Integrand size = 98, antiderivative size = 26 \[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\frac {2^{-1+\frac {2}{x}} \left (\frac {1}{(1-x)^4}\right )^{\frac {1}{x}}}{1+x} \]
\[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx \]
Integrate[(4^x^(-1)*((1 - 4*x + 6*x^2 - 4*x^3 + x^4)^(-1))^x^(-1)*(-4*x - 3*x^2 - x^3 + (1 - x^2)*Log[4/(1 - 4*x + 6*x^2 - 4*x^3 + x^4)]))/(-2*x^2 - 2*x^3 + 2*x^4 + 2*x^5),x]
Integrate[(4^x^(-1)*((1 - 4*x + 6*x^2 - 4*x^3 + x^4)^(-1))^x^(-1)*(-4*x - 3*x^2 - x^3 + (1 - x^2)*Log[4/(1 - 4*x + 6*x^2 - 4*x^3 + x^4)]))/(-2*x^2 - 2*x^3 + 2*x^4 + 2*x^5), x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {4^{\frac {1}{x}} \left (\frac {1}{x^4-4 x^3+6 x^2-4 x+1}\right )^{\frac {1}{x}} \left (-x^3-3 x^2+\left (1-x^2\right ) \log \left (\frac {4}{x^4-4 x^3+6 x^2-4 x+1}\right )-4 x\right )}{2 x^5+2 x^4-2 x^3-2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {4^{\frac {1}{x}} \left (\frac {1}{x^4-4 x^3+6 x^2-4 x+1}\right )^{\frac {1}{x}} \left (-x^3-3 x^2+\left (1-x^2\right ) \log \left (\frac {4}{x^4-4 x^3+6 x^2-4 x+1}\right )-4 x\right )}{x^2 \left (2 x^3+2 x^2-2 x-2\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {4^{\frac {1}{x}-1} \left (\frac {1}{x^4-4 x^3+6 x^2-4 x+1}\right )^{\frac {1}{x}} \left (-x^3-3 x^2+\left (1-x^2\right ) \log \left (\frac {4}{x^4-4 x^3+6 x^2-4 x+1}\right )-4 x\right )}{x^2 \left (x^2-1\right )}-\frac {4^{\frac {1}{x}-1} \left (\frac {1}{x^4-4 x^3+6 x^2-4 x+1}\right )^{\frac {1}{x}} \left (-x^3-3 x^2+\left (1-x^2\right ) \log \left (\frac {4}{x^4-4 x^3+6 x^2-4 x+1}\right )-4 x\right )}{x^2 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {\int \frac {4^{\frac {1}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x^2}dx}{x-1}dx-2 \log \left (\frac {4}{(1-x)^4}\right ) \int \frac {4^{\frac {1}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x^2}dx+\int \frac {4^{\frac {1}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{-x-1}dx-\int \frac {4^{\frac {1}{x}} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x-1}dx+2 \int \frac {4^{\frac {1}{x}} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x}dx-\int \frac {2^{\frac {2}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{(x+1)^2}dx-3 \int \frac {4^{\frac {1}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x+1}dx+4 \int \frac {\int \frac {2^{\frac {2}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x}dx}{x-1}dx-4 \int \frac {\int \frac {2^{\frac {2}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x+1}dx}{x-1}dx+\log \left (\frac {4}{(1-x)^4}\right ) \int \frac {2^{\frac {2}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x}dx-\log \left (\frac {4}{(1-x)^4}\right ) \int \frac {2^{\frac {2}{x}-1} \left (\frac {1}{(x-1)^4}\right )^{\frac {1}{x}}}{x+1}dx\) |
Int[(4^x^(-1)*((1 - 4*x + 6*x^2 - 4*x^3 + x^4)^(-1))^x^(-1)*(-4*x - 3*x^2 - x^3 + (1 - x^2)*Log[4/(1 - 4*x + 6*x^2 - 4*x^3 + x^4)]))/(-2*x^2 - 2*x^3 + 2*x^4 + 2*x^5),x]
3.4.48.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.66 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {\left (\frac {4}{x^{4}-4 x^{3}+6 x^{2}-4 x +1}\right )^{\frac {1}{x}}}{2+2 x}\) | \(34\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {\ln \left (\frac {4}{x^{4}-4 x^{3}+6 x^{2}-4 x +1}\right )}{x}}}{2+2 x}\) | \(36\) |
int(((-x^2+1)*ln(4/(x^4-4*x^3+6*x^2-4*x+1))-x^3-3*x^2-4*x)*exp(ln(4/(x^4-4 *x^3+6*x^2-4*x+1))/x)/(2*x^5+2*x^4-2*x^3-2*x^2),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\frac {\left (\frac {4}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right )^{\left (\frac {1}{x}\right )}}{2 \, {\left (x + 1\right )}} \]
integrate(((-x^2+1)*log(4/(x^4-4*x^3+6*x^2-4*x+1))-x^3-3*x^2-4*x)*exp(log( 4/(x^4-4*x^3+6*x^2-4*x+1))/x)/(2*x^5+2*x^4-2*x^3-2*x^2),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\frac {e^{\frac {\log {\left (\frac {4}{x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1} \right )}}{x}}}{2 x + 2} \]
integrate(((-x**2+1)*ln(4/(x**4-4*x**3+6*x**2-4*x+1))-x**3-3*x**2-4*x)*exp (ln(4/(x**4-4*x**3+6*x**2-4*x+1))/x)/(2*x**5+2*x**4-2*x**3-2*x**2),x)
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\frac {e^{\left (\frac {2 \, \log \left (2\right )}{x} - \frac {4 \, \log \left (x - 1\right )}{x}\right )}}{2 \, {\left (x + 1\right )}} \]
integrate(((-x^2+1)*log(4/(x^4-4*x^3+6*x^2-4*x+1))-x^3-3*x^2-4*x)*exp(log( 4/(x^4-4*x^3+6*x^2-4*x+1))/x)/(2*x^5+2*x^4-2*x^3-2*x^2),x, algorithm=\
\[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\int { -\frac {{\left (x^{3} + 3 \, x^{2} + {\left (x^{2} - 1\right )} \log \left (\frac {4}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + 4 \, x\right )} \left (\frac {4}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right )^{\left (\frac {1}{x}\right )}}{2 \, {\left (x^{5} + x^{4} - x^{3} - x^{2}\right )}} \,d x } \]
integrate(((-x^2+1)*log(4/(x^4-4*x^3+6*x^2-4*x+1))-x^3-3*x^2-4*x)*exp(log( 4/(x^4-4*x^3+6*x^2-4*x+1))/x)/(2*x^5+2*x^4-2*x^3-2*x^2),x, algorithm=\
integrate(-1/2*(x^3 + 3*x^2 + (x^2 - 1)*log(4/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) + 4*x)*(4/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1))^(1/x)/(x^5 + x^4 - x^3 - x ^2), x)
Time = 9.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {4^{\frac {1}{x}} \left (\frac {1}{1-4 x+6 x^2-4 x^3+x^4}\right )^{\frac {1}{x}} \left (-4 x-3 x^2-x^3+\left (1-x^2\right ) \log \left (\frac {4}{1-4 x+6 x^2-4 x^3+x^4}\right )\right )}{-2 x^2-2 x^3+2 x^4+2 x^5} \, dx=\frac {{\left (\frac {4}{x^4-4\,x^3+6\,x^2-4\,x+1}\right )}^{1/x}}{2\,\left (x+1\right )} \]