3.96.56 \(\int \frac {126-153 x-18 x^2}{e (64+160 x+160 x^2+80 x^3+20 x^4+2 x^5)} \, dx\)

Optimal. Leaf size=16 \[ \frac {9 x (7+x)}{2 e (2+x)^4} \]

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Rubi [B]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 2.19, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 2074} \begin {gather*} \frac {9}{2 e (x+2)^2}+\frac {27}{2 e (x+2)^3}-\frac {45}{e (x+2)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(126 - 153*x - 18*x^2)/(E*(64 + 160*x + 160*x^2 + 80*x^3 + 20*x^4 + 2*x^5)),x]

[Out]

-45/(E*(2 + x)^4) + 27/(2*E*(2 + x)^3) + 9/(2*E*(2 + x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {126-153 x-18 x^2}{64+160 x+160 x^2+80 x^3+20 x^4+2 x^5} \, dx}{e}\\ &=\frac {\int \left (\frac {180}{(2+x)^5}-\frac {81}{2 (2+x)^4}-\frac {9}{(2+x)^3}\right ) \, dx}{e}\\ &=-\frac {45}{e (2+x)^4}+\frac {27}{2 e (2+x)^3}+\frac {9}{2 e (2+x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {9 x (7+x)}{2 e (2+x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(126 - 153*x - 18*x^2)/(E*(64 + 160*x + 160*x^2 + 80*x^3 + 20*x^4 + 2*x^5)),x]

[Out]

(9*x*(7 + x))/(2*E*(2 + x)^4)

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fricas [B]  time = 0.59, size = 31, normalized size = 1.94 \begin {gather*} \frac {9 \, {\left (x^{2} + 7 \, x\right )} e^{\left (-1\right )}}{2 \, {\left (x^{4} + 8 \, x^{3} + 24 \, x^{2} + 32 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x^2-153*x+126)/(2*x^5+20*x^4+80*x^3+160*x^2+160*x+64)/exp(1),x, algorithm="fricas")

[Out]

9/2*(x^2 + 7*x)*e^(-1)/(x^4 + 8*x^3 + 24*x^2 + 32*x + 16)

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giac [A]  time = 0.16, size = 16, normalized size = 1.00 \begin {gather*} \frac {9 \, {\left (x^{2} + 7 \, x\right )} e^{\left (-1\right )}}{2 \, {\left (x + 2\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x^2-153*x+126)/(2*x^5+20*x^4+80*x^3+160*x^2+160*x+64)/exp(1),x, algorithm="giac")

[Out]

9/2*(x^2 + 7*x)*e^(-1)/(x + 2)^4

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maple [A]  time = 0.05, size = 24, normalized size = 1.50

method	result	size
norman	\(\frac {\frac {9 x^{2} {\mathrm e}^{-1}}{2}+\frac {63 \,{\mathrm e}^{-1} x}{2}}{\left (2+x \right )^{4}}\)	\(24\)
default	\(\frac {9 \,{\mathrm e}^{-1} \left (-\frac {10}{\left (2+x \right )^{4}}+\frac {3}{\left (2+x \right )^{3}}+\frac {1}{\left (2+x \right )^{2}}\right )}{2}\)	\(27\)
gosper	\(\frac {9 \left (x +7\right ) x \,{\mathrm e}^{-1}}{2 \left (x^{4}+8 x^{3}+24 x^{2}+32 x +16\right )}\)	\(31\)
risch	\(\frac {{\mathrm e}^{-1} \left (\frac {9}{2} x^{2}+\frac {63}{2} x \right )}{x^{4}+8 x^{3}+24 x^{2}+32 x +16}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-18*x^2-153*x+126)/(2*x^5+20*x^4+80*x^3+160*x^2+160*x+64)/exp(1),x,method=_RETURNVERBOSE)

[Out]

(9/2*x^2/exp(1)+63/2*x/exp(1))/(2+x)^4

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maxima [B]  time = 0.37, size = 31, normalized size = 1.94 \begin {gather*} \frac {9 \, {\left (x^{2} + 7 \, x\right )} e^{\left (-1\right )}}{2 \, {\left (x^{4} + 8 \, x^{3} + 24 \, x^{2} + 32 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x^2-153*x+126)/(2*x^5+20*x^4+80*x^3+160*x^2+160*x+64)/exp(1),x, algorithm="maxima")

[Out]

9/2*(x^2 + 7*x)*e^(-1)/(x^4 + 8*x^3 + 24*x^2 + 32*x + 16)

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mupad [B]  time = 0.15, size = 28, normalized size = 1.75 \begin {gather*} \frac {9\,{\mathrm {e}}^{-1}}{2\,{\left (x+2\right )}^2}+\frac {27\,{\mathrm {e}}^{-1}}{2\,{\left (x+2\right )}^3}-\frac {45\,{\mathrm {e}}^{-1}}{{\left (x+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*(153*x + 18*x^2 - 126))/(160*x + 160*x^2 + 80*x^3 + 20*x^4 + 2*x^5 + 64),x)

[Out]

(9*exp(-1))/(2*(x + 2)^2) + (27*exp(-1))/(2*(x + 2)^3) - (45*exp(-1))/(x + 2)^4

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sympy [B]  time = 0.22, size = 48, normalized size = 3.00 \begin {gather*} - \frac {- 9 x^{2} - 63 x}{2 e x^{4} + 16 e x^{3} + 48 e x^{2} + 64 e x + 32 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*x**2-153*x+126)/(2*x**5+20*x**4+80*x**3+160*x**2+160*x+64)/exp(1),x)

[Out]

-(-9*x**2 - 63*x)/(2*E*x**4 + 16*E*x**3 + 48*E*x**2 + 64*E*x + 32*E)

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