Integrand size = 32, antiderivative size = 312 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\frac {1+x^2+2 x^4+x \left (-1-x^2\right ) \left (x^2+\sqrt {1+x^4}\right )+\sqrt {1+x^4} \left (1+2 x^2-x \left (x^2+\sqrt {1+x^4}\right )\right )}{2 \left (-1+x^2\right ) \left (x^2+\sqrt {1+x^4}\right )^{3/2}}+\frac {\arctan \left (\frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {-1+\sqrt {2}}}\right )}{2 \sqrt {-1+\sqrt {2}}}-\frac {1}{2} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+\sqrt {2}}}\right )}{2 \sqrt {1+\sqrt {2}}}+\frac {1}{2} \sqrt {-1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right ) \] Output:
1/2*(1+x^2+2*x^4+x*(-x^2-1)*(x^2+(x^4+1)^(1/2))+(x^4+1)^(1/2)*(1+2*x^2-x*( x^2+(x^4+1)^(1/2))))/(x^2-1)/(x^2+(x^4+1)^(1/2))^(3/2)+1/2*arctan((x^2+(x^ 4+1)^(1/2))^(1/2)/(2^(1/2)-1)^(1/2))/(2^(1/2)-1)^(1/2)-1/2*(1+2^(1/2))^(1/ 2)*arctan((-2+2*2^(1/2))^(1/2)*x*(x^2+(x^4+1)^(1/2))^(1/2)/(1+x^2+(x^4+1)^ (1/2)))-1/2*arctanh((x^2+(x^4+1)^(1/2))^(1/2)/(1+2^(1/2))^(1/2))/(1+2^(1/2 ))^(1/2)+1/2*(2^(1/2)-1)^(1/2)*arctanh((2+2*2^(1/2))^(1/2)*x*(x^2+(x^4+1)^ (1/2))^(1/2)/(1+x^2+(x^4+1)^(1/2)))
Time = 2.44 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\frac {1}{2} \left (\frac {-1-2 x^4-\sqrt {1+x^4}-x^2 \left (1+2 \sqrt {1+x^4}\right )}{(1+x) \left (x^2+\sqrt {1+x^4}\right )^{3/2}}+\frac {\arctan \left (\sqrt {1+\sqrt {2}} \sqrt {x^2+\sqrt {1+x^4}}\right )}{\sqrt {-1+\sqrt {2}}}-\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )-\frac {\text {arctanh}\left (\sqrt {-1+\sqrt {2}} \sqrt {x^2+\sqrt {1+x^4}}\right )}{\sqrt {1+\sqrt {2}}}+\sqrt {-1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )\right ) \] Input:
Integrate[Sqrt[x^2 + Sqrt[1 + x^4]]/((1 + x)^2*Sqrt[1 + x^4]),x]
Output:
((-1 - 2*x^4 - Sqrt[1 + x^4] - x^2*(1 + 2*Sqrt[1 + x^4]))/((1 + x)*(x^2 + Sqrt[1 + x^4])^(3/2)) + ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[x^2 + Sqrt[1 + x^4]] ]/Sqrt[-1 + Sqrt[2]] - Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(-1 + Sqrt[2])]*x* Sqrt[x^2 + Sqrt[1 + x^4]])/(1 + x^2 + Sqrt[1 + x^4])] - ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[x^2 + Sqrt[1 + x^4]]]/Sqrt[1 + Sqrt[2]] + Sqrt[-1 + Sqrt[2]] *ArcTanh[(Sqrt[2*(1 + Sqrt[2])]*x*Sqrt[x^2 + Sqrt[1 + x^4]])/(1 + x^2 + Sq rt[1 + x^4])])/2
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.48, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2558, 491, 488, 219}
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.
\(\displaystyle \int \frac {\sqrt {\sqrt {x^4+1}+x^2}}{(x+1)^2 \sqrt {x^4+1}} \, dx\) |
\(\Big \downarrow \) 2558 |
\(\displaystyle \left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{(x+1)^2 \sqrt {1-i x^2}}dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{(x+1)^2 \sqrt {i x^2+1}}dx\) |
\(\Big \downarrow \) 491 |
\(\displaystyle \left (\frac {1}{2}-\frac {i}{2}\right ) \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{(x+1) \sqrt {1-i x^2}}dx-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1-i x^2}}{x+1}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{(x+1) \sqrt {i x^2+1}}dx-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+i x^2}}{x+1}\right )\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \left (\frac {1}{2}-\frac {i}{2}\right ) \left (\left (-\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{(1-i)-\frac {(i x+1)^2}{1-i x^2}}d\frac {i x+1}{\sqrt {1-i x^2}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1-i x^2}}{x+1}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{(1+i)-\frac {(1-i x)^2}{i x^2+1}}d\frac {1-i x}{\sqrt {i x^2+1}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+i x^2}}{x+1}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \left (\frac {1}{2}-\frac {i}{2}\right ) \left (-\frac {1}{2} \sqrt {1-i} \text {arctanh}\left (\frac {1+i x}{\sqrt {1-i} \sqrt {1-i x^2}}\right )-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1-i x^2}}{x+1}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) \left (-\frac {1}{2} \sqrt {1+i} \text {arctanh}\left (\frac {1-i x}{\sqrt {1+i} \sqrt {1+i x^2}}\right )-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+i x^2}}{x+1}\right )\) |
Input:
Int[Sqrt[x^2 + Sqrt[1 + x^4]]/((1 + x)^2*Sqrt[1 + x^4]),x]
Output:
(1/2 - I/2)*(((-1/2 - I/2)*Sqrt[1 - I*x^2])/(1 + x) - (Sqrt[1 - I]*ArcTanh [(1 + I*x)/(Sqrt[1 - I]*Sqrt[1 - I*x^2])])/2) + (1/2 + I/2)*(((-1/2 + I/2) *Sqrt[1 + I*x^2])/(1 + x) - (Sqrt[1 + I]*ArcTanh[(1 - I*x)/(Sqrt[1 + I]*Sq rt[1 + I*x^2])])/2)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + S imp[b*(c/(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n + 2*p + 3, 0]
Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^ 4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_Symbol] :> Simp[(1 - I)/2 Int[(c + d*x) ^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Simp[(1 + I)/2 Int[(c + d*x)^m/Sqrt[ Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && G tQ[a, 0]
\[\int \frac {\sqrt {x^{2}+\sqrt {x^{4}+1}}}{\left (1+x \right )^{2} \sqrt {x^{4}+1}}d x\]
Input:
int((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x)
Output:
int((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x)
Time = 1.04 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\frac {2 \, {\left (x + 1\right )} \sqrt {\sqrt {2} + 1} \arctan \left (\frac {{\left (x^{3} + x^{2} - \sqrt {2} {\left (x^{3} + 1\right )} + \sqrt {x^{4} + 1} {\left (\sqrt {2} x - x - 1\right )} - x + 1\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \sqrt {\sqrt {2} + 1}}{x^{2} - 2 \, x + 1}\right ) + {\left (x + 1\right )} \sqrt {\sqrt {2} - 1} \log \left (-\frac {{\left (2 \, x^{3} - \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} + \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + {\left (\sqrt {2} {\left (x^{2} + 1\right )} + 2 \, \sqrt {x^{4} + 1}\right )} \sqrt {\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) - {\left (x + 1\right )} \sqrt {\sqrt {2} - 1} \log \left (-\frac {{\left (2 \, x^{3} - \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} + \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} - {\left (\sqrt {2} {\left (x^{2} + 1\right )} + 2 \, \sqrt {x^{4} + 1}\right )} \sqrt {\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) + 4 \, \sqrt {x^{2} + \sqrt {x^{4} + 1}} {\left (x^{2} - \sqrt {x^{4} + 1} - 1\right )}}{8 \, {\left (x + 1\right )}} \] Input:
integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="fr icas")
Output:
1/8*(2*(x + 1)*sqrt(sqrt(2) + 1)*arctan((x^3 + x^2 - sqrt(2)*(x^3 + 1) + s qrt(x^4 + 1)*(sqrt(2)*x - x - 1) - x + 1)*sqrt(x^2 + sqrt(x^4 + 1))*sqrt(s qrt(2) + 1)/(x^2 - 2*x + 1)) + (x + 1)*sqrt(sqrt(2) - 1)*log(-((2*x^3 - sq rt(2)*(x^3 - x^2 - x - 1) + sqrt(x^4 + 1)*(sqrt(2)*(x - 1) - 2*x) - 2)*sqr t(x^2 + sqrt(x^4 + 1)) + (sqrt(2)*(x^2 + 1) + 2*sqrt(x^4 + 1))*sqrt(sqrt(2 ) - 1))/(x^2 + 2*x + 1)) - (x + 1)*sqrt(sqrt(2) - 1)*log(-((2*x^3 - sqrt(2 )*(x^3 - x^2 - x - 1) + sqrt(x^4 + 1)*(sqrt(2)*(x - 1) - 2*x) - 2)*sqrt(x^ 2 + sqrt(x^4 + 1)) - (sqrt(2)*(x^2 + 1) + 2*sqrt(x^4 + 1))*sqrt(sqrt(2) - 1))/(x^2 + 2*x + 1)) + 4*sqrt(x^2 + sqrt(x^4 + 1))*(x^2 - sqrt(x^4 + 1) - 1))/(x + 1)
\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\left (x + 1\right )^{2} \sqrt {x^{4} + 1}}\, dx \] Input:
integrate((x**2+(x**4+1)**(1/2))**(1/2)/(1+x)**2/(x**4+1)**(1/2),x)
Output:
Integral(sqrt(x**2 + sqrt(x**4 + 1))/((x + 1)**2*sqrt(x**4 + 1)), x)
\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\sqrt {x^{4} + 1} {\left (x + 1\right )}^{2}} \,d x } \] Input:
integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="ma xima")
Output:
integrate(sqrt(x^2 + sqrt(x^4 + 1))/(sqrt(x^4 + 1)*(x + 1)^2), x)
\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\sqrt {x^{4} + 1} {\left (x + 1\right )}^{2}} \,d x } \] Input:
integrate((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x, algorithm="gi ac")
Output:
integrate(sqrt(x^2 + sqrt(x^4 + 1))/(sqrt(x^4 + 1)*(x + 1)^2), x)
Timed out. \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int \frac {\sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}\,{\left (x+1\right )}^2} \,d x \] Input:
int(((x^4 + 1)^(1/2) + x^2)^(1/2)/((x^4 + 1)^(1/2)*(x + 1)^2),x)
Output:
int(((x^4 + 1)^(1/2) + x^2)^(1/2)/((x^4 + 1)^(1/2)*(x + 1)^2), x)
\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int \frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, \sqrt {x^{4}+1}}{x^{6}+2 x^{5}+x^{4}+x^{2}+2 x +1}d x \] Input:
int((x^2+(x^4+1)^(1/2))^(1/2)/(1+x)^2/(x^4+1)^(1/2),x)
Output:
int((sqrt(sqrt(x**4 + 1) + x**2)*sqrt(x**4 + 1))/(x**6 + 2*x**5 + x**4 + x **2 + 2*x + 1),x)