3.1.76 \(\int \frac {e^5 (-432 e x^2-64 e^2 x^6)+288 e^6 x^2 \log (2)-48 e^6 x^2 \log ^2(2)}{81-72 e x^4+16 e^2 x^8+(-108+48 e x^4) \log (2)+(54-8 e x^4) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 e^5}{x-\frac {(-3+\log (2))^2}{4 e x^3}} \]

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Rubi [A]  time = 0.10, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6, 1994, 28, 1588} \begin {gather*} \frac {16 e^6 x^3}{4 e x^4-(3-\log (2))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(-432*E*x^2 - 64*E^2*x^6) + 288*E^6*x^2*Log[2] - 48*E^6*x^2*Log[2]^2)/(81 - 72*E*x^4 + 16*E^2*x^8 + (
-108 + 48*E*x^4)*Log[2] + (54 - 8*E*x^4)*Log[2]^2 - 12*Log[2]^3 + Log[2]^4),x]

[Out]

(16*E^6*x^3)/(4*E*x^4 - (3 - Log[2])^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 1994

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && TrinomialQ
[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+e^6 x^2 \left (288 \log (2)-48 \log ^2(2)\right )}{81-72 e x^4+16 e^2 x^8+\left (-108+48 e x^4\right ) \log (2)+\left (54-8 e x^4\right ) \log ^2(2)-12 \log ^3(2)+\log ^4(2)} \, dx\\ &=\int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+e^6 x^2 \left (288 \log (2)-48 \log ^2(2)\right )}{16 e^2 x^8-8 e x^4 (3-\log (2))^2+(-3+\log (2))^4} \, dx\\ &=\left (16 e^2\right ) \int \frac {e^5 \left (-432 e x^2-64 e^2 x^6\right )+e^6 x^2 \left (288 \log (2)-48 \log ^2(2)\right )}{\left (16 e^2 x^4-4 e (3-\log (2))^2\right )^2} \, dx\\ &=\frac {16 e^6 x^3}{4 e x^4-(3-\log (2))^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 54, normalized size = 2.16 \begin {gather*} \frac {4 e^6 x^3 \left (36-18 \log (2)+4 \log ^2(2)-\log (64)\right )}{\left (9+\log ^2(2)-\log (64)\right ) \left (-9+4 e x^4-\log ^2(2)+\log (64)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-432*E*x^2 - 64*E^2*x^6) + 288*E^6*x^2*Log[2] - 48*E^6*x^2*Log[2]^2)/(81 - 72*E*x^4 + 16*E^2*x
^8 + (-108 + 48*E*x^4)*Log[2] + (54 - 8*E*x^4)*Log[2]^2 - 12*Log[2]^3 + Log[2]^4),x]

[Out]

(4*E^6*x^3*(36 - 18*Log[2] + 4*Log[2]^2 - Log[64]))/((9 + Log[2]^2 - Log[64])*(-9 + 4*E*x^4 - Log[2]^2 + Log[6
4]))

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fricas [A]  time = 0.49, size = 28, normalized size = 1.12 \begin {gather*} \frac {16 \, x^{3} e^{6}}{4 \, x^{4} e - \log \relax (2)^{2} + 6 \, \log \relax (2) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-64*x^6*exp(1)^2-432*x^2*exp(1))*exp(5
))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1)+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+
81),x, algorithm="fricas")

[Out]

16*x^3*e^6/(4*x^4*e - log(2)^2 + 6*log(2) - 9)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-64*x^6*exp(1)^2-432*x^2*exp(1))*exp(5
))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1)+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+
81),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.13, size = 28, normalized size = 1.12

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{6} x^{3}}{x^{4} {\mathrm e}-\frac {\ln \relax (2)^{2}}{4}+\frac {3 \ln \relax (2)}{2}-\frac {9}{4}}\)	\(28\)
gosper	\(\frac {16 x^{3} {\mathrm e}^{6}}{4 x^{4} {\mathrm e}-\ln \relax (2)^{2}+6 \ln \relax (2)-9}\)	\(29\)
norman	\(\frac {16 x^{3} {\mathrm e} \,{\mathrm e}^{5}}{4 x^{4} {\mathrm e}-\ln \relax (2)^{2}+6 \ln \relax (2)-9}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*x^2*exp(1)*exp(5)*ln(2)^2+288*x^2*exp(1)*exp(5)*ln(2)+(-64*x^6*exp(1)^2-432*x^2*exp(1))*exp(5))/(ln(2
)^4-12*ln(2)^3+(-8*x^4*exp(1)+54)*ln(2)^2+(48*x^4*exp(1)-108)*ln(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+81),x,method
=_RETURNVERBOSE)

[Out]

4*exp(6)*x^3/(x^4*exp(1)-1/4*ln(2)^2+3/2*ln(2)-9/4)

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maxima [A]  time = 0.49, size = 28, normalized size = 1.12 \begin {gather*} \frac {16 \, x^{3} e^{6}}{4 \, x^{4} e - \log \relax (2)^{2} + 6 \, \log \relax (2) - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*exp(1)*exp(5)*log(2)^2+288*x^2*exp(1)*exp(5)*log(2)+(-64*x^6*exp(1)^2-432*x^2*exp(1))*exp(5
))/(log(2)^4-12*log(2)^3+(-8*x^4*exp(1)+54)*log(2)^2+(48*x^4*exp(1)-108)*log(2)+16*x^8*exp(1)^2-72*x^4*exp(1)+
81),x, algorithm="maxima")

[Out]

16*x^3*e^6/(4*x^4*e - log(2)^2 + 6*log(2) - 9)

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mupad [B]  time = 0.27, size = 26, normalized size = 1.04 \begin {gather*} \frac {16\,x^3\,{\mathrm {e}}^6}{4\,\mathrm {e}\,x^4+\ln \left (64\right )-{\ln \relax (2)}^2-9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(432*x^2*exp(1) + 64*x^6*exp(2)) - 288*x^2*exp(6)*log(2) + 48*x^2*exp(6)*log(2)^2)/(16*x^8*exp(2)
- 72*x^4*exp(1) - log(2)^2*(8*x^4*exp(1) - 54) + log(2)*(48*x^4*exp(1) - 108) - 12*log(2)^3 + log(2)^4 + 81),
x)

[Out]

(16*x^3*exp(6))/(log(64) + 4*x^4*exp(1) - log(2)^2 - 9)

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sympy [A]  time = 1.30, size = 27, normalized size = 1.08 \begin {gather*} \frac {16 x^{3} e^{6}}{4 e x^{4} - 9 - \log {\relax (2 )}^{2} + 6 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x**2*exp(1)*exp(5)*ln(2)**2+288*x**2*exp(1)*exp(5)*ln(2)+(-64*x**6*exp(1)**2-432*x**2*exp(1))*e
xp(5))/(ln(2)**4-12*ln(2)**3+(-8*x**4*exp(1)+54)*ln(2)**2+(48*x**4*exp(1)-108)*ln(2)+16*x**8*exp(1)**2-72*x**4
*exp(1)+81),x)

[Out]

16*x**3*exp(6)/(4*E*x**4 - 9 - log(2)**2 + 6*log(2))

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