\(\int \frac {4 x^4-x^6-90 x^7+(5 x^2-2 x^4-180 x^5) \log (4)+(3-x^2-90 x^3) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx\) [8105]

Optimal result

Integrand size = 74, antiderivative size = 26 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=5-\frac {3}{x}-x-45 x^2-\frac {x}{x^2+\log (4)} \]

[Out]

5-x/(x^2+2*ln(2))-x-45*x^2-3/x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1608, 28, 1819, 1600, 14} \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-45 x^2-\frac {x}{x^2+\log (4)}-\frac {3}{x}-\frac {x \log (16)}{2 \log (4)} \]

[In]

Int[(4*x^4 - x^6 - 90*x^7 + (5*x^2 - 2*x^4 - 180*x^5)*Log[4] + (3 - x^2 - 90*x^3)*Log[4]^2)/(x^6 + 2*x^4*Log[4
] + x^2*Log[4]^2),x]

[Out]

-3/x - 45*x^2 - x/(x^2 + Log[4]) - (x*Log[16])/(2*Log[4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 1600

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 1608

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 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^2 \left (x^4+2 x^2 \log (4)+\log ^2(4)\right )} \, dx \\ & = \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^2 \left (x^2+\log (4)\right )^2} \, dx \\ & = -\frac {x}{x^2+\log (4)}-\frac {\int \frac {2 x^4 \log (4)+180 x^5 \log (4)-2 x^2 (3-\log (4)) \log (4)-6 \log ^2(4)+180 x^3 \log ^2(4)}{x^2 \left (x^2+\log (4)\right )} \, dx}{2 \log (4)} \\ & = -\frac {x}{x^2+\log (4)}-\frac {\int \frac {-6 \log (4)+2 x^2 \log (4)+180 x^3 \log (4)}{x^2} \, dx}{2 \log (4)} \\ & = -\frac {x}{x^2+\log (4)}-\frac {\int \left (-\frac {6 \log (4)}{x^2}+180 x \log (4)+\log (16)\right ) \, dx}{2 \log (4)} \\ & = -\frac {3}{x}-45 x^2-\frac {x}{x^2+\log (4)}-\frac {x \log (16)}{2 \log (4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-\frac {3}{x}-x-45 x^2-\frac {x}{x^2+\log (4)} \]

[In]

Integrate[(4*x^4 - x^6 - 90*x^7 + (5*x^2 - 2*x^4 - 180*x^5)*Log[4] + (3 - x^2 - 90*x^3)*Log[4]^2)/(x^6 + 2*x^4
*Log[4] + x^2*Log[4]^2),x]

[Out]

-3/x - x - 45*x^2 - x/(x^2 + Log[4])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08

method	result	size
default	\(-x -45 x^{2}-\frac {x}{2 \left (\frac {x^{2}}{2}+\ln \left (2\right )\right )}-\frac {3}{x}\)	\(28\)
risch	\(-45 x^{2}-x +\frac {-4 x^{2}-6 \ln \left (2\right )}{x \left (x^{2}+2 \ln \left (2\right )\right )}\)	\(35\)
norman	\(\frac {180 x \ln \left (2\right )^{2}+\left (-2 \ln \left (2\right )-4\right ) x^{2}-x^{4}-45 x^{5}-6 \ln \left (2\right )}{x \left (x^{2}+2 \ln \left (2\right )\right )}\)	\(47\)
gosper	\(\frac {-45 x^{5}-x^{4}+180 x \ln \left (2\right )^{2}-2 x^{2} \ln \left (2\right )-4 x^{2}-6 \ln \left (2\right )}{x \left (x^{2}+2 \ln \left (2\right )\right )}\)	\(49\)
parallelrisch	\(\frac {-45 x^{5}-x^{4}+180 x \ln \left (2\right )^{2}-2 x^{2} \ln \left (2\right )-4 x^{2}-6 \ln \left (2\right )}{x \left (x^{2}+2 \ln \left (2\right )\right )}\)	\(49\)

[In]

int((4*(-90*x^3-x^2+3)*ln(2)^2+2*(-180*x^5-2*x^4+5*x^2)*ln(2)-90*x^7-x^6+4*x^4)/(4*x^2*ln(2)^2+4*x^4*ln(2)+x^6
),x,method=_RETURNVERBOSE)

[Out]

-x-45*x^2-1/2*x/(1/2*x^2+ln(2))-3/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-\frac {45 \, x^{5} + x^{4} + 4 \, x^{2} + 2 \, {\left (45 \, x^{3} + x^{2} + 3\right )} \log \left (2\right )}{x^{3} + 2 \, x \log \left (2\right )} \]

[In]

integrate((4*(-90*x^3-x^2+3)*log(2)^2+2*(-180*x^5-2*x^4+5*x^2)*log(2)-90*x^7-x^6+4*x^4)/(4*x^2*log(2)^2+4*x^4*
log(2)+x^6),x, algorithm="fricas")

[Out]

-(45*x^5 + x^4 + 4*x^2 + 2*(45*x^3 + x^2 + 3)*log(2))/(x^3 + 2*x*log(2))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=- 45 x^{2} - x - \frac {4 x^{2} + 6 \log {\left (2 \right )}}{x^{3} + 2 x \log {\left (2 \right )}} \]

[In]

integrate((4*(-90*x**3-x**2+3)*ln(2)**2+2*(-180*x**5-2*x**4+5*x**2)*ln(2)-90*x**7-x**6+4*x**4)/(4*x**2*ln(2)**
2+4*x**4*ln(2)+x**6),x)

[Out]

-45*x**2 - x - (4*x**2 + 6*log(2))/(x**3 + 2*x*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-45 \, x^{2} - x - \frac {2 \, {\left (2 \, x^{2} + 3 \, \log \left (2\right )\right )}}{x^{3} + 2 \, x \log \left (2\right )} \]

[In]

integrate((4*(-90*x^3-x^2+3)*log(2)^2+2*(-180*x^5-2*x^4+5*x^2)*log(2)-90*x^7-x^6+4*x^4)/(4*x^2*log(2)^2+4*x^4*
log(2)+x^6),x, algorithm="maxima")

[Out]

-45*x^2 - x - 2*(2*x^2 + 3*log(2))/(x^3 + 2*x*log(2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-45 \, x^{2} - x - \frac {2 \, {\left (2 \, x^{2} + 3 \, \log \left (2\right )\right )}}{x^{3} + 2 \, x \log \left (2\right )} \]

[In]

integrate((4*(-90*x^3-x^2+3)*log(2)^2+2*(-180*x^5-2*x^4+5*x^2)*log(2)-90*x^7-x^6+4*x^4)/(4*x^2*log(2)^2+4*x^4*
log(2)+x^6),x, algorithm="giac")

[Out]

-45*x^2 - x - 2*(2*x^2 + 3*log(2))/(x^3 + 2*x*log(2))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {4 x^4-x^6-90 x^7+\left (5 x^2-2 x^4-180 x^5\right ) \log (4)+\left (3-x^2-90 x^3\right ) \log ^2(4)}{x^6+2 x^4 \log (4)+x^2 \log ^2(4)} \, dx=-x-\frac {4\,x^2+6\,\ln \left (2\right )}{x^3+2\,\ln \left (2\right )\,x}-45\,x^2 \]

[In]

int(-(4*log(2)^2*(x^2 + 90*x^3 - 3) + 2*log(2)*(2*x^4 - 5*x^2 + 180*x^5) - 4*x^4 + x^6 + 90*x^7)/(4*x^2*log(2)
^2 + 4*x^4*log(2) + x^6),x)

[Out]

- x - (6*log(2) + 4*x^2)/(2*x*log(2) + x^3) - 45*x^2