Integrand size = 11, antiderivative size = 173 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=-\frac {1}{16} \sqrt {\frac {1}{5} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )}-2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {1}{16} \sqrt {\frac {1}{5} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )}+2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}}+\frac {1}{16} \sqrt {\frac {1}{5} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {2 \left (-2+\sqrt {5}\right )} x}{\sqrt {5}+x^2}\right ) \] Output:
-1/80*(5+5*5^(1/2))^(1/2)*arctan(((-4+2*5^(1/2))^(1/2)-2*x)/(4+2*5^(1/2))^ (1/2))+1/80*(5+5*5^(1/2))^(1/2)*arctan(((-4+2*5^(1/2))^(1/2)+2*x)/(4+2*5^( 1/2))^(1/2))+1/16*arctanh(2^(1/2)*x/(x^2+1))*2^(1/2)+1/80*(-5+5*5^(1/2))^( 1/2)*arctanh((-4+2*5^(1/2))^(1/2)*x/(5^(1/2)+x^2))
Result contains complex when optimal does not.
Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.49 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=\frac {1}{160} \left ((6-2 i) \sqrt {2-i} \arctan \left (\frac {x}{\sqrt {2-i}}\right )+(6+2 i) \sqrt {2+i} \arctan \left (\frac {x}{\sqrt {2+i}}\right )+5 \sqrt {2} \left (-\log \left (-1+\sqrt {2} x-x^2\right )+\log \left (1+\sqrt {2} x+x^2\right )\right )\right ) \] Input:
Integrate[(4 + (1 + x^2)^4)^(-1),x]
Output:
((6 - 2*I)*Sqrt[2 - I]*ArcTan[x/Sqrt[2 - I]] + (6 + 2*I)*Sqrt[2 + I]*ArcTa n[x/Sqrt[2 + I]] + 5*Sqrt[2]*(-Log[-1 + Sqrt[2]*x - x^2] + Log[1 + Sqrt[2] *x + x^2]))/160
Time = 0.69 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.37, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2460, 2009}
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.
| \(\displaystyle \int \frac {1}{\left (x^2+1\right )^4+4} \, dx\) |
| \(\Big \downarrow \) 2460 |
| \(\displaystyle \int \left (\frac {1-x^2}{8 \left (x^4+1\right )}+\frac {x^2+3}{8 \left (x^4+4 x^2+5\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{16} \sqrt {\frac {1}{5} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {2 \left (\sqrt {5}-2\right )}-2 x}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )+\frac {1}{16} \sqrt {\frac {1}{5} \left (1+\sqrt {5}\right )} \arctan \left (\frac {2 x+\sqrt {2 \left (\sqrt {5}-2\right )}}{\sqrt {2 \left (2+\sqrt {5}\right )}}\right )-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {1}{32} \sqrt {\frac {1}{5} \left (\sqrt {5}-1\right )} \log \left (x^2-\sqrt {2 \left (\sqrt {5}-2\right )} x+\sqrt {5}\right )+\frac {1}{32} \sqrt {\frac {1}{5} \left (\sqrt {5}-1\right )} \log \left (x^2+\sqrt {2 \left (\sqrt {5}-2\right )} x+\sqrt {5}\right )\) |
Input:
Int[(4 + (1 + x^2)^4)^(-1),x]
Output:
-1/16*(Sqrt[(1 + Sqrt[5])/5]*ArcTan[(Sqrt[2*(-2 + Sqrt[5])] - 2*x)/Sqrt[2* (2 + Sqrt[5])]]) + (Sqrt[(1 + Sqrt[5])/5]*ArcTan[(Sqrt[2*(-2 + Sqrt[5])] + 2*x)/Sqrt[2*(2 + Sqrt[5])]])/16 - Log[1 - Sqrt[2]*x + x^2]/(16*Sqrt[2]) + Log[1 + Sqrt[2]*x + x^2]/(16*Sqrt[2]) - (Sqrt[(-1 + Sqrt[5])/5]*Log[Sqrt[ 5] - Sqrt[2*(-2 + Sqrt[5])]*x + x^2])/32 + (Sqrt[(-1 + Sqrt[5])/5]*Log[Sqr t[5] + Sqrt[2*(-2 + Sqrt[5])]*x + x^2])/32
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, Int[ExpandIntegrand[u*(Qx /. x -> x^2)^p, x], x] /; !SumQ[NonfreeFactors[Q x, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0] && RationalFunctionQ[u, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.39
| method | result | size |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (20 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (10 \textit {\_R}^{3}+7 \textit {\_R} +2 x \right )\right )}{16}+\frac {\sqrt {2}\, \ln \left (x^{2}+x \sqrt {2}+1\right )}{32}-\frac {\sqrt {2}\, \ln \left (x^{2}-x \sqrt {2}+1\right )}{32}\) | \(68\) |
| default | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}+x \sqrt {2}+1}{x^{2}-x \sqrt {2}+1}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{64}-\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-x \sqrt {2}+1}{x^{2}+x \sqrt {2}+1}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{64}-\frac {\left (\sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+5 \sqrt {-4+2 \sqrt {5}}\right ) \ln \left (x^{2}-x \sqrt {-4+2 \sqrt {5}}+\sqrt {5}\right )}{320}-\frac {\left (-6 \sqrt {5}+\frac {\left (\sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+5 \sqrt {-4+2 \sqrt {5}}\right ) \sqrt {-4+2 \sqrt {5}}}{2}\right ) \arctan \left (\frac {2 x -\sqrt {-4+2 \sqrt {5}}}{\sqrt {4+2 \sqrt {5}}}\right )}{80 \sqrt {4+2 \sqrt {5}}}+\frac {\left (\sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+5 \sqrt {-4+2 \sqrt {5}}\right ) \ln \left (x^{2}+x \sqrt {-4+2 \sqrt {5}}+\sqrt {5}\right )}{320}+\frac {\left (6 \sqrt {5}-\frac {\left (\sqrt {-4+2 \sqrt {5}}\, \sqrt {5}+5 \sqrt {-4+2 \sqrt {5}}\right ) \sqrt {-4+2 \sqrt {5}}}{2}\right ) \arctan \left (\frac {\sqrt {-4+2 \sqrt {5}}+2 x}{\sqrt {4+2 \sqrt {5}}}\right )}{80 \sqrt {4+2 \sqrt {5}}}\) | \(353\) |
Input:
int(1/(4+(x^2+1)^4),x,method=_RETURNVERBOSE)
Output:
1/16*sum(_R*ln(10*_R^3+7*_R+2*x),_R=RootOf(20*_Z^4+4*_Z^2+1))+1/32*2^(1/2) *ln(x^2+x*2^(1/2)+1)-1/32*2^(1/2)*ln(x^2-x*2^(1/2)+1)
Time = 0.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.26 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=\frac {1}{16} \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{5}} \arctan \left (\frac {1}{2} \, {\left (3 \, \sqrt {5} x + 5 \, {\left (\sqrt {5} - 2\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} - 5 \, x\right )} \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{5}}\right ) - \frac {1}{16} \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{5}} \arctan \left (-\frac {1}{2} \, {\left (3 \, \sqrt {5} x - 5 \, {\left (\sqrt {5} - 2\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} - 5 \, x\right )} \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{5}}\right ) - \frac {1}{32} \, \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} \log \left (2 \, x^{2} + {\left (\sqrt {5} x - 5 \, x\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} + 2 \, \sqrt {5}\right ) + \frac {1}{32} \, \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} \log \left (2 \, x^{2} - {\left (\sqrt {5} x - 5 \, x\right )} \sqrt {\frac {1}{5} \, \sqrt {5} - \frac {1}{5}} + 2 \, \sqrt {5}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} + 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \] Input:
integrate(1/(4+(x^2+1)^4),x, algorithm="fricas")
Output:
1/16*sqrt(1/5*sqrt(5) + 1/5)*arctan(1/2*(3*sqrt(5)*x + 5*(sqrt(5) - 2)*sqr t(1/5*sqrt(5) - 1/5) - 5*x)*sqrt(1/5*sqrt(5) + 1/5)) - 1/16*sqrt(1/5*sqrt( 5) + 1/5)*arctan(-1/2*(3*sqrt(5)*x - 5*(sqrt(5) - 2)*sqrt(1/5*sqrt(5) - 1/ 5) - 5*x)*sqrt(1/5*sqrt(5) + 1/5)) - 1/32*sqrt(1/5*sqrt(5) - 1/5)*log(2*x^ 2 + (sqrt(5)*x - 5*x)*sqrt(1/5*sqrt(5) - 1/5) + 2*sqrt(5)) + 1/32*sqrt(1/5 *sqrt(5) - 1/5)*log(2*x^2 - (sqrt(5)*x - 5*x)*sqrt(1/5*sqrt(5) - 1/5) + 2* sqrt(5)) + 1/32*sqrt(2)*log((x^4 + 4*x^2 + 2*sqrt(2)*(x^3 + x) + 1)/(x^4 + 1))
Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.38 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \operatorname {RootSum} {\left (1310720 t^{4} + 1024 t^{2} + 1, \left ( t \mapsto t \log {\left (20480 t^{3} + 56 t + x \right )} \right )\right )} \] Input:
integrate(1/(4+(x**2+1)**4),x)
Output:
-sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 + sqrt(2)*log(x**2 + sqrt(2)*x + 1)/ 32 + RootSum(1310720*_t**4 + 1024*_t**2 + 1, Lambda(_t, _t*log(20480*_t**3 + 56*_t + x)))
\[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=\int { \frac {1}{{\left (x^{2} + 1\right )}^{4} + 4} \,d x } \] Input:
integrate(1/(4+(x^2+1)^4),x, algorithm="maxima")
Output:
1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 1/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/8*integrate((x^2 + 3)/(x^4 + 4*x^2 + 5), x)
Time = 0.26 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=-\frac {1}{80} \, \sqrt {5 \, \sqrt {5} - 11} \arctan \left (\frac {5^{\frac {3}{4}} {\left (x + 5^{\frac {1}{4}} \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}\right )}}{5 \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{80} \, \sqrt {5 \, \sqrt {5} - 11} \arctan \left (\frac {5^{\frac {3}{4}} {\left (x - 5^{\frac {1}{4}} \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}\right )}}{5 \, \sqrt {\frac {1}{5} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{160} \, \sqrt {5 \, \sqrt {5} + 11} \log \left (x^{2} + 2 \cdot 5^{\frac {1}{4}} x \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}} + \sqrt {5}\right ) + \frac {1}{160} \, \sqrt {5 \, \sqrt {5} + 11} \log \left (x^{2} - 2 \cdot 5^{\frac {1}{4}} x \sqrt {-\frac {1}{5} \, \sqrt {5} + \frac {1}{2}} + \sqrt {5}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \] Input:
integrate(1/(4+(x^2+1)^4),x, algorithm="giac")
Output:
-1/80*sqrt(5*sqrt(5) - 11)*arctan(1/5*5^(3/4)*(x + 5^(1/4)*sqrt(-1/5*sqrt( 5) + 1/2))/sqrt(1/5*sqrt(5) + 1/2)) - 1/80*sqrt(5*sqrt(5) - 11)*arctan(1/5 *5^(3/4)*(x - 5^(1/4)*sqrt(-1/5*sqrt(5) + 1/2))/sqrt(1/5*sqrt(5) + 1/2)) - 1/160*sqrt(5*sqrt(5) + 11)*log(x^2 + 2*5^(1/4)*x*sqrt(-1/5*sqrt(5) + 1/2) + sqrt(5)) + 1/160*sqrt(5*sqrt(5) + 11)*log(x^2 - 2*5^(1/4)*x*sqrt(-1/5*s qrt(5) + 1/2) + sqrt(5)) + 1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 1/32*sq rt(2)*log(x^2 - sqrt(2)*x + 1)
Time = 11.31 (sec) , antiderivative size = 1069, normalized size of antiderivative = 6.18 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=\text {Too large to display} \] Input:
int(1/((x^2 + 1)^4 + 4),x)
Output:
atan((((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*(8 *x + ((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*((2 097152*x - (2147483648*x*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2)) - 67108864)*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5 120 - 1/5120)^(1/2)))*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1 /5120)^(1/2))^3 - 256))*1i + ((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5 120 - 1/5120)^(1/2))*(8*x + ((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/51 20 - 1/5120)^(1/2))*((2097152*x - (2147483648*x*((- 5^(1/2)/5120 - 1/5120) ^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2)) + 67108864)*((- 5^(1/2)/5120 - 1/5 120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2)))*((- 5^(1/2)/5120 - 1/5120)^(1 /2) + (5^(1/2)/5120 - 1/5120)^(1/2))^3 + 256))*1i)/(((- 5^(1/2)/5120 - 1/5 120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*(8*x + ((- 5^(1/2)/5120 - 1/51 20)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*((2097152*x - (2147483648*x*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2)) - 67108864) *((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2)))*((- 5^ (1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))^3 - 256)) - (( - 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*(8*x + ((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120)^(1/2))*((2097152*x - (2147483648*x*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 - 1/5120 )^(1/2)) + 67108864)*((- 5^(1/2)/5120 - 1/5120)^(1/2) + (5^(1/2)/5120 -...
Time = 0.27 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.50 \[ \int \frac {1}{4+\left (1+x^2\right )^4} \, dx=\frac {\sqrt {2}\, \left (2 \sqrt {\sqrt {5}+2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}-2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-10 \sqrt {\sqrt {5}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}-2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-2 \sqrt {\sqrt {5}+2}\, \sqrt {5}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}+2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )+10 \sqrt {\sqrt {5}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {5}-2}\, \sqrt {2}+2 x}{\sqrt {\sqrt {5}+2}\, \sqrt {2}}\right )-\sqrt {\sqrt {5}-2}\, \sqrt {5}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )+\sqrt {\sqrt {5}-2}\, \sqrt {5}\, \mathrm {log}\left (\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )-5 \sqrt {\sqrt {5}-2}\, \mathrm {log}\left (-\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )+5 \sqrt {\sqrt {5}-2}\, \mathrm {log}\left (\sqrt {\sqrt {5}-2}\, \sqrt {2}\, x +\sqrt {5}+x^{2}\right )-10 \,\mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )+10 \,\mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )\right )}{320} \] Input:
int(1/(4+(x^2+1)^4),x)
Output:
(sqrt(2)*(2*sqrt(sqrt(5) + 2)*sqrt(5)*atan((sqrt(sqrt(5) - 2)*sqrt(2) - 2* x)/(sqrt(sqrt(5) + 2)*sqrt(2))) - 10*sqrt(sqrt(5) + 2)*atan((sqrt(sqrt(5) - 2)*sqrt(2) - 2*x)/(sqrt(sqrt(5) + 2)*sqrt(2))) - 2*sqrt(sqrt(5) + 2)*sqr t(5)*atan((sqrt(sqrt(5) - 2)*sqrt(2) + 2*x)/(sqrt(sqrt(5) + 2)*sqrt(2))) + 10*sqrt(sqrt(5) + 2)*atan((sqrt(sqrt(5) - 2)*sqrt(2) + 2*x)/(sqrt(sqrt(5) + 2)*sqrt(2))) - sqrt(sqrt(5) - 2)*sqrt(5)*log( - sqrt(sqrt(5) - 2)*sqrt( 2)*x + sqrt(5) + x**2) + sqrt(sqrt(5) - 2)*sqrt(5)*log(sqrt(sqrt(5) - 2)*s qrt(2)*x + sqrt(5) + x**2) - 5*sqrt(sqrt(5) - 2)*log( - sqrt(sqrt(5) - 2)* sqrt(2)*x + sqrt(5) + x**2) + 5*sqrt(sqrt(5) - 2)*log(sqrt(sqrt(5) - 2)*sq rt(2)*x + sqrt(5) + x**2) - 10*log( - sqrt(2)*x + x**2 + 1) + 10*log(sqrt( 2)*x + x**2 + 1)))/320