3.61.34 \(\int \frac {e^4 (18+72 x-63 x^2+9 x^3)}{32-32 x-24 x^2+32 x^3-8 x^5+2 x^6} \, dx\)

Optimal. Leaf size=26 \[ 1+\frac {9 e^4 (-4+x)}{(2-x) \left (-8+4 x^2\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 2074, 639, 207} \begin {gather*} \frac {9 e^4 (x+3)}{4 \left (2-x^2\right )}-\frac {9 e^4}{4 (2-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(18 + 72*x - 63*x^2 + 9*x^3))/(32 - 32*x - 24*x^2 + 32*x^3 - 8*x^5 + 2*x^6),x]

[Out]

(-9*E^4)/(4*(2 - x)) + (9*E^4*(3 + x))/(4*(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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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} &=e^4 \int \frac {18+72 x-63 x^2+9 x^3}{32-32 x-24 x^2+32 x^3-8 x^5+2 x^6} \, dx\\ &=e^4 \int \left (-\frac {9}{4 (-2+x)^2}+\frac {9 (2+3 x)}{2 \left (-2+x^2\right )^2}+\frac {9}{4 \left (-2+x^2\right )}\right ) \, dx\\ &=-\frac {9 e^4}{4 (2-x)}+\frac {1}{4} \left (9 e^4\right ) \int \frac {1}{-2+x^2} \, dx+\frac {1}{2} \left (9 e^4\right ) \int \frac {2+3 x}{\left (-2+x^2\right )^2} \, dx\\ &=-\frac {9 e^4}{4 (2-x)}+\frac {9 e^4 (3+x)}{4 \left (2-x^2\right )}-\frac {9 e^4 \tanh ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1}{4} \left (9 e^4\right ) \int \frac {1}{-2+x^2} \, dx\\ &=-\frac {9 e^4}{4 (2-x)}+\frac {9 e^4 (3+x)}{4 \left (2-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.04 \begin {gather*} \frac {9 e^4 (4-x)}{4 \left (4-2 x-2 x^2+x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(18 + 72*x - 63*x^2 + 9*x^3))/(32 - 32*x - 24*x^2 + 32*x^3 - 8*x^5 + 2*x^6),x]

[Out]

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

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fricas [A]  time = 0.71, size = 22, normalized size = 0.85 \begin {gather*} -\frac {9 \, {\left (x - 4\right )} e^{4}}{4 \, {\left (x^{3} - 2 \, x^{2} - 2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^3-63*x^2+72*x+18)*exp(2)^2/(2*x^6-8*x^5+32*x^3-24*x^2-32*x+32),x, algorithm="fricas")

[Out]

-9/4*(x - 4)*e^4/(x^3 - 2*x^2 - 2*x + 4)

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giac [A]  time = 0.13, size = 22, normalized size = 0.85 \begin {gather*} -\frac {9 \, {\left (x - 4\right )} e^{4}}{4 \, {\left (x^{3} - 2 \, x^{2} - 2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^3-63*x^2+72*x+18)*exp(2)^2/(2*x^6-8*x^5+32*x^3-24*x^2-32*x+32),x, algorithm="giac")

[Out]

-9/4*(x - 4)*e^4/(x^3 - 2*x^2 - 2*x + 4)

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maple [A]  time = 0.04, size = 24, normalized size = 0.92

method	result	size
risch	\(\frac {{\mathrm e}^{4} \left (-\frac {9 x}{4}+9\right )}{x^{3}-2 x^{2}-2 x +4}\)	\(24\)
gosper	\(-\frac {9 \left (x -4\right ) {\mathrm e}^{4}}{4 \left (x^{3}-2 x^{2}-2 x +4\right )}\)	\(25\)
default	\(\frac {9 \,{\mathrm e}^{4} \left (\frac {-3-x}{2 x^{2}-4}+\frac {1}{2 x -4}\right )}{2}\)	\(29\)
norman	\(\frac {-\frac {9 x \,{\mathrm e}^{4}}{4}+9 \,{\mathrm e}^{4}}{x^{3}-2 x^{2}-2 x +4}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x^3-63*x^2+72*x+18)*exp(2)^2/(2*x^6-8*x^5+32*x^3-24*x^2-32*x+32),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.34, size = 22, normalized size = 0.85 \begin {gather*} -\frac {9 \, {\left (x - 4\right )} e^{4}}{4 \, {\left (x^{3} - 2 \, x^{2} - 2 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x^3-63*x^2+72*x+18)*exp(2)^2/(2*x^6-8*x^5+32*x^3-24*x^2-32*x+32),x, algorithm="maxima")

[Out]

-9/4*(x - 4)*e^4/(x^3 - 2*x^2 - 2*x + 4)

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mupad [B]  time = 0.08, size = 19, normalized size = 0.73 \begin {gather*} -\frac {9\,{\mathrm {e}}^4\,\left (x-4\right )}{4\,\left (x^2-2\right )\,\left (x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4)*(72*x - 63*x^2 + 9*x^3 + 18))/(32*x + 24*x^2 - 32*x^3 + 8*x^5 - 2*x^6 - 32),x)

[Out]

-(9*exp(4)*(x - 4))/(4*(x^2 - 2)*(x - 2))

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sympy [A]  time = 0.25, size = 26, normalized size = 1.00 \begin {gather*} \frac {- 9 x e^{4} + 36 e^{4}}{4 x^{3} - 8 x^{2} - 8 x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*x**3-63*x**2+72*x+18)*exp(2)**2/(2*x**6-8*x**5+32*x**3-24*x**2-32*x+32),x)

[Out]

(-9*x*exp(4) + 36*exp(4))/(4*x**3 - 8*x**2 - 8*x + 16)

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