3.26.66 \(\int \frac {-900+14300 x+25500 x^2+18500 x^3+5000 x^4+e^{2 x} (200 x+800 x^2+1200 x^3+800 x^4+200 x^5)+e^x (-100+3200 x+9600 x^2+11000 x^3+5700 x^4+1000 x^5)}{1+3 x+3 x^2+x^3} \, dx\)

Optimal. Leaf size=19 \[ \left (-5+10 x \left (5+e^x+\frac {4}{1+x}\right )\right )^2 \]

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Rubi [B]  time = 0.58, antiderivative size = 72, normalized size of antiderivative = 3.79, number of steps used = 29, number of rules used = 11, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {6688, 12, 6742, 37, 43, 2196, 2176, 2194, 2199, 2177, 2178} \begin {gather*} 1000 e^x x^2+100 e^{2 x} x^2+\frac {7150 x^2}{(x+1)^2}+2500 x^2+700 e^x x+3500 x-800 e^x+\frac {800 e^x}{x+1}+\frac {15500}{x+1}-\frac {5550}{(x+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-900 + 14300*x + 25500*x^2 + 18500*x^3 + 5000*x^4 + E^(2*x)*(200*x + 800*x^2 + 1200*x^3 + 800*x^4 + 200*x
^5) + E^x*(-100 + 3200*x + 9600*x^2 + 11000*x^3 + 5700*x^4 + 1000*x^5))/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

-800*E^x + 3500*x + 700*E^x*x + 2500*x^2 + 1000*E^x*x^2 + 100*E^(2*x)*x^2 - 5550/(1 + x)^2 + (7150*x^2)/(1 + x
)^2 + 15500/(1 + x) + (800*E^x)/(1 + 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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 {100 \left (-9+143 x+255 x^2+185 x^3+50 x^4+2 e^{2 x} x (1+x)^4+e^x \left (-1+32 x+96 x^2+110 x^3+57 x^4+10 x^5\right )\right )}{(1+x)^3} \, dx\\ &=100 \int \frac {-9+143 x+255 x^2+185 x^3+50 x^4+2 e^{2 x} x (1+x)^4+e^x \left (-1+32 x+96 x^2+110 x^3+57 x^4+10 x^5\right )}{(1+x)^3} \, dx\\ &=100 \int \left (-\frac {9}{(1+x)^3}+\frac {143 x}{(1+x)^3}+\frac {255 x^2}{(1+x)^3}+\frac {185 x^3}{(1+x)^3}+\frac {50 x^4}{(1+x)^3}+2 e^{2 x} x (1+x)+\frac {e^x \left (-1+33 x+63 x^2+47 x^3+10 x^4\right )}{(1+x)^2}\right ) \, dx\\ &=\frac {450}{(1+x)^2}+100 \int \frac {e^x \left (-1+33 x+63 x^2+47 x^3+10 x^4\right )}{(1+x)^2} \, dx+200 \int e^{2 x} x (1+x) \, dx+5000 \int \frac {x^4}{(1+x)^3} \, dx+14300 \int \frac {x}{(1+x)^3} \, dx+18500 \int \frac {x^3}{(1+x)^3} \, dx+25500 \int \frac {x^2}{(1+x)^3} \, dx\\ &=\frac {450}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+100 \int \left (-e^x+27 e^x x+10 e^x x^2-\frac {8 e^x}{(1+x)^2}+\frac {8 e^x}{1+x}\right ) \, dx+200 \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx+5000 \int \left (-3+x+\frac {1}{(1+x)^3}-\frac {4}{(1+x)^2}+\frac {6}{1+x}\right ) \, dx+18500 \int \left (1-\frac {1}{(1+x)^3}+\frac {3}{(1+x)^2}-\frac {3}{1+x}\right ) \, dx+25500 \int \left (\frac {1}{(1+x)^3}-\frac {2}{(1+x)^2}+\frac {1}{1+x}\right ) \, dx\\ &=3500 x+2500 x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}-100 \int e^x \, dx+200 \int e^{2 x} x \, dx+200 \int e^{2 x} x^2 \, dx-800 \int \frac {e^x}{(1+x)^2} \, dx+800 \int \frac {e^x}{1+x} \, dx+1000 \int e^x x^2 \, dx+2700 \int e^x x \, dx\\ &=-100 e^x+3500 x+2700 e^x x+100 e^{2 x} x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x}+\frac {800 \text {Ei}(1+x)}{e}-100 \int e^{2 x} \, dx-200 \int e^{2 x} x \, dx-800 \int \frac {e^x}{1+x} \, dx-2000 \int e^x x \, dx-2700 \int e^x \, dx\\ &=-2800 e^x-50 e^{2 x}+3500 x+700 e^x x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x}+100 \int e^{2 x} \, dx+2000 \int e^x \, dx\\ &=-800 e^x+3500 x+700 e^x x+2500 x^2+1000 e^x x^2+100 e^{2 x} x^2-\frac {5550}{(1+x)^2}+\frac {7150 x^2}{(1+x)^2}+\frac {15500}{1+x}+\frac {800 e^x}{1+x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.25, size = 52, normalized size = 2.74 \begin {gather*} 100 \left (35 x+25 x^2+e^{2 x} x^2+\frac {4 (7+3 x)}{(1+x)^2}+\frac {e^x x \left (-1+17 x+10 x^2\right )}{1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-900 + 14300*x + 25500*x^2 + 18500*x^3 + 5000*x^4 + E^(2*x)*(200*x + 800*x^2 + 1200*x^3 + 800*x^4 +
200*x^5) + E^x*(-100 + 3200*x + 9600*x^2 + 11000*x^3 + 5700*x^4 + 1000*x^5))/(1 + 3*x + 3*x^2 + x^3),x]

[Out]

100*(35*x + 25*x^2 + E^(2*x)*x^2 + (4*(7 + 3*x))/(1 + x)^2 + (E^x*x*(-1 + 17*x + 10*x^2))/(1 + x))

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fricas [B]  time = 0.98, size = 71, normalized size = 3.74 \begin {gather*} \frac {100 \, {\left (25 \, x^{4} + 85 \, x^{3} + 95 \, x^{2} + {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + {\left (10 \, x^{4} + 27 \, x^{3} + 16 \, x^{2} - x\right )} e^{x} + 47 \, x + 28\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5+800*x^4+1200*x^3+800*x^2+200*x)*exp(x)^2+(1000*x^5+5700*x^4+11000*x^3+9600*x^2+3200*x-100)
*exp(x)+5000*x^4+18500*x^3+25500*x^2+14300*x-900)/(x^3+3*x^2+3*x+1),x, algorithm="fricas")

[Out]

100*(25*x^4 + 85*x^3 + 95*x^2 + (x^4 + 2*x^3 + x^2)*e^(2*x) + (10*x^4 + 27*x^3 + 16*x^2 - x)*e^x + 47*x + 28)/
(x^2 + 2*x + 1)

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giac [B]  time = 0.21, size = 83, normalized size = 4.37 \begin {gather*} \frac {100 \, {\left (x^{4} e^{\left (2 \, x\right )} + 10 \, x^{4} e^{x} + 25 \, x^{4} + 2 \, x^{3} e^{\left (2 \, x\right )} + 27 \, x^{3} e^{x} + 85 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} e^{x} + 95 \, x^{2} - x e^{x} + 47 \, x + 28\right )}}{x^{2} + 2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5+800*x^4+1200*x^3+800*x^2+200*x)*exp(x)^2+(1000*x^5+5700*x^4+11000*x^3+9600*x^2+3200*x-100)
*exp(x)+5000*x^4+18500*x^3+25500*x^2+14300*x-900)/(x^3+3*x^2+3*x+1),x, algorithm="giac")

[Out]

100*(x^4*e^(2*x) + 10*x^4*e^x + 25*x^4 + 2*x^3*e^(2*x) + 27*x^3*e^x + 85*x^3 + x^2*e^(2*x) + 16*x^2*e^x + 95*x
^2 - x*e^x + 47*x + 28)/(x^2 + 2*x + 1)

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maple [B]  time = 0.08, size = 55, normalized size = 2.89

method	result	size
risch	\(2500 x^{2}+3500 x +\frac {1200 x +2800}{x^{2}+2 x +1}+100 \,{\mathrm e}^{2 x} x^{2}+\frac {100 x \left (10 x^{2}+17 x -1\right ) {\mathrm e}^{x}}{x +1}\)	\(55\)
default	\(\frac {1600}{\left (x +1\right )^{2}}+\frac {1200}{x +1}+3500 x +2500 x^{2}+\frac {800 \,{\mathrm e}^{x}}{x +1}+1000 \,{\mathrm e}^{x} x^{2}+700 \,{\mathrm e}^{x} x -800 \,{\mathrm e}^{x}+100 \,{\mathrm e}^{2 x} x^{2}\)	\(58\)
norman	\(\frac {-14300 x +8500 x^{3}+2500 x^{4}-100 \,{\mathrm e}^{x} x +1600 \,{\mathrm e}^{x} x^{2}+2700 \,{\mathrm e}^{x} x^{3}+1000 \,{\mathrm e}^{x} x^{4}+100 \,{\mathrm e}^{2 x} x^{2}+200 \,{\mathrm e}^{2 x} x^{3}+100 \,{\mathrm e}^{2 x} x^{4}-6700}{\left (x +1\right )^{2}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((200*x^5+800*x^4+1200*x^3+800*x^2+200*x)*exp(x)^2+(1000*x^5+5700*x^4+11000*x^3+9600*x^2+3200*x-100)*exp(x
)+5000*x^4+18500*x^3+25500*x^2+14300*x-900)/(x^3+3*x^2+3*x+1),x,method=_RETURNVERBOSE)

[Out]

2500*x^2+3500*x+(1200*x+2800)/(x^2+2*x+1)+100*exp(2*x)*x^2+100*x*(10*x^2+17*x-1)/(x+1)*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2500 \, x^{2} + 3500 \, x + \frac {2500 \, {\left (8 \, x + 7\right )}}{x^{2} + 2 \, x + 1} - \frac {9250 \, {\left (6 \, x + 5\right )}}{x^{2} + 2 \, x + 1} + \frac {12750 \, {\left (4 \, x + 3\right )}}{x^{2} + 2 \, x + 1} - \frac {7150 \, {\left (2 \, x + 1\right )}}{x^{2} + 2 \, x + 1} + \frac {100 \, {\left ({\left (x^{3} + x^{2}\right )} e^{\left (2 \, x\right )} + {\left (10 \, x^{3} + 17 \, x^{2} - x\right )} e^{x}\right )}}{x + 1} + \frac {100 \, e^{\left (-1\right )} E_{3}\left (-x - 1\right )}{{\left (x + 1\right )}^{2}} + \frac {450}{x^{2} + 2 \, x + 1} + 100 \, \int \frac {e^{x}}{x^{3} + 3 \, x^{2} + 3 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5+800*x^4+1200*x^3+800*x^2+200*x)*exp(x)^2+(1000*x^5+5700*x^4+11000*x^3+9600*x^2+3200*x-100)
*exp(x)+5000*x^4+18500*x^3+25500*x^2+14300*x-900)/(x^3+3*x^2+3*x+1),x, algorithm="maxima")

[Out]

2500*x^2 + 3500*x + 2500*(8*x + 7)/(x^2 + 2*x + 1) - 9250*(6*x + 5)/(x^2 + 2*x + 1) + 12750*(4*x + 3)/(x^2 + 2
*x + 1) - 7150*(2*x + 1)/(x^2 + 2*x + 1) + 100*((x^3 + x^2)*e^(2*x) + (10*x^3 + 17*x^2 - x)*e^x)/(x + 1) + 100
*e^(-1)*exp_integral_e(3, -x - 1)/(x + 1)^2 + 450/(x^2 + 2*x + 1) + 100*integrate(e^x/(x^3 + 3*x^2 + 3*x + 1),
x)

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mupad [B]  time = 1.57, size = 49, normalized size = 2.58 \begin {gather*} x^2\,\left (100\,{\mathrm {e}}^{2\,x}+1000\,{\mathrm {e}}^x+2500\right )-800\,{\mathrm {e}}^x+x\,\left (700\,{\mathrm {e}}^x+3500\right )+\frac {800\,{\mathrm {e}}^x+x\,\left (800\,{\mathrm {e}}^x+1200\right )+2800}{{\left (x+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((14300*x + exp(x)*(3200*x + 9600*x^2 + 11000*x^3 + 5700*x^4 + 1000*x^5 - 100) + exp(2*x)*(200*x + 800*x^2
+ 1200*x^3 + 800*x^4 + 200*x^5) + 25500*x^2 + 18500*x^3 + 5000*x^4 - 900)/(3*x + 3*x^2 + x^3 + 1),x)

[Out]

x^2*(100*exp(2*x) + 1000*exp(x) + 2500) - 800*exp(x) + x*(700*exp(x) + 3500) + (800*exp(x) + x*(800*exp(x) + 1
200) + 2800)/(x + 1)^2

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sympy [B]  time = 0.19, size = 56, normalized size = 2.95 \begin {gather*} 2500 x^{2} + 3500 x + \frac {1200 x + 2800}{x^{2} + 2 x + 1} + \frac {\left (100 x^{3} + 100 x^{2}\right ) e^{2 x} + \left (1000 x^{3} + 1700 x^{2} - 100 x\right ) e^{x}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x**5+800*x**4+1200*x**3+800*x**2+200*x)*exp(x)**2+(1000*x**5+5700*x**4+11000*x**3+9600*x**2+32
00*x-100)*exp(x)+5000*x**4+18500*x**3+25500*x**2+14300*x-900)/(x**3+3*x**2+3*x+1),x)

[Out]

2500*x**2 + 3500*x + (1200*x + 2800)/(x**2 + 2*x + 1) + ((100*x**3 + 100*x**2)*exp(2*x) + (1000*x**3 + 1700*x*
*2 - 100*x)*exp(x))/(x + 1)

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