3.22.7 \(\int \frac {(2+x) (-1-x+x^2) \sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}}{2 x (1+2 x+x^2+x^4)} \, dx\)

Optimal. Leaf size=153 \[ -\frac {1}{4} \text {RootSum}\left [17 \text {$\#$1}^8+14 \text {$\#$1}^4+5\& ,\frac {13 \text {$\#$1}^4 \log \left (\sqrt [4]{\frac {-2 x^2+x+1}{4 x^2+x+1}}-\text {$\#$1}\right )+5 \log \left (\sqrt [4]{\frac {-2 x^2+x+1}{4 x^2+x+1}}-\text {$\#$1}\right )}{17 \text {$\#$1}^7+7 \text {$\#$1}^3}\& \right ]+\tan ^{-1}\left (\sqrt [4]{\frac {-2 x^2+x+1}{4 x^2+x+1}}\right )+\tanh ^{-1}\left (\sqrt [4]{\frac {-2 x^2+x+1}{4 x^2+x+1}}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 5.32, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(2+x) \left (-1-x+x^2\right ) \sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}}{2 x \left (1+2 x+x^2+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((2 + x)*(-1 - x + x^2)*((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4))/(2*x*(1 + 2*x + x^2 + x^4)),x]

[Out]

-((((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)*(1 + x + 4*x^2)^(1/4)*Defer[Int][(1 + x - 2*x^2)^(1/4)/(x*(1 + x +
4*x^2)^(1/4)), x])/(1 + x - 2*x^2)^(1/4)) + (((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)*(1 + x + 4*x^2)^(1/4)*Def
er[Int][(1 + x - 2*x^2)^(1/4)/((1 + x + 4*x^2)^(1/4)*(1 + 2*x + x^2 + x^4)), x])/(2*(1 + x - 2*x^2)^(1/4)) + (
3*((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)*(1 + x + 4*x^2)^(1/4)*Defer[Int][(x*(1 + x - 2*x^2)^(1/4))/((1 + x +
4*x^2)^(1/4)*(1 + 2*x + x^2 + x^4)), x])/(2*(1 + x - 2*x^2)^(1/4)) + (((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)
*(1 + x + 4*x^2)^(1/4)*Defer[Int][(x^2*(1 + x - 2*x^2)^(1/4))/((1 + x + 4*x^2)^(1/4)*(1 + 2*x + x^2 + x^4)), x
])/(2*(1 + x - 2*x^2)^(1/4)) + (((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)*(1 + x + 4*x^2)^(1/4)*Defer[Int][(x^3*
(1 + x - 2*x^2)^(1/4))/((1 + x + 4*x^2)^(1/4)*(1 + 2*x + x^2 + x^4)), x])/(1 + x - 2*x^2)^(1/4)

Rubi steps

\begin {align*} \int \frac {(2+x) \left (-1-x+x^2\right ) \sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}}{2 x \left (1+2 x+x^2+x^4\right )} \, dx &=\frac {1}{2} \int \frac {(2+x) \left (-1-x+x^2\right ) \sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}}{x \left (1+2 x+x^2+x^4\right )} \, dx\\ &=\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {(2+x) \sqrt [4]{1+x-2 x^2} \left (-1-x+x^2\right )}{x \sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{2 \sqrt [4]{1+x-2 x^2}}\\ &=\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \left (-\frac {2 \sqrt [4]{1+x-2 x^2}}{x \sqrt [4]{1+x+4 x^2}}+\frac {\sqrt [4]{1+x-2 x^2} \left (1+3 x+x^2+2 x^3\right )}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )}\right ) \, dx}{2 \sqrt [4]{1+x-2 x^2}}\\ &=\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {\sqrt [4]{1+x-2 x^2} \left (1+3 x+x^2+2 x^3\right )}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{2 \sqrt [4]{1+x-2 x^2}}-\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {\sqrt [4]{1+x-2 x^2}}{x \sqrt [4]{1+x+4 x^2}} \, dx}{\sqrt [4]{1+x-2 x^2}}\\ &=\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \left (\frac {\sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )}+\frac {3 x \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )}+\frac {x^2 \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )}+\frac {2 x^3 \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )}\right ) \, dx}{2 \sqrt [4]{1+x-2 x^2}}-\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {\sqrt [4]{1+x-2 x^2}}{x \sqrt [4]{1+x+4 x^2}} \, dx}{\sqrt [4]{1+x-2 x^2}}\\ &=\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {\sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{2 \sqrt [4]{1+x-2 x^2}}+\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {x^2 \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{2 \sqrt [4]{1+x-2 x^2}}-\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {\sqrt [4]{1+x-2 x^2}}{x \sqrt [4]{1+x+4 x^2}} \, dx}{\sqrt [4]{1+x-2 x^2}}+\frac {\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {x^3 \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{\sqrt [4]{1+x-2 x^2}}+\frac {\left (3 \sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}} \sqrt [4]{1+x+4 x^2}\right ) \int \frac {x \sqrt [4]{1+x-2 x^2}}{\sqrt [4]{1+x+4 x^2} \left (1+2 x+x^2+x^4\right )} \, dx}{2 \sqrt [4]{1+x-2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.74, size = 57, normalized size = 0.37 \begin {gather*} \frac {1}{2} \int \frac {(x+2) \left (x^2-x-1\right ) \sqrt [4]{\frac {-2 x^2+x+1}{4 x^2+x+1}}}{x \left (x^4+x^2+2 x+1\right )} \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + x)*(-1 - x + x^2)*((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4))/(2*x*(1 + 2*x + x^2 + x^4)),x]

[Out]

Integrate[((2 + x)*(-1 - x + x^2)*((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4))/(x*(1 + 2*x + x^2 + x^4)), x]/2

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.37, size = 153, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}\right )+\tanh ^{-1}\left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}\right )-\frac {1}{4} \text {RootSum}\left [5+14 \text {$\#$1}^4+17 \text {$\#$1}^8\&,\frac {5 \log \left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}-\text {$\#$1}\right )+13 \log \left (\sqrt [4]{\frac {1+x-2 x^2}{1+x+4 x^2}}-\text {$\#$1}\right ) \text {$\#$1}^4}{7 \text {$\#$1}^3+17 \text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 + x)*(-1 - x + x^2)*((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4))/(2*x*(1 + 2*x + x^2 + x^4)
),x]

[Out]

ArcTan[((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)] + ArcTanh[((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4)] - RootSum[5
+ 14*#1^4 + 17*#1^8 & , (5*Log[((1 + x - 2*x^2)/(1 + x + 4*x^2))^(1/4) - #1] + 13*Log[((1 + x - 2*x^2)/(1 + x
+ 4*x^2))^(1/4) - #1]*#1^4)/(7*#1^3 + 17*#1^7) & ]/4

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2+x)*(x^2-x-1)*((-2*x^2+x+1)/(4*x^2+x+1))^(1/4)/x/(x^4+x^2+2*x+1),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - x - 1\right )} {\left (x + 2\right )} \left (-\frac {2 \, x^{2} - x - 1}{4 \, x^{2} + x + 1}\right )^{\frac {1}{4}}}{2 \, {\left (x^{4} + x^{2} + 2 \, x + 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2+x)*(x^2-x-1)*((-2*x^2+x+1)/(4*x^2+x+1))^(1/4)/x/(x^4+x^2+2*x+1),x, algorithm="giac")

[Out]

integrate(1/2*(x^2 - x - 1)*(x + 2)*(-(2*x^2 - x - 1)/(4*x^2 + x + 1))^(1/4)/((x^4 + x^2 + 2*x + 1)*x), x)

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (2+x \right ) \left (x^{2}-x -1\right ) \left (\frac {-2 x^{2}+x +1}{4 x^{2}+x +1}\right )^{\frac {1}{4}}}{2 x \left (x^{4}+x^{2}+2 x +1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2+x)*(x^2-x-1)*((-2*x^2+x+1)/(4*x^2+x+1))^(1/4)/x/(x^4+x^2+2*x+1),x)

[Out]

int(1/2*(2+x)*(x^2-x-1)*((-2*x^2+x+1)/(4*x^2+x+1))^(1/4)/x/(x^4+x^2+2*x+1),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, \int \frac {{\left (x^{2} - x - 1\right )} {\left (x + 2\right )} \left (-\frac {2 \, x^{2} - x - 1}{4 \, x^{2} + x + 1}\right )^{\frac {1}{4}}}{{\left (x^{4} + x^{2} + 2 \, x + 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2+x)*(x^2-x-1)*((-2*x^2+x+1)/(4*x^2+x+1))^(1/4)/x/(x^4+x^2+2*x+1),x, algorithm="maxima")

[Out]

1/2*integrate((x^2 - x - 1)*(x + 2)*(-(2*x^2 - x - 1)/(4*x^2 + x + 1))^(1/4)/((x^4 + x^2 + 2*x + 1)*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {\left (x+2\right )\,{\left (\frac {-2\,x^2+x+1}{4\,x^2+x+1}\right )}^{1/4}\,\left (-x^2+x+1\right )}{2\,x\,\left (x^4+x^2+2\,x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x + 2)*((x - 2*x^2 + 1)/(x + 4*x^2 + 1))^(1/4)*(x - x^2 + 1))/(2*x*(2*x + x^2 + x^4 + 1)),x)

[Out]

int(-((x + 2)*((x - 2*x^2 + 1)/(x + 4*x^2 + 1))^(1/4)*(x - x^2 + 1))/(2*x*(2*x + x^2 + x^4 + 1)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2+x)*(x**2-x-1)*((-2*x**2+x+1)/(4*x**2+x+1))**(1/4)/x/(x**4+x**2+2*x+1),x)

[Out]

Timed out

________________________________________________________________________________________