3.22.22 \(\int \frac {e^5 (-6-12 x-6 x^2+\frac {7 x (2 x^2+2 x^3+x^4)}{e^5})}{7 x (x^2+2 x^3+x^4)} \, dx\)

Optimal. Leaf size=20 \[ -4+\frac {3 e^5}{7 x^2}+x-\frac {1}{1+x} \]

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Rubi [A]  time = 0.08, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 1585, 27, 1620} \begin {gather*} \frac {3 e^5}{7 x^2}+x-\frac {1}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(-6 - 12*x - 6*x^2 + (7*x*(2*x^2 + 2*x^3 + x^4))/E^5))/(7*x*(x^2 + 2*x^3 + x^4)),x]

[Out]

(3*E^5)/(7*x^2) + x - (1 + x)^(-1)

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 1585

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

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{7} e^5 \int \frac {-6-12 x-6 x^2+\frac {7 x \left (2 x^2+2 x^3+x^4\right )}{e^5}}{x \left (x^2+2 x^3+x^4\right )} \, dx\\ &=\frac {1}{7} e^5 \int \frac {-6-12 x-6 x^2+\frac {7 x \left (2 x^2+2 x^3+x^4\right )}{e^5}}{x^3 \left (1+2 x+x^2\right )} \, dx\\ &=\frac {1}{7} e^5 \int \frac {-6-12 x-6 x^2+\frac {7 x \left (2 x^2+2 x^3+x^4\right )}{e^5}}{x^3 (1+x)^2} \, dx\\ &=\frac {1}{7} e^5 \int \left (\frac {7}{e^5}-\frac {6}{x^3}+\frac {7}{e^5 (1+x)^2}\right ) \, dx\\ &=\frac {3 e^5}{7 x^2}+x-\frac {1}{1+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.95 \begin {gather*} \frac {3 e^5}{7 x^2}+x-\frac {1}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-6 - 12*x - 6*x^2 + (7*x*(2*x^2 + 2*x^3 + x^4))/E^5))/(7*x*(x^2 + 2*x^3 + x^4)),x]

[Out]

(3*E^5)/(7*x^2) + x - (1 + x)^(-1)

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fricas [A]  time = 0.79, size = 40, normalized size = 2.00 \begin {gather*} \frac {{\left ({\left (x^{4} + x^{3} - x^{2}\right )} e^{\left (\log \relax (7) - 5\right )} + 3 \, x + 3\right )} e^{\left (-\log \relax (7) + 5\right )}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+2*x^3+2*x^2)*exp(log(x)+log(7)-5)-6*x^2-12*x-6)/(x^4+2*x^3+x^2)/exp(log(x)+log(7)-5),x, algori
thm="fricas")

[Out]

((x^4 + x^3 - x^2)*e^(log(7) - 5) + 3*x + 3)*e^(-log(7) + 5)/(x^3 + x^2)

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giac [A]  time = 0.26, size = 16, normalized size = 0.80 \begin {gather*} x - \frac {1}{x + 1} + \frac {3 \, e^{5}}{7 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+2*x^3+2*x^2)*exp(log(x)+log(7)-5)-6*x^2-12*x-6)/(x^4+2*x^3+x^2)/exp(log(x)+log(7)-5),x, algori
thm="giac")

[Out]

x - 1/(x + 1) + 3/7*e^5/x^2

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maple [A]  time = 0.06, size = 17, normalized size = 0.85

method	result	size
default	\(x -\frac {1}{x +1}+\frac {3 \,{\mathrm e}^{5}}{7 x^{2}}\)	\(17\)
risch	\(x +\frac {{\mathrm e}^{5} \left (-7 x^{2} {\mathrm e}^{-5}+3 x +3\right )}{7 \left (x +1\right ) x^{2}}\)	\(27\)
norman	\(\frac {x^{4}-2 x^{2}+\frac {3 x \,{\mathrm e}^{5}}{7}+\frac {3 \,{\mathrm e}^{5}}{7}}{\left (x +1\right ) x^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4+2*x^3+2*x^2)*exp(ln(x)+ln(7)-5)-6*x^2-12*x-6)/(x^4+2*x^3+x^2)/exp(ln(x)+ln(7)-5),x,method=_RETURNVER
BOSE)

[Out]

x-1/(x+1)+3/7*exp(5)/x^2

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maxima [A]  time = 0.37, size = 42, normalized size = 2.10 \begin {gather*} \frac {1}{7} \, {\left (7 \, x e^{\left (-5\right )} - \frac {7 \, x^{2} - 3 \, x e^{5} - 3 \, e^{5}}{x^{3} e^{5} + x^{2} e^{5}}\right )} e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+2*x^3+2*x^2)*exp(log(x)+log(7)-5)-6*x^2-12*x-6)/(x^4+2*x^3+x^2)/exp(log(x)+log(7)-5),x, algori
thm="maxima")

[Out]

1/7*(7*x*e^(-5) - (7*x^2 - 3*x*e^5 - 3*e^5)/(x^3*e^5 + x^2*e^5))*e^5

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mupad [B]  time = 0.10, size = 16, normalized size = 0.80 \begin {gather*} x-\frac {1}{x+1}+\frac {3\,{\mathrm {e}}^5}{7\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5 - log(x) - log(7))*(12*x - exp(log(7) + log(x) - 5)*(2*x^2 + 2*x^3 + x^4) + 6*x^2 + 6))/(x^2 + 2*x
^3 + x^4),x)

[Out]

x - 1/(x + 1) + (3*exp(5))/(7*x^2)

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sympy [A]  time = 0.39, size = 27, normalized size = 1.35 \begin {gather*} x + \frac {- 7 x^{2} + 3 x e^{5} + 3 e^{5}}{7 x^{3} + 7 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4+2*x**3+2*x**2)*exp(ln(x)+ln(7)-5)-6*x**2-12*x-6)/(x**4+2*x**3+x**2)/exp(ln(x)+ln(7)-5),x)

[Out]

x + (-7*x**2 + 3*x*exp(5) + 3*exp(5))/(7*x**3 + 7*x**2)

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