3.1.12 \(\int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx\) [12]

3.1.12.1 Optimal result

Integrand size = 32, antiderivative size = 125 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=-\frac {\sqrt {1-i x^2}}{2 (1+x)}-\frac {\sqrt {1+i x^2}}{2 (1+x)}-\frac {1}{4} (1-i)^{3/2} \text {arctanh}\left (\frac {1+i x}{\sqrt {1-i} \sqrt {1-i x^2}}\right )-\frac {1}{4} (1+i)^{3/2} \text {arctanh}\left (\frac {1-i x}{\sqrt {1+i} \sqrt {1+i x^2}}\right ) \]

output

-1/4*(1-I)^(3/2)*arctanh((1+I*x)/(1-I)^(1/2)/(1-I*x^2)^(1/2))-1/4*(1+I)^(3 
/2)*arctanh((1-I*x)/(1+I)^(1/2)/(1+I*x^2)^(1/2))-1/2*(1-I*x^2)^(1/2)/(1+x) 
-1/2*(1+I*x^2)^(1/2)/(1+x)
 

3.1.12.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(125)=250\).

Time = 2.37 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.18 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\frac {1}{2} \left (\frac {-1-2 x^4-\sqrt {1+x^4}-x^2 \left (1+2 \sqrt {1+x^4}\right )}{(1+x) \left (x^2+\sqrt {1+x^4}\right )^{3/2}}+\frac {\arctan \left (\sqrt {1+\sqrt {2}} \sqrt {x^2+\sqrt {1+x^4}}\right )}{\sqrt {-1+\sqrt {2}}}-\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )-\frac {\text {arctanh}\left (\sqrt {-1+\sqrt {2}} \sqrt {x^2+\sqrt {1+x^4}}\right )}{\sqrt {1+\sqrt {2}}}+\sqrt {-1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )\right ) \]

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
 

3.1.12.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.21, 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)
 

3.1.12.3.1 Defintions of rubi rules used

rule 219

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])
 

rule 488

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]
 

rule 491

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]
 

rule 2558

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]
 

3.1.12.4 Maple [F]

\[\int \frac {\sqrt {x^{2}+\sqrt {x^{4}+1}}}{\left (1+x \right )^{2} \sqrt {x^{4}+1}}d x\]

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)
 

3.1.12.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (81) = 162\).

Time = 1.17 (sec) , antiderivative size = 502, normalized size of antiderivative = 4.02 \[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=-\frac {{\left (x + 1\right )} \sqrt {-\sqrt {2} - 1} \log \left (-\frac {\sqrt {2} {\left (x^{2} + 1\right )} \sqrt {-\sqrt {2} - 1} + {\left (2 \, x^{3} + \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} - \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} + 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} - 2 \, \sqrt {x^{4} + 1} \sqrt {-\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) - {\left (x + 1\right )} \sqrt {-\sqrt {2} - 1} \log \left (\frac {\sqrt {2} {\left (x^{2} + 1\right )} \sqrt {-\sqrt {2} - 1} - {\left (2 \, x^{3} + \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} - \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} + 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} - 2 \, \sqrt {x^{4} + 1} \sqrt {-\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) - {\left (x + 1\right )} \sqrt {\sqrt {2} - 1} \log \left (-\frac {{\left (2 \, x^{3} - \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} + \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + {\left (\sqrt {2} {\left (x^{2} + 1\right )} + 2 \, \sqrt {x^{4} + 1}\right )} \sqrt {\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) + {\left (x + 1\right )} \sqrt {\sqrt {2} - 1} \log \left (-\frac {{\left (2 \, x^{3} - \sqrt {2} {\left (x^{3} - x^{2} - x - 1\right )} + \sqrt {x^{4} + 1} {\left (\sqrt {2} {\left (x - 1\right )} - 2 \, x\right )} - 2\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} - {\left (\sqrt {2} {\left (x^{2} + 1\right )} + 2 \, \sqrt {x^{4} + 1}\right )} \sqrt {\sqrt {2} - 1}}{x^{2} + 2 \, x + 1}\right ) - 4 \, \sqrt {x^{2} + \sqrt {x^{4} + 1}} {\left (x^{2} - \sqrt {x^{4} + 1} - 1\right )}}{8 \, {\left (x + 1\right )}} \]

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*((x + 1)*sqrt(-sqrt(2) - 1)*log(-(sqrt(2)*(x^2 + 1)*sqrt(-sqrt(2) - 1 
) + (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)) - 2*sqrt(x^4 + 1)*sqrt(-sqrt(2) - 1) 
)/(x^2 + 2*x + 1)) - (x + 1)*sqrt(-sqrt(2) - 1)*log((sqrt(2)*(x^2 + 1)*sqr 
t(-sqrt(2) - 1) - (2*x^3 + sqrt(2)*(x^3 - x^2 - x - 1) - sqrt(x^4 + 1)*(sq 
rt(2)*(x - 1) + 2*x) - 2)*sqrt(x^2 + sqrt(x^4 + 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(sq 
rt(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)) - 4*sqrt(x^2 + sqrt(x^4 + 1))*(x^2 - sqrt(x^4 + 1 
) - 1))/(x + 1)
 

3.1.12.6 Sympy [F]

\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\left (x + 1\right )^{2} \sqrt {x^{4} + 1}}\, dx \]

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)
 

3.1.12.7 Maxima [F]

\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\sqrt {x^{4} + 1} {\left (x + 1\right )}^{2}} \,d x } \]

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)
 

3.1.12.8 Giac [F]

\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\sqrt {x^{4} + 1} {\left (x + 1\right )}^{2}} \,d x } \]

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)
 

3.1.12.9 Mupad [F(-1)]

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)
 

3.1.12.10 Reduce [F]

\[ \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x)^2 \sqrt {1+x^4}} \, dx=\int \frac {\sqrt {\sqrt {x^{4}+1}+x^{2}}\, \sqrt {x^{4}+1}}{x^{6}+2 x^{5}+x^{4}+x^{2}+2 x +1}d x \]

input

int(sqrt(sqrt(x**4 + 1) + x**2)/(sqrt(x**4 + 1)*(x**2 + 2*x + 1)),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)