3.31.51 \(\int \frac {-2 e^4+256 e^4 x^3 \log ^2(\log (2))}{x+(2704 x^3+256 x^4) \log ^2(\log (2))} \, dx\)

Optimal. Leaf size=22 \[ e^4 \log \left (\frac {169}{16}+x+\frac {1}{256 x^2 \log ^2(\log (2))}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 35, normalized size of antiderivative = 1.59, 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 = {2074, 1587} \begin {gather*} e^4 \log \left (256 x^3 \log ^2(\log (2))+2704 x^2 \log ^2(\log (2))+1\right )-2 e^4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^4 + 256*E^4*x^3*Log[Log[2]]^2)/(x + (2704*x^3 + 256*x^4)*Log[Log[2]]^2),x]

[Out]

-2*E^4*Log[x] + E^4*Log[1 + 2704*x^2*Log[Log[2]]^2 + 256*x^3*Log[Log[2]]^2]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, 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} &=\int \left (-\frac {2 e^4}{x}+\frac {32 e^4 x (169+24 x) \log ^2(\log (2))}{1+2704 x^2 \log ^2(\log (2))+256 x^3 \log ^2(\log (2))}\right ) \, dx\\ &=-2 e^4 \log (x)+\left (32 e^4 \log ^2(\log (2))\right ) \int \frac {x (169+24 x)}{1+2704 x^2 \log ^2(\log (2))+256 x^3 \log ^2(\log (2))} \, dx\\ &=-2 e^4 \log (x)+e^4 \log \left (1+2704 x^2 \log ^2(\log (2))+256 x^3 \log ^2(\log (2))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 1.45 \begin {gather*} e^4 \left (-2 \log (x)+\log \left (1+2704 x^2 \log ^2(\log (2))+256 x^3 \log ^2(\log (2))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^4 + 256*E^4*x^3*Log[Log[2]]^2)/(x + (2704*x^3 + 256*x^4)*Log[Log[2]]^2),x]

[Out]

E^4*(-2*Log[x] + Log[1 + 2704*x^2*Log[Log[2]]^2 + 256*x^3*Log[Log[2]]^2])

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fricas [A]  time = 0.64, size = 31, normalized size = 1.41 \begin {gather*} e^{4} \log \left (16 \, {\left (16 \, x^{3} + 169 \, x^{2}\right )} \log \left (\log \relax (2)\right )^{2} + 1\right ) - 2 \, e^{4} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^3*exp(4)*log(log(2))^2-2*exp(4))/((256*x^4+2704*x^3)*log(log(2))^2+x),x, algorithm="fricas")

[Out]

e^4*log(16*(16*x^3 + 169*x^2)*log(log(2))^2 + 1) - 2*e^4*log(x)

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giac [B]  time = 0.20, size = 35, normalized size = 1.59 \begin {gather*} e^{4} \log \left ({\left | 256 \, x^{3} \log \left (\log \relax (2)\right )^{2} + 2704 \, x^{2} \log \left (\log \relax (2)\right )^{2} + 1 \right |}\right ) - 2 \, e^{4} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^3*exp(4)*log(log(2))^2-2*exp(4))/((256*x^4+2704*x^3)*log(log(2))^2+x),x, algorithm="giac")

[Out]

e^4*log(abs(256*x^3*log(log(2))^2 + 2704*x^2*log(log(2))^2 + 1)) - 2*e^4*log(abs(x))

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maple [A]  time = 0.08, size = 34, normalized size = 1.55

method	result	size
norman	\({\mathrm e}^{4} \ln \left (256 x^{3} \ln \left (\ln \relax (2)\right )^{2}+2704 x^{2} \ln \left (\ln \relax (2)\right )^{2}+1\right )-2 \,{\mathrm e}^{4} \ln \relax (x )\)	\(34\)
risch	\(-2 \,{\mathrm e}^{4} \ln \relax (x )+{\mathrm e}^{4} \ln \left (-256 x^{3} \ln \left (\ln \relax (2)\right )^{2}-2704 x^{2} \ln \left (\ln \relax (2)\right )^{2}-1\right )\)	\(34\)
default	\(2 \,{\mathrm e}^{4} \left (-\ln \relax (x )+\frac {\ln \left (256 x^{3} \ln \left (\ln \relax (2)\right )^{2}+2704 x^{2} \ln \left (\ln \relax (2)\right )^{2}+1\right )}{2}\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((256*x^3*exp(4)*ln(ln(2))^2-2*exp(4))/((256*x^4+2704*x^3)*ln(ln(2))^2+x),x,method=_RETURNVERBOSE)

[Out]

exp(4)*ln(256*x^3*ln(ln(2))^2+2704*x^2*ln(ln(2))^2+1)-2*exp(4)*ln(x)

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maxima [A]  time = 0.41, size = 33, normalized size = 1.50 \begin {gather*} e^{4} \log \left (256 \, x^{3} \log \left (\log \relax (2)\right )^{2} + 2704 \, x^{2} \log \left (\log \relax (2)\right )^{2} + 1\right ) - 2 \, e^{4} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x^3*exp(4)*log(log(2))^2-2*exp(4))/((256*x^4+2704*x^3)*log(log(2))^2+x),x, algorithm="maxima")

[Out]

e^4*log(256*x^3*log(log(2))^2 + 2704*x^2*log(log(2))^2 + 1) - 2*e^4*log(x)

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mupad [B]  time = 0.18, size = 33, normalized size = 1.50 \begin {gather*} {\mathrm {e}}^4\,\ln \left (256\,{\ln \left (\ln \relax (2)\right )}^2\,x^3+2704\,{\ln \left (\ln \relax (2)\right )}^2\,x^2+1\right )-2\,{\mathrm {e}}^4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(4) - 256*x^3*exp(4)*log(log(2))^2)/(x + log(log(2))^2*(2704*x^3 + 256*x^4)),x)

[Out]

exp(4)*log(2704*x^2*log(log(2))^2 + 256*x^3*log(log(2))^2 + 1) - 2*exp(4)*log(x)

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sympy [A]  time = 0.72, size = 32, normalized size = 1.45 \begin {gather*} - 2 e^{4} \log {\relax (x )} + e^{4} \log {\left (x^{3} + \frac {169 x^{2}}{16} + \frac {1}{256 \log {\left (\log {\relax (2 )} \right )}^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((256*x**3*exp(4)*ln(ln(2))**2-2*exp(4))/((256*x**4+2704*x**3)*ln(ln(2))**2+x),x)

[Out]

-2*exp(4)*log(x) + exp(4)*log(x**3 + 169*x**2/16 + 1/(256*log(log(2))**2))

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