3.18.25 \(\int \frac {\frac {e^4 x (2 x+3 x^3)}{2+x^2}+(4 x+2 x^3) \log ^2(2)}{\frac {e^8 x^2 (8+4 x^2)}{(2+x^2)^2}+\frac {e^4 x (16+8 x^2) \log ^2(2)}{2+x^2}+(8+4 x^2) \log ^4(2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {x^2}{4 \left (\frac {e^4}{\frac {2}{x}+x}+\log ^2(2)\right )} \]

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Rubi [B]  time = 0.33, antiderivative size = 99, normalized size of antiderivative = 3.67, number of steps used = 5, number of rules used = 4, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6688, 12, 1644, 1586} \begin {gather*} \frac {x^2}{4 \log ^2(2)}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{4 \left (e^8-8 \log ^4(2)\right ) \left (x^2 \log ^2(2)+e^4 x+2 \log ^2(2)\right )}-\frac {e^4 x}{4 \log ^4(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((E^4*x*(2*x + 3*x^3))/(2 + x^2) + (4*x + 2*x^3)*Log[2]^2)/((E^8*x^2*(8 + 4*x^2))/(2 + x^2)^2 + (E^4*x*(16
+ 8*x^2)*Log[2]^2)/(2 + x^2) + (8 + 4*x^2)*Log[2]^4),x]

[Out]

-1/4*(E^4*x)/Log[2]^4 + x^2/(4*Log[2]^2) - (E^4*x*(E^4*x*(8 - E^8/Log[2]^4) - (2*(E^8 - 8*Log[2]^4))/Log[2]^2)
)/(4*(E^4*x + 2*Log[2]^2 + x^2*Log[2]^2)*(E^8 - 8*Log[2]^4))

Rule 12

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1644

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g +
(2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x
+ c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Q + g*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c
*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[p] ||  !IntegerQ[m
] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2,
0]))

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} &=\int \frac {x \left (2 e^4 x+3 e^4 x^3+8 \log ^2(2)+8 x^2 \log ^2(2)+2 x^4 \log ^2(2)\right )}{4 \left (e^4 x+2 \log ^2(2)+x^2 \log ^2(2)\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {x \left (2 e^4 x+3 e^4 x^3+8 \log ^2(2)+8 x^2 \log ^2(2)+2 x^4 \log ^2(2)\right )}{\left (e^4 x+2 \log ^2(2)+x^2 \log ^2(2)\right )^2} \, dx\\ &=-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{4 \left (e^4 x+2 \log ^2(2)+x^2 \log ^2(2)\right ) \left (e^8-8 \log ^4(2)\right )}-\frac {\int \frac {-x \left (12 e^8-\frac {e^{16}}{\log ^4(2)}-32 \log ^4(2)\right )-2 x^3 \left (e^8-8 \log ^4(2)\right )+\frac {2 e^4 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}-\frac {e^4 x^2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}}{e^4 x+2 \log ^2(2)+x^2 \log ^2(2)} \, dx}{4 \left (e^8-8 \log ^4(2)\right )}\\ &=-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{4 \left (e^4 x+2 \log ^2(2)+x^2 \log ^2(2)\right ) \left (e^8-8 \log ^4(2)\right )}-\frac {\int \left (-8 e^4+\frac {e^{12}}{\log ^4(2)}+x \left (-\frac {2 e^8}{\log ^2(2)}+16 \log ^2(2)\right )\right ) \, dx}{4 \left (e^8-8 \log ^4(2)\right )}\\ &=-\frac {e^4 x}{4 \log ^4(2)}+\frac {x^2}{4 \log ^2(2)}-\frac {e^4 x \left (e^4 x \left (8-\frac {e^8}{\log ^4(2)}\right )-\frac {2 \left (e^8-8 \log ^4(2)\right )}{\log ^2(2)}\right )}{4 \left (e^4 x+2 \log ^2(2)+x^2 \log ^2(2)\right ) \left (e^8-8 \log ^4(2)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 60, normalized size = 2.22 \begin {gather*} \frac {-e^{12} x-e^8 \left (2+x^2\right ) \log ^2(2)+x^2 \left (2+x^2\right ) \log ^6(2)}{4 \left (e^4 x \log ^6(2)+\left (2+x^2\right ) \log ^8(2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^4*x*(2*x + 3*x^3))/(2 + x^2) + (4*x + 2*x^3)*Log[2]^2)/((E^8*x^2*(8 + 4*x^2))/(2 + x^2)^2 + (E^4
*x*(16 + 8*x^2)*Log[2]^2)/(2 + x^2) + (8 + 4*x^2)*Log[2]^4),x]

[Out]

(-(E^12*x) - E^8*(2 + x^2)*Log[2]^2 + x^2*(2 + x^2)*Log[2]^6)/(4*(E^4*x*Log[2]^6 + (2 + x^2)*Log[2]^8))

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fricas [B]  time = 0.69, size = 56, normalized size = 2.07 \begin {gather*} \frac {{\left (x^{4} + 2 \, x^{2}\right )} \log \relax (2)^{6} - {\left (x^{2} + 2\right )} e^{8} \log \relax (2)^{2} - x e^{12}}{4 \, {\left ({\left (x^{2} + 2\right )} \log \relax (2)^{8} + x e^{4} \log \relax (2)^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2
+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+(4*x^2+8)*log(2)^4),x, algorithm="fricas")

[Out]

1/4*((x^4 + 2*x^2)*log(2)^6 - (x^2 + 2)*e^8*log(2)^2 - x*e^12)/((x^2 + 2)*log(2)^8 + x*e^4*log(2)^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{3} + 2 \, x\right )} \log \relax (2)^{2} + {\left (3 \, x^{3} + 2 \, x\right )} e^{\left (-\log \left (\frac {x^{2} + 2}{x}\right ) + 4\right )}}{4 \, {\left ({\left (x^{2} + 2\right )} \log \relax (2)^{4} + 2 \, {\left (x^{2} + 2\right )} e^{\left (-\log \left (\frac {x^{2} + 2}{x}\right ) + 4\right )} \log \relax (2)^{2} + {\left (x^{2} + 2\right )} e^{\left (-2 \, \log \left (\frac {x^{2} + 2}{x}\right ) + 8\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2
+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+(4*x^2+8)*log(2)^4),x, algorithm="giac")

[Out]

integrate(1/4*(2*(x^3 + 2*x)*log(2)^2 + (3*x^3 + 2*x)*e^(-log((x^2 + 2)/x) + 4))/((x^2 + 2)*log(2)^4 + 2*(x^2
+ 2)*e^(-log((x^2 + 2)/x) + 4)*log(2)^2 + (x^2 + 2)*e^(-2*log((x^2 + 2)/x) + 8)), x)

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maple [B]  time = 0.22, size = 62, normalized size = 2.30

method	result	size
norman	\(\frac {-3 x^{2}-\frac {{\mathrm e}^{4} x^{3}}{\ln \relax (2)^{2}}+\frac {x^{6}}{4}-\frac {2 \,{\mathrm e}^{4} x}{\ln \relax (2)^{2}}-4}{\left (x^{2}+2\right ) \left (x^{2} \ln \relax (2)^{2}+x \,{\mathrm e}^{4}+2 \ln \relax (2)^{2}\right )}\)	\(62\)
risch	\(\frac {x^{2}}{4 \ln \relax (2)^{2}}-\frac {x \,{\mathrm e}^{4}}{4 \ln \relax (2)^{4}}+\frac {-\frac {\left (-2 \ln \relax (2)^{4}+{\mathrm e}^{8}\right ) {\mathrm e}^{4} x}{\ln \relax (2)^{2}}-2 \,{\mathrm e}^{8}}{4 \ln \relax (2)^{4} \left (x^{2} \ln \relax (2)^{2}+x \,{\mathrm e}^{4}+2 \ln \relax (2)^{2}\right )}\)	\(70\)
default	\(-\frac {-x^{2} \ln \relax (2)^{2}+x \,{\mathrm e}^{4}}{4 \ln \relax (2)^{4}}+\frac {\munderset {\textit {\_R} =\RootOf \left (\ln \relax (2)^{4} \textit {\_Z}^{4}+2 \,{\mathrm e}^{4} \ln \relax (2)^{2} \textit {\_Z}^{3}+\left (4 \ln \relax (2)^{4}+{\mathrm e}^{8}\right ) \textit {\_Z}^{2}+4 \ln \relax (2)^{2} \textit {\_Z} \,{\mathrm e}^{4}+4 \ln \relax (2)^{4}\right )}{\sum }\frac {\left ({\mathrm e}^{4} \left (-2 \ln \relax (2)^{4}+{\mathrm e}^{8}\right ) \textit {\_R}^{2}+4 \,{\mathrm e}^{8} \textit {\_R} \ln \relax (2)^{2}+4 \,{\mathrm e}^{4} \ln \relax (2)^{4}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} \ln \relax (2)^{4}+3 \,{\mathrm e}^{4} \ln \relax (2)^{2} \textit {\_R}^{2}+4 \textit {\_R} \ln \relax (2)^{4}+2 \,{\mathrm e}^{4} \ln \relax (2)^{2}+\textit {\_R} \,{\mathrm e}^{8}}}{8 \ln \relax (2)^{4}}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3+2*x)*exp(-ln((x^2+2)/x)+4)+(2*x^3+4*x)*ln(2)^2)/((4*x^2+8)*exp(-ln((x^2+2)/x)+4)^2+(8*x^2+16)*ln(2
)^2*exp(-ln((x^2+2)/x)+4)+(4*x^2+8)*ln(2)^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.38, size = 73, normalized size = 2.70 \begin {gather*} -\frac {2 \, e^{8} \log \relax (2)^{2} - {\left (2 \, e^{4} \log \relax (2)^{4} - e^{12}\right )} x}{4 \, {\left (x^{2} \log \relax (2)^{8} + x e^{4} \log \relax (2)^{6} + 2 \, \log \relax (2)^{8}\right )}} + \frac {x^{2} \log \relax (2)^{2} - x e^{4}}{4 \, \log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+2*x)*exp(-log((x^2+2)/x)+4)+(2*x^3+4*x)*log(2)^2)/((4*x^2+8)*exp(-log((x^2+2)/x)+4)^2+(8*x^2
+16)*log(2)^2*exp(-log((x^2+2)/x)+4)+(4*x^2+8)*log(2)^4),x, algorithm="maxima")

[Out]

-1/4*(2*e^8*log(2)^2 - (2*e^4*log(2)^4 - e^12)*x)/(x^2*log(2)^8 + x*e^4*log(2)^6 + 2*log(2)^8) + 1/4*(x^2*log(
2)^2 - x*e^4)/log(2)^4

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mupad [B]  time = 0.57, size = 33, normalized size = 1.22 \begin {gather*} \frac {x^4+2\,x^2}{4\,{\ln \relax (2)}^2\,x^2+4\,{\mathrm {e}}^4\,x+8\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)^2*(4*x + 2*x^3) + exp(4 - log((x^2 + 2)/x))*(2*x + 3*x^3))/(log(2)^4*(4*x^2 + 8) + exp(8 - 2*log((
x^2 + 2)/x))*(4*x^2 + 8) + exp(4 - log((x^2 + 2)/x))*log(2)^2*(8*x^2 + 16)),x)

[Out]

(2*x^2 + x^4)/(4*x^2*log(2)^2 + 4*x*exp(4) + 8*log(2)^2)

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sympy [B]  time = 0.72, size = 75, normalized size = 2.78 \begin {gather*} \frac {x^{2}}{4 \log {\relax (2 )}^{2}} - \frac {x e^{4}}{4 \log {\relax (2 )}^{4}} + \frac {x \left (- e^{12} + 2 e^{4} \log {\relax (2 )}^{4}\right ) - 2 e^{8} \log {\relax (2 )}^{2}}{4 x^{2} \log {\relax (2 )}^{8} + 4 x e^{4} \log {\relax (2 )}^{6} + 8 \log {\relax (2 )}^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3+2*x)*exp(-ln((x**2+2)/x)+4)+(2*x**3+4*x)*ln(2)**2)/((4*x**2+8)*exp(-ln((x**2+2)/x)+4)**2+(8
*x**2+16)*ln(2)**2*exp(-ln((x**2+2)/x)+4)+(4*x**2+8)*ln(2)**4),x)

[Out]

x**2/(4*log(2)**2) - x*exp(4)/(4*log(2)**4) + (x*(-exp(12) + 2*exp(4)*log(2)**4) - 2*exp(8)*log(2)**2)/(4*x**2
*log(2)**8 + 4*x*exp(4)*log(2)**6 + 8*log(2)**8)

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