3.50.64 \(\int \frac {2960 x+1865 x^2+300 x^3+15 x^4+e^x (640 x+480 x^2+90 x^3+5 x^4)}{64+16 x+x^2} \, dx\)

Optimal. Leaf size=22 \[ 5 x^2 \left (4+e^x+x+\frac {5+2 x}{8+x}\right ) \]

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Rubi [A]  time = 0.34, antiderivative size = 29, normalized size of antiderivative = 1.32, number of steps used = 20, number of rules used = 8, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.151, Rules used = {27, 6688, 12, 6742, 2196, 2176, 2194, 43} \begin {gather*} 5 x^3+5 e^x x^2+30 x^2-55 x-\frac {3520}{x+8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2960*x + 1865*x^2 + 300*x^3 + 15*x^4 + E^x*(640*x + 480*x^2 + 90*x^3 + 5*x^4))/(64 + 16*x + x^2),x]

[Out]

-55*x + 30*x^2 + 5*E^x*x^2 + 5*x^3 - 3520/(8 + 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 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 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 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 {2960 x+1865 x^2+300 x^3+15 x^4+e^x \left (640 x+480 x^2+90 x^3+5 x^4\right )}{(8+x)^2} \, dx\\ &=\int \frac {5 x \left (592+373 x+60 x^2+3 x^3+e^x (2+x) (8+x)^2\right )}{(8+x)^2} \, dx\\ &=5 \int \frac {x \left (592+373 x+60 x^2+3 x^3+e^x (2+x) (8+x)^2\right )}{(8+x)^2} \, dx\\ &=5 \int \left (e^x x (2+x)+\frac {592 x}{(8+x)^2}+\frac {373 x^2}{(8+x)^2}+\frac {60 x^3}{(8+x)^2}+\frac {3 x^4}{(8+x)^2}\right ) \, dx\\ &=5 \int e^x x (2+x) \, dx+15 \int \frac {x^4}{(8+x)^2} \, dx+300 \int \frac {x^3}{(8+x)^2} \, dx+1865 \int \frac {x^2}{(8+x)^2} \, dx+2960 \int \frac {x}{(8+x)^2} \, dx\\ &=5 \int \left (2 e^x x+e^x x^2\right ) \, dx+15 \int \left (192-16 x+x^2+\frac {4096}{(8+x)^2}-\frac {2048}{8+x}\right ) \, dx+300 \int \left (-16+x-\frac {512}{(8+x)^2}+\frac {192}{8+x}\right ) \, dx+1865 \int \left (1+\frac {64}{(8+x)^2}-\frac {16}{8+x}\right ) \, dx+2960 \int \left (-\frac {8}{(8+x)^2}+\frac {1}{8+x}\right ) \, dx\\ &=-55 x+30 x^2+5 x^3-\frac {3520}{8+x}+5 \int e^x x^2 \, dx+10 \int e^x x \, dx\\ &=-55 x+10 e^x x+30 x^2+5 e^x x^2+5 x^3-\frac {3520}{8+x}-10 \int e^x \, dx-10 \int e^x x \, dx\\ &=-10 e^x-55 x+30 x^2+5 e^x x^2+5 x^3-\frac {3520}{8+x}+10 \int e^x \, dx\\ &=-55 x+30 x^2+5 e^x x^2+5 x^3-\frac {3520}{8+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 25, normalized size = 1.14 \begin {gather*} 5 \left (-11 x+\left (6+e^x\right ) x^2+x^3-\frac {704}{8+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2960*x + 1865*x^2 + 300*x^3 + 15*x^4 + E^x*(640*x + 480*x^2 + 90*x^3 + 5*x^4))/(64 + 16*x + x^2),x]

[Out]

5*(-11*x + (6 + E^x)*x^2 + x^3 - 704/(8 + x))

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fricas [A]  time = 0.96, size = 37, normalized size = 1.68 \begin {gather*} \frac {5 \, {\left (x^{4} + 14 \, x^{3} + 37 \, x^{2} + {\left (x^{3} + 8 \, x^{2}\right )} e^{x} - 88 \, x - 704\right )}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^3+480*x^2+640*x)*exp(x)+15*x^4+300*x^3+1865*x^2+2960*x)/(x^2+16*x+64),x, algorithm="fri
cas")

[Out]

5*(x^4 + 14*x^3 + 37*x^2 + (x^3 + 8*x^2)*e^x - 88*x - 704)/(x + 8)

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giac [A]  time = 0.13, size = 38, normalized size = 1.73 \begin {gather*} \frac {5 \, {\left (x^{4} + x^{3} e^{x} + 14 \, x^{3} + 8 \, x^{2} e^{x} + 37 \, x^{2} - 88 \, x - 704\right )}}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^3+480*x^2+640*x)*exp(x)+15*x^4+300*x^3+1865*x^2+2960*x)/(x^2+16*x+64),x, algorithm="gia
c")

[Out]

5*(x^4 + x^3*e^x + 14*x^3 + 8*x^2*e^x + 37*x^2 - 88*x - 704)/(x + 8)

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maple [A]  time = 0.16, size = 29, normalized size = 1.32

method	result	size
default	\(-\frac {3520}{x +8}-55 x +30 x^{2}+5 x^{3}+5 \,{\mathrm e}^{x} x^{2}\)	\(29\)
risch	\(-\frac {3520}{x +8}-55 x +30 x^{2}+5 x^{3}+5 \,{\mathrm e}^{x} x^{2}\)	\(29\)
norman	\(\frac {185 x^{2}+70 x^{3}+5 x^{4}+40 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} x^{3}}{x +8}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^4+90*x^3+480*x^2+640*x)*exp(x)+15*x^4+300*x^3+1865*x^2+2960*x)/(x^2+16*x+64),x,method=_RETURNVERBOSE
)

[Out]

-3520/(x+8)-55*x+30*x^2+5*x^3+5*exp(x)*x^2

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maxima [A]  time = 0.42, size = 28, normalized size = 1.27 \begin {gather*} 5 \, x^{3} + 5 \, x^{2} e^{x} + 30 \, x^{2} - 55 \, x - \frac {3520}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^4+90*x^3+480*x^2+640*x)*exp(x)+15*x^4+300*x^3+1865*x^2+2960*x)/(x^2+16*x+64),x, algorithm="max
ima")

[Out]

5*x^3 + 5*x^2*e^x + 30*x^2 - 55*x - 3520/(x + 8)

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mupad [B]  time = 0.09, size = 26, normalized size = 1.18 \begin {gather*} x^2\,\left (5\,{\mathrm {e}}^x+30\right )-\frac {3520}{x+8}-55\,x+5\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2960*x + exp(x)*(640*x + 480*x^2 + 90*x^3 + 5*x^4) + 1865*x^2 + 300*x^3 + 15*x^4)/(16*x + x^2 + 64),x)

[Out]

x^2*(5*exp(x) + 30) - 3520/(x + 8) - 55*x + 5*x^3

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sympy [A]  time = 0.14, size = 26, normalized size = 1.18 \begin {gather*} 5 x^{3} + 5 x^{2} e^{x} + 30 x^{2} - 55 x - \frac {3520}{x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**4+90*x**3+480*x**2+640*x)*exp(x)+15*x**4+300*x**3+1865*x**2+2960*x)/(x**2+16*x+64),x)

[Out]

5*x**3 + 5*x**2*exp(x) + 30*x**2 - 55*x - 3520/(x + 8)

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