}{-4+x} \]

[In]

Integrate[(-52 + E^((3*x)/4)*(-16 + 3*x))/(64 - 32*x + 4*x^2),x]

[Out]

(13 + E^((3*x)/4))/(-4 + x)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {{\mathrm e}^{\frac {3 x}{4}}+13}{x -4}\)	\(13\)
parallelrisch	\(\frac {52+4 \,{\mathrm e}^{\frac {3 x}{4}}}{4 x -16}\)	\(16\)
risch	\(\frac {13}{x -4}+\frac {{\mathrm e}^{\frac {3 x}{4}}}{x -4}\)	\(19\)
parts	\(\frac {13}{x -4}+\frac {3 \,{\mathrm e}^{\frac {3 x}{4}}}{4 \left (\frac {3 x}{4}-3\right )}\)	\(22\)
derivativedivides	\(\frac {39}{4 \left (\frac {3 x}{4}-3\right )}+\frac {3 \,{\mathrm e}^{\frac {3 x}{4}}}{4 \left (\frac {3 x}{4}-3\right )}\)	\(24\)
default	\(\frac {39}{4 \left (\frac {3 x}{4}-3\right )}+\frac {3 \,{\mathrm e}^{\frac {3 x}{4}}}{4 \left (\frac {3 x}{4}-3\right )}\)	\(24\)

[In]

int(((3*x-16)*exp(3/4*x)-52)/(4*x^2-32*x+64),x,method=_RETURNVERBOSE)

[Out]

(exp(3/4*x)+13)/(x-4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {e^{\left (\frac {3}{4} \, x\right )} + 13}{x - 4} \]

[In]

integrate(((3*x-16)*exp(3/4*x)-52)/(4*x^2-32*x+64),x, algorithm="fricas")

[Out]

(e^(3/4*x) + 13)/(x - 4)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {e^{\frac {3 x}{4}}}{x - 4} + \frac {13}{x - 4} \]

[In]

integrate(((3*x-16)*exp(3/4*x)-52)/(4*x**2-32*x+64),x)

[Out]

exp(3*x/4)/(x - 4) + 13/(x - 4)

Maxima [F]

\[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\int { \frac {{\left (3 \, x - 16\right )} e^{\left (\frac {3}{4} \, x\right )} - 52}{4 \, {\left (x^{2} - 8 \, x + 16\right )}} \,d x } \]

[In]

integrate(((3*x-16)*exp(3/4*x)-52)/(4*x^2-32*x+64),x, algorithm="maxima")

[Out]

x*e^(3/4*x)/(x^2 - 8*x + 16) + 4*e^3*exp_integral_e(2, -3/4*x + 3)/(x - 4) + 13/(x - 4) + 3/4*integrate(4/3*(x
+ 4)*e^(3/4*x)/(x^3 - 12*x^2 + 48*x - 64), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {e^{\left (\frac {3}{4} \, x\right )} + 13}{x - 4} \]

[In]

integrate(((3*x-16)*exp(3/4*x)-52)/(4*x^2-32*x+64),x, algorithm="giac")

[Out]

(e^(3/4*x) + 13)/(x - 4)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-52+e^{3 x/4} (-16+3 x)}{64-32 x+4 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {3\,x}{4}}+13}{x-4} \]

[In]

int((exp((3*x)/4)*(3*x - 16) - 52)/(4*x^2 - 32*x + 64),x)

[Out]

(exp((3*x)/4) + 13)/(x - 4)