3.103.34 \(\int \frac {-4 x+x^6 (3+x)+x^3 (3+2 x) \log (2)+x \log ^2(2)}{-4 x+x^7+2 x^4 \log (2)+x \log ^2(2)} \, dx\)

Optimal. Leaf size=20 \[ 64+x+\frac {1}{2} \left (5+\log \left (-4+\left (x^3+\log (2)\right )^2\right )\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 6, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6, 1594, 1586, 1790, 1468, 628} \begin {gather*} \frac {1}{2} \log \left (-x^6-2 x^3 \log (2)+4-\log ^2(2)\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x + x^6*(3 + x) + x^3*(3 + 2*x)*Log[2] + x*Log[2]^2)/(-4*x + x^7 + 2*x^4*Log[2] + x*Log[2]^2),x]

[Out]

x + Log[4 - x^6 - 2*x^3*Log[2] - Log[2]^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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1468

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x]
&& EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1790

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[x^j*Sum[Coeff[Pq, x, j + k*n]*x^(k*n), {k, 0, (q - j)/n + 1}]*(a + b*x^n + c*x^(2*n))^p, {j, 0, n - 1}], x
]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] &&  !PolyQ[P
q, x^n]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x+x^6 (3+x)+x^3 (3+2 x) \log (2)+x \log ^2(2)}{x^7+2 x^4 \log (2)+x \left (-4+\log ^2(2)\right )} \, dx\\ &=\int \frac {x^6 (3+x)+x^3 (3+2 x) \log (2)+x \left (-4+\log ^2(2)\right )}{x^7+2 x^4 \log (2)+x \left (-4+\log ^2(2)\right )} \, dx\\ &=\int \frac {x^6 (3+x)+x^3 (3+2 x) \log (2)+x \left (-4+\log ^2(2)\right )}{x \left (-4+x^6+2 x^3 \log (2)+\log ^2(2)\right )} \, dx\\ &=\int \frac {-4+3 x^5+x^6+3 x^2 \log (2)+2 x^3 \log (2)+\log ^2(2)}{-4+x^6+2 x^3 \log (2)+\log ^2(2)} \, dx\\ &=\int \left (1+\frac {x^2 \left (3 x^3+3 \log (2)\right )}{-4+x^6+2 x^3 \log (2)+\log ^2(2)}\right ) \, dx\\ &=x+\int \frac {x^2 \left (3 x^3+3 \log (2)\right )}{-4+x^6+2 x^3 \log (2)+\log ^2(2)} \, dx\\ &=x+\frac {1}{3} \operatorname {Subst}\left (\int \frac {3 x+3 \log (2)}{-4+x^2+2 x \log (2)+\log ^2(2)} \, dx,x,x^3\right )\\ &=x+\frac {1}{2} \log \left (4-x^6-2 x^3 \log (2)-\log ^2(2)\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.03, size = 62, normalized size = 3.10 \begin {gather*} x+\frac {1}{3} \text {RootSum}\left [-4+\log ^2(2)+\log (4) \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {\log (8) \log (x-\text {$\#$1})+3 \log (x-\text {$\#$1}) \text {$\#$1}^3}{\log (4)+2 \text {$\#$1}^3}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + x^6*(3 + x) + x^3*(3 + 2*x)*Log[2] + x*Log[2]^2)/(-4*x + x^7 + 2*x^4*Log[2] + x*Log[2]^2),x]

[Out]

x + RootSum[-4 + Log[2]^2 + Log[4]*#1^3 + #1^6 & , (Log[8]*Log[x - #1] + 3*Log[x - #1]*#1^3)/(Log[4] + 2*#1^3)
& ]/3

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fricas [A]  time = 0.74, size = 21, normalized size = 1.05 \begin {gather*} x + \frac {1}{2} \, \log \left (x^{6} + 2 \, x^{3} \log \relax (2) + \log \relax (2)^{2} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3+x)*x^6+(2*x+3)*log(2)*x^3+x*log(2)^2-4*x)/(x^7+2*x^4*log(2)+x*log(2)^2-4*x),x, algorithm="fricas
")

[Out]

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

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giac [A]  time = 0.14, size = 24, normalized size = 1.20 \begin {gather*} x + \frac {1}{2} \, \log \left ({\left | x^{3} + \log \relax (2) + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x^{3} + \log \relax (2) - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3+x)*x^6+(2*x+3)*log(2)*x^3+x*log(2)^2-4*x)/(x^7+2*x^4*log(2)+x*log(2)^2-4*x),x, algorithm="giac")

[Out]

x + 1/2*log(abs(x^3 + log(2) + 2)) + 1/2*log(abs(x^3 + log(2) - 2))

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maple [A]  time = 0.09, size = 22, normalized size = 1.10

method	result	size
risch	\(x +\frac {\ln \left (x^{6}+2 x^{3} \ln \relax (2)+\ln \relax (2)^{2}-4\right )}{2}\)	\(22\)
default	\(x +\frac {\ln \left (x^{3}+\ln \relax (2)+2\right )}{2}+\frac {\ln \left (x^{3}+\ln \relax (2)-2\right )}{2}\)	\(23\)
norman	\(x +\frac {\ln \left (x^{3}+\ln \relax (2)+2\right )}{2}+\frac {\ln \left (x^{3}+\ln \relax (2)-2\right )}{2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3+x)*x^6+(2*x+3)*ln(2)*x^3+x*ln(2)^2-4*x)/(x^7+2*x^4*ln(2)+x*ln(2)^2-4*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 22, normalized size = 1.10 \begin {gather*} x + \frac {1}{2} \, \log \left (x^{3} + \log \relax (2) + 2\right ) + \frac {1}{2} \, \log \left (x^{3} + \log \relax (2) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3+x)*x^6+(2*x+3)*log(2)*x^3+x*log(2)^2-4*x)/(x^7+2*x^4*log(2)+x*log(2)^2-4*x),x, algorithm="maxima
")

[Out]

x + 1/2*log(x^3 + log(2) + 2) + 1/2*log(x^3 + log(2) - 2)

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mupad [B]  time = 5.99, size = 21, normalized size = 1.05 \begin {gather*} x+\frac {\ln \left (x^6+2\,\ln \relax (2)\,x^3+{\ln \relax (2)}^2-4\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(x + 3) - 4*x + x*log(2)^2 + x^3*log(2)*(2*x + 3))/(x*log(2)^2 - 4*x + 2*x^4*log(2) + x^7),x)

[Out]

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

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sympy [A]  time = 0.44, size = 22, normalized size = 1.10 \begin {gather*} x + \frac {\log {\left (x^{6} + 2 x^{3} \log {\relax (2 )} - 4 + \log {\relax (2 )}^{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3+x)*x**6+(2*x+3)*ln(2)*x**3+x*ln(2)**2-4*x)/(x**7+2*x**4*ln(2)+x*ln(2)**2-4*x),x)

[Out]

x + log(x**6 + 2*x**3*log(2) - 4 + log(2)**2)/2

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