\(\int \frac {e^{2+\frac {1}{4} (9+8 x)} (160 x+152 x^2-16 x^3)}{100-20 x+x^2} \, dx\) [3917]

Optimal result

Integrand size = 38, antiderivative size = 21 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=\frac {8 e^{\frac {17}{4}+2 x} x^2}{10-x} \]

[Out]

8*exp(2)*x^2*exp(2*x+9/4)/(10-x)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {27, 1608, 2230, 2225, 2208, 2209, 2207} \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-8 e^{2 x+\frac {17}{4}} x-80 e^{2 x+\frac {17}{4}}+\frac {800 e^{2 x+\frac {17}{4}}}{10-x} \]

[In]

Int[(E^(2 + (9 + 8*x)/4)*(160*x + 152*x^2 - 16*x^3))/(100 - 20*x + x^2),x]

[Out]

-80*E^(17/4 + 2*x) + (800*E^(17/4 + 2*x))/(10 - x) - 8*E^(17/4 + 2*x)*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 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

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 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{(-10+x)^2} \, dx \\ & = \int \frac {e^{2+\frac {1}{4} (9+8 x)} x \left (160+152 x-16 x^2\right )}{(-10+x)^2} \, dx \\ & = \int \left (-168 e^{\frac {17}{4}+2 x}+\frac {800 e^{\frac {17}{4}+2 x}}{(-10+x)^2}-\frac {1600 e^{\frac {17}{4}+2 x}}{-10+x}-16 e^{\frac {17}{4}+2 x} x\right ) \, dx \\ & = -\left (16 \int e^{\frac {17}{4}+2 x} x \, dx\right )-168 \int e^{\frac {17}{4}+2 x} \, dx+800 \int \frac {e^{\frac {17}{4}+2 x}}{(-10+x)^2} \, dx-1600 \int \frac {e^{\frac {17}{4}+2 x}}{-10+x} \, dx \\ & = -84 e^{\frac {17}{4}+2 x}+\frac {800 e^{\frac {17}{4}+2 x}}{10-x}-8 e^{\frac {17}{4}+2 x} x-1600 e^{97/4} \text {Ei}(-2 (10-x))+8 \int e^{\frac {17}{4}+2 x} \, dx+1600 \int \frac {e^{\frac {17}{4}+2 x}}{-10+x} \, dx \\ & = -80 e^{\frac {17}{4}+2 x}+\frac {800 e^{\frac {17}{4}+2 x}}{10-x}-8 e^{\frac {17}{4}+2 x} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-\frac {8 e^{\frac {17}{4}+2 x} x^2}{-10+x} \]

[In]

Integrate[(E^(2 + (9 + 8*x)/4)*(160*x + 152*x^2 - 16*x^3))/(100 - 20*x + x^2),x]

[Out]

(-8*E^(17/4 + 2*x)*x^2)/(-10 + x)

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
gosper	\(-\frac {8 x^{2} {\mathrm e}^{\frac {17}{4}+2 x}}{x -10}\)	\(17\)
risch	\(-\frac {8 x^{2} {\mathrm e}^{\frac {17}{4}+2 x}}{x -10}\)	\(17\)
norman	\(-\frac {8 x^{2} {\mathrm e}^{2} {\mathrm e}^{2 x +\frac {9}{4}}}{x -10}\)	\(19\)
parallelrisch	\(-\frac {8 x^{2} {\mathrm e}^{2} {\mathrm e}^{2 x +\frac {9}{4}}}{x -10}\)	\(19\)
derivativedivides	\(4 \,{\mathrm e}^{2} \left (-\frac {400 \,{\mathrm e}^{2 x +\frac {9}{4}}}{2 x -20}-\frac {71 \,{\mathrm e}^{2 x +\frac {9}{4}}}{4}-{\mathrm e}^{2 x +\frac {9}{4}} \left (2 x +\frac {9}{4}\right )\right )\)	\(42\)
default	\(8 \,{\mathrm e}^{2} \left (-\frac {200 \,{\mathrm e}^{2 x +\frac {9}{4}}}{2 x -20}-\frac {71 \,{\mathrm e}^{2 x +\frac {9}{4}}}{8}-\frac {{\mathrm e}^{2 x +\frac {9}{4}} \left (2 x +\frac {9}{4}\right )}{2}\right )\)	\(42\)

[In]

int((-16*x^3+152*x^2+160*x)*exp(2)*exp(2*x+9/4)/(x^2-20*x+100),x,method=_RETURNVERBOSE)

[Out]

-8*x^2*exp(17/4+2*x)/(x-10)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \]

[In]

integrate((-16*x^3+152*x^2+160*x)*exp(2)*exp(2*x+9/4)/(x^2-20*x+100),x, algorithm="fricas")

[Out]

-8*x^2*e^(2*x + 17/4)/(x - 10)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=- \frac {8 x^{2} e^{2} e^{2 x + \frac {9}{4}}}{x - 10} \]

[In]

integrate((-16*x**3+152*x**2+160*x)*exp(2)*exp(2*x+9/4)/(x**2-20*x+100),x)

[Out]

-8*x**2*exp(2)*exp(2*x + 9/4)/(x - 10)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \]

[In]

integrate((-16*x^3+152*x^2+160*x)*exp(2)*exp(2*x+9/4)/(x^2-20*x+100),x, algorithm="maxima")

[Out]

-8*x^2*e^(2*x + 17/4)/(x - 10)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \]

[In]

integrate((-16*x^3+152*x^2+160*x)*exp(2)*exp(2*x+9/4)/(x^2-20*x+100),x, algorithm="giac")

[Out]

-8*x^2*e^(2*x + 17/4)/(x - 10)

Mupad [B] (verification not implemented)

Time = 9.50 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+\frac {1}{4} (9+8 x)} \left (160 x+152 x^2-16 x^3\right )}{100-20 x+x^2} \, dx=-\frac {8\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{17/4}}{x-10} \]

[In]

int((exp(2)*exp(2*x + 9/4)*(160*x + 152*x^2 - 16*x^3))/(x^2 - 20*x + 100),x)

[Out]

-(8*x^2*exp(2*x)*exp(17/4))/(x - 10)