∫ e8(−75−50x−25x3+x2(108x2+108x3+36x4+4x5) e8 ) _______________________________________________________________

3.47.1 \(\int \frac {e^8 (-75-50 x-25 x^3+\frac {x^2 (108 x^2+108 x^3+36 x^4+4 x^5)}{e^8})}{x^2 (54 x+54 x^2+18 x^3+2 x^4)} \, dx\)

Optimal. Leaf size=24 \[ x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \]

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Rubi [A] time = 0.22, antiderivative size = 48, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 6688, 1620} \begin {gather*} x^2+\frac {25 e^8}{36 x^2}+\frac {25 e^8}{54 (x+3)}+\frac {125 e^8}{18 (x+3)^2}-\frac {25 e^8}{54 x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(E^8*(-75 - 50*x - 25*x^3 + (x^2*(108*x^2 + 108*x^3 + 36*x^4 + 4*x^5))/E^8))/(x^2*(54*x + 54*x^2 + 18*x^3

+ 2*x^4)),x]

[Out]

(25*E^8)/(36*x^2) - (25*E^8)/(54*x) + x^2 + (125*E^8)/(18*(3 + x)^2) + (25*E^8)/(54*(3 + 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)

^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&

GtQ[Expon[Px, x], 2]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^8 \int \frac {-75-50 x-25 x^3+\frac {x^2 \left (108 x^2+108 x^3+36 x^4+4 x^5\right )}{e^8}}{x^2 \left (54 x+54 x^2+18 x^3+2 x^4\right )} \, dx\\ &=e^8 \int \left (\frac {2 x}{e^8}-\frac {25 \left (3+2 x+x^3\right )}{2 x^3 (3+x)^3}\right ) \, dx\\ &=x^2-\frac {1}{2} \left (25 e^8\right ) \int \frac {3+2 x+x^3}{x^3 (3+x)^3} \, dx\\ &=x^2-\frac {1}{2} \left (25 e^8\right ) \int \left (\frac {1}{9 x^3}-\frac {1}{27 x^2}+\frac {10}{9 (3+x)^3}+\frac {1}{27 (3+x)^2}\right ) \, dx\\ &=\frac {25 e^8}{36 x^2}-\frac {25 e^8}{54 x}+x^2+\frac {125 e^8}{18 (3+x)^2}+\frac {25 e^8}{54 (3+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.02, size = 24, normalized size = 1.00 \begin {gather*} x^2+\frac {25 e^8 \left (1+x^2\right )}{4 x^2 (3+x)^2} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^8*(-75 - 50*x - 25*x^3 + (x^2*(108*x^2 + 108*x^3 + 36*x^4 + 4*x^5))/E^8))/(x^2*(54*x + 54*x^2 + 1

8*x^3 + 2*x^4)),x]

[Out]

x^2 + (25*E^8*(1 + x^2))/(4*x^2*(3 + x)^2)

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fricas [A] time = 0.54, size = 43, normalized size = 1.79 \begin {gather*} \frac {4 \, x^{6} + 24 \, x^{5} + 36 \, x^{4} + 25 \, {\left (x^{2} + 1\right )} e^{8}}{4 \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/(2*x^4+18*x^3+54*x^2+54*x)/exp(log(x

)-4)^2,x, algorithm="fricas")

[Out]

1/4*(4*x^6 + 24*x^5 + 36*x^4 + 25*(x^2 + 1)*e^8)/(x^4 + 6*x^3 + 9*x^2)

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giac [A] time = 0.12, size = 24, normalized size = 1.00 \begin {gather*} x^{2} + \frac {25 \, {\left (x^{2} e^{8} + e^{8}\right )}}{4 \, {\left (x^{2} + 3 \, x\right )}^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/(2*x^4+18*x^3+54*x^2+54*x)/exp(log(x

)-4)^2,x, algorithm="giac")

[Out]

x^2 + 25/4*(x^2*e^8 + e^8)/(x^2 + 3*x)^2

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maple [A] time = 0.07, size = 28, normalized size = 1.17


method	result	size

risch	\(x^{2}+\frac {{\mathrm e}^{8} \left (\frac {25 x^{2}}{4}+\frac {25}{4}\right )}{x^{2} \left (x^{2}+6 x +9\right )}\)	\(28\)
default	\(x^{2}-\frac {25 \,{\mathrm e}^{8} \left (-\frac {5}{9 \left (3+x \right )^{2}}-\frac {1}{27 \left (3+x \right )}-\frac {1}{18 x^{2}}+\frac {1}{27 x}\right )}{2}\)	\(34\)
norman	\(\frac {\left (x^{6} {\mathrm e}^{4}-54 x^{3} {\mathrm e}^{4}+\frac {{\mathrm e}^{4} \left (25 \,{\mathrm e}^{8}-324\right ) x^{2}}{4}+\frac {25 \,{\mathrm e}^{12}}{4}+6 x^{5} {\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{x^{2} \left (3+x \right )^{2}}\)	\(56\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(ln(x)-4)^2-25*x^3-50*x-75)/(2*x^4+18*x^3+54*x^2+54*x)/exp(ln(x)-4)^2,x

,method=_RETURNVERBOSE)

[Out]

x^2+exp(8)*(25/4*x^2+25/4)/x^2/(x^2+6*x+9)

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maxima [A] time = 0.36, size = 35, normalized size = 1.46 \begin {gather*} \frac {1}{4} \, {\left (4 \, x^{2} e^{\left (-8\right )} + \frac {25 \, {\left (x^{2} + 1\right )}}{x^{4} + 6 \, x^{3} + 9 \, x^{2}}\right )} e^{8} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^5+36*x^4+108*x^3+108*x^2)*exp(log(x)-4)^2-25*x^3-50*x-75)/(2*x^4+18*x^3+54*x^2+54*x)/exp(log(x

)-4)^2,x, algorithm="maxima")

[Out]

1/4*(4*x^2*e^(-8) + 25*(x^2 + 1)/(x^4 + 6*x^3 + 9*x^2))*e^8

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mupad [B] time = 3.36, size = 25, normalized size = 1.04 \begin {gather*} x^2+\frac {\frac {25\,{\mathrm {e}}^8\,x^2}{4}+\frac {25\,{\mathrm {e}}^8}{4}}{x^2\,{\left (x+3\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8 - 2*log(x))*(50*x - exp(2*log(x) - 8)*(108*x^2 + 108*x^3 + 36*x^4 + 4*x^5) + 25*x^3 + 75))/(54*x +

54*x^2 + 18*x^3 + 2*x^4),x)

[Out]

x^2 + ((25*exp(8))/4 + (25*x^2*exp(8))/4)/(x^2*(x + 3)^2)

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sympy [A] time = 0.24, size = 31, normalized size = 1.29 \begin {gather*} x^{2} + \frac {25 x^{2} e^{8} + 25 e^{8}}{4 x^{4} + 24 x^{3} + 36 x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**5+36*x**4+108*x**3+108*x**2)*exp(ln(x)-4)**2-25*x**3-50*x-75)/(2*x**4+18*x**3+54*x**2+54*x)/e

xp(ln(x)-4)**2,x)

[Out]

x**2 + (25*x**2*exp(8) + 25*exp(8))/(4*x**4 + 24*x**3 + 36*x**2)

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