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

Optimal. Leaf size=21 \[ \frac {8 e^{\frac {17}{4}+2 x} x^2}{10-x} \]

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Rubi [A]  time = 0.23, antiderivative size = 42, normalized size of antiderivative = 2.00, number of steps used = 10, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {27, 1594, 2199, 2194, 2177, 2178, 2176} \begin {gather*} -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} \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 2176

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] &&  !$UseGamma === True

Rule 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

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 2199

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] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.90 \begin {gather*} -\frac {8 e^{\frac {17}{4}+2 x} x^2}{-10+x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.84, size = 16, normalized size = 0.76 \begin {gather*} -\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 16, normalized size = 0.76 \begin {gather*} -\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.09, size = 17, normalized size = 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	\(-\frac {8 x^{2} {\mathrm e}^{2} {\mathrm e}^{2 x +\frac {9}{4}}}{x -10}\)	\(19\)
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	\(8 \,{\mathrm e}^{2} \left (-\frac {200 \,{\mathrm e}^{2 x +\frac {9}{4}}}{2 x -20}-\frac {71 \,{\mathrm e}^{2 x +\frac {9}{4}}}{8}-\frac {{\mathrm e}^{2 x +\frac {9}{4}} \left (2 x +\frac {9}{4}\right )}{2}\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.39, size = 16, normalized size = 0.76 \begin {gather*} -\frac {8 \, x^{2} e^{\left (2 \, x + \frac {17}{4}\right )}}{x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.26, size = 16, normalized size = 0.76 \begin {gather*} -\frac {8\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{17/4}}{x-10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.95 \begin {gather*} - \frac {8 x^{2} e^{2} e^{2 x + \frac {9}{4}}}{x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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