\(\int \frac {1}{1+\sin ^4(x)} \, dx\) [238]

Optimal result

Integrand size = 8, antiderivative size = 309 \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\frac {x}{2 \sqrt {-1+\sqrt {2}}}+\frac {\arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)-\left (-2+\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}+\left (-2+\sqrt {2}\right ) \cos ^2(x)-2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )}{4 \sqrt {-1+\sqrt {2}}}-\frac {\arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)+\left (-2+\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}+\left (-2+\sqrt {2}\right ) \cos ^2(x)+2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )}{4 \sqrt {-1+\sqrt {2}}}-\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (\sqrt {2}-2 \sqrt {-1+\sqrt {2}} \tan (x)+2 \tan ^2(x)\right )+\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (1+\sqrt {2 \left (-1+\sqrt {2}\right )} \tan (x)+\sqrt {2} \tan ^2(x)\right ) \]

[Out]

1/2*x/(2^(1/2)-1)^(1/2)+1/4*arctan((-cos(x)*sin(x)*(-2+2^(1/2))+(2^(1/2)-1)^(1/2)-2*cos(x)^2*(2^(1/2)-1)^(1/2)
)/(2+cos(x)^2*(-2+2^(1/2))-2*cos(x)*sin(x)*(2^(1/2)-1)^(1/2)+(1+2^(1/2))^(1/2)))/(2^(1/2)-1)^(1/2)-1/4*arctan(
(cos(x)*sin(x)*(-2+2^(1/2))+(2^(1/2)-1)^(1/2)-2*cos(x)^2*(2^(1/2)-1)^(1/2))/(2+cos(x)^2*(-2+2^(1/2))+2*cos(x)*
sin(x)*(2^(1/2)-1)^(1/2)+(1+2^(1/2))^(1/2)))/(2^(1/2)-1)^(1/2)-1/8*ln(2^(1/2)-2*(2^(1/2)-1)^(1/2)*tan(x)+2*tan
(x)^2)*(2^(1/2)-1)^(1/2)+1/8*ln(1+(-2+2*2^(1/2))^(1/2)*tan(x)+2^(1/2)*tan(x)^2)*(2^(1/2)-1)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3288, 1183, 648, 632, 210, 642} \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\frac {\arctan \left (\frac {-2 \sqrt {\sqrt {2}-1} \cos ^2(x)-\left (\sqrt {2}-2\right ) \sin (x) \cos (x)+\sqrt {\sqrt {2}-1}}{\left (\sqrt {2}-2\right ) \cos ^2(x)-2 \sqrt {\sqrt {2}-1} \sin (x) \cos (x)+\sqrt {1+\sqrt {2}}+2}\right )}{4 \sqrt {\sqrt {2}-1}}-\frac {\arctan \left (\frac {-2 \sqrt {\sqrt {2}-1} \cos ^2(x)+\left (\sqrt {2}-2\right ) \sin (x) \cos (x)+\sqrt {\sqrt {2}-1}}{\left (\sqrt {2}-2\right ) \cos ^2(x)+2 \sqrt {\sqrt {2}-1} \sin (x) \cos (x)+\sqrt {1+\sqrt {2}}+2}\right )}{4 \sqrt {\sqrt {2}-1}}+\frac {x}{2 \sqrt {\sqrt {2}-1}}-\frac {1}{8} \sqrt {\sqrt {2}-1} \log \left (2 \tan ^2(x)-2 \sqrt {\sqrt {2}-1} \tan (x)+\sqrt {2}\right )+\frac {1}{8} \sqrt {\sqrt {2}-1} \log \left (\sqrt {2} \tan ^2(x)+\sqrt {2 \left (\sqrt {2}-1\right )} \tan (x)+1\right ) \]

[In]

Int[(1 + Sin[x]^4)^(-1),x]

[Out]

x/(2*Sqrt[-1 + Sqrt[2]]) + ArcTan[(Sqrt[-1 + Sqrt[2]] - 2*Sqrt[-1 + Sqrt[2]]*Cos[x]^2 - (-2 + Sqrt[2])*Cos[x]*
Sin[x])/(2 + Sqrt[1 + Sqrt[2]] + (-2 + Sqrt[2])*Cos[x]^2 - 2*Sqrt[-1 + Sqrt[2]]*Cos[x]*Sin[x])]/(4*Sqrt[-1 + S
qrt[2]]) - ArcTan[(Sqrt[-1 + Sqrt[2]] - 2*Sqrt[-1 + Sqrt[2]]*Cos[x]^2 + (-2 + Sqrt[2])*Cos[x]*Sin[x])/(2 + Sqr
t[1 + Sqrt[2]] + (-2 + Sqrt[2])*Cos[x]^2 + 2*Sqrt[-1 + Sqrt[2]]*Cos[x]*Sin[x])]/(4*Sqrt[-1 + Sqrt[2]]) - (Sqrt
[-1 + Sqrt[2]]*Log[Sqrt[2] - 2*Sqrt[-1 + Sqrt[2]]*Tan[x] + 2*Tan[x]^2])/8 + (Sqrt[-1 + Sqrt[2]]*Log[1 + Sqrt[2
*(-1 + Sqrt[2])]*Tan[x] + Sqrt[2]*Tan[x]^2])/8

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1183

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 3288

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1+x^2}{1+2 x^2+2 x^4} \, dx,x,\tan (x)\right ) \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {-1+\sqrt {2}}-\left (1-\frac {1}{\sqrt {2}}\right ) x}{\frac {1}{\sqrt {2}}-\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right )}{2 \sqrt {2 \left (-1+\sqrt {2}\right )}}+\frac {\text {Subst}\left (\int \frac {\sqrt {-1+\sqrt {2}}+\left (1-\frac {1}{\sqrt {2}}\right ) x}{\frac {1}{\sqrt {2}}+\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right )}{2 \sqrt {2 \left (-1+\sqrt {2}\right )}} \\ & = -\left (\frac {1}{8} \sqrt {-1+\sqrt {2}} \text {Subst}\left (\int \frac {-\sqrt {-1+\sqrt {2}}+2 x}{\frac {1}{\sqrt {2}}-\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right )\right )+\frac {1}{8} \sqrt {-1+\sqrt {2}} \text {Subst}\left (\int \frac {\sqrt {-1+\sqrt {2}}+2 x}{\frac {1}{\sqrt {2}}+\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right )+\frac {1}{8} \sqrt {3+2 \sqrt {2}} \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {2}}-\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right )+\frac {1}{8} \sqrt {3+2 \sqrt {2}} \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {2}}+\sqrt {-1+\sqrt {2}} x+x^2} \, dx,x,\tan (x)\right ) \\ & = -\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (\sqrt {2}-2 \sqrt {-1+\sqrt {2}} \tan (x)+2 \tan ^2(x)\right )+\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (1+\sqrt {2 \left (-1+\sqrt {2}\right )} \tan (x)+\sqrt {2} \tan ^2(x)\right )-\frac {1}{4} \sqrt {3+2 \sqrt {2}} \text {Subst}\left (\int \frac {1}{-1-\sqrt {2}-x^2} \, dx,x,-\sqrt {-1+\sqrt {2}}+2 \tan (x)\right )-\frac {1}{4} \sqrt {3+2 \sqrt {2}} \text {Subst}\left (\int \frac {1}{-1-\sqrt {2}-x^2} \, dx,x,\sqrt {-1+\sqrt {2}}+2 \tan (x)\right ) \\ & = \frac {1}{2} \sqrt {1+\sqrt {2}} x+\frac {1}{4} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)+\left (2-\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}-\left (2-\sqrt {2}\right ) \cos ^2(x)-2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )-\frac {1}{4} \sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)-\left (2-\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}-\left (2-\sqrt {2}\right ) \cos ^2(x)+2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )-\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (\sqrt {2}-2 \sqrt {-1+\sqrt {2}} \tan (x)+2 \tan ^2(x)\right )+\frac {1}{8} \sqrt {-1+\sqrt {2}} \log \left (1+\sqrt {2 \left (-1+\sqrt {2}\right )} \tan (x)+\sqrt {2} \tan ^2(x)\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.15 \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{2 \sqrt {1-i}}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{2 \sqrt {1+i}} \]

[In]

Integrate[(1 + Sin[x]^4)^(-1),x]

[Out]

ArcTan[Sqrt[1 - I]*Tan[x]]/(2*Sqrt[1 - I]) + ArcTan[Sqrt[1 + I]*Tan[x]]/(2*Sqrt[1 + I])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.41

method	result	size
risch	\(\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2+2 i}+\sqrt {-2+2 i}-1+2 i\right )}{8}-\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2+2 i}-\sqrt {-2+2 i}-1+2 i\right )}{8}+\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2-2 i}-\sqrt {-2-2 i}-1-2 i\right )}{8}-\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2-2 i}+\sqrt {-2-2 i}-1-2 i\right )}{8}\)	\(126\)
default	\(\frac {\sqrt {2}\, \left (-\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (-\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \left (\tan ^{2}\left (x \right )\right )+\sqrt {2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{4}+\frac {\sqrt {2}\, \left (\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (\sqrt {2}+\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \left (\tan ^{2}\left (x \right )\right )\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{4}\)	\(190\)

[In]

int(1/(1+sin(x)^4),x,method=_RETURNVERBOSE)

[Out]

1/8*(-2+2*I)^(1/2)*ln(exp(2*I*x)+I*(-2+2*I)^(1/2)+(-2+2*I)^(1/2)-1+2*I)-1/8*(-2+2*I)^(1/2)*ln(exp(2*I*x)-I*(-2
+2*I)^(1/2)-(-2+2*I)^(1/2)-1+2*I)+1/8*(-2-2*I)^(1/2)*ln(exp(2*I*x)+I*(-2-2*I)^(1/2)-(-2-2*I)^(1/2)-1-2*I)-1/8*
(-2-2*I)^(1/2)*ln(exp(2*I*x)-I*(-2-2*I)^(1/2)+(-2-2*I)^(1/2)-1-2*I)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.38 \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\frac {1}{16} \, \sqrt {2} \sqrt {i - 1} \log \left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i - 1\right ) \, \cos \left (x\right )^{2} - i + 1\right ) - \frac {1}{16} \, \sqrt {2} \sqrt {i - 1} \log \left (\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i - 1\right ) \, \cos \left (x\right )^{2} - i + 1\right ) - \frac {1}{16} \, \sqrt {2} \sqrt {-i - 1} \log \left (\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i + 1\right ) \, \cos \left (x\right )^{2} - i - 1\right ) + \frac {1}{16} \, \sqrt {2} \sqrt {-i - 1} \log \left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i + 1\right ) \, \cos \left (x\right )^{2} - i - 1\right ) \]

[In]

integrate(1/(1+sin(x)^4),x, algorithm="fricas")

[Out]

1/16*sqrt(2)*sqrt(I - 1)*log(-(I - 1)*sqrt(2)*sqrt(I - 1)*cos(x)*sin(x) + (2*I - 1)*cos(x)^2 - I + 1) - 1/16*s
qrt(2)*sqrt(I - 1)*log((I - 1)*sqrt(2)*sqrt(I - 1)*cos(x)*sin(x) + (2*I - 1)*cos(x)^2 - I + 1) - 1/16*sqrt(2)*
sqrt(-I - 1)*log((I + 1)*sqrt(2)*sqrt(-I - 1)*cos(x)*sin(x) + (2*I + 1)*cos(x)^2 - I - 1) + 1/16*sqrt(2)*sqrt(
-I - 1)*log(-(I + 1)*sqrt(2)*sqrt(-I - 1)*cos(x)*sin(x) + (2*I + 1)*cos(x)^2 - I - 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\text {Timed out} \]

[In]

integrate(1/(1+sin(x)**4),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(1/(1+sin(x)^4),x, algorithm="maxima")

[Out]

integrate(1/(sin(x)^4 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.58 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.55 \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\frac {1}{4} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{4} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{8} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \left (x\right )^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \left (x\right )^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right ) \]

[In]

integrate(1/(1+sin(x)^4),x, algorithm="giac")

[Out]

1/4*(pi*floor(x/pi + 1/2) + arctan(2*(1/2)^(3/4)*((1/2)^(1/4)*sqrt(-sqrt(2) + 2) + 2*tan(x))/sqrt(sqrt(2) + 2)
))*sqrt(sqrt(2) + 1) + 1/4*(pi*floor(x/pi + 1/2) + arctan(-2*(1/2)^(3/4)*((1/2)^(1/4)*sqrt(-sqrt(2) + 2) - 2*t
an(x))/sqrt(sqrt(2) + 2)))*sqrt(sqrt(2) + 1) + 1/8*sqrt(sqrt(2) - 1)*log(tan(x)^2 + (1/2)^(1/4)*sqrt(-sqrt(2)
+ 2)*tan(x) + sqrt(1/2)) - 1/8*sqrt(sqrt(2) - 1)*log(tan(x)^2 - (1/2)^(1/4)*sqrt(-sqrt(2) + 2)*tan(x) + sqrt(1
/2))

Mupad [B] (verification not implemented)

Time = 14.51 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.76 \[ \int \frac {1}{1+\sin ^4(x)} \, dx=\mathrm {atanh}\left (\frac {\mathrm {tan}\left (x\right )}{8\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}}-\frac {\mathrm {tan}\left (x\right )}{8\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}}+\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{16\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}}+\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{16\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}-2\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}\right )+\mathrm {atanh}\left (\frac {\mathrm {tan}\left (x\right )}{8\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}}+\frac {\mathrm {tan}\left (x\right )}{8\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}}+\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{16\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}}-\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{16\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}}\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}+2\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}\right )-\frac {\left (x-\mathrm {atan}\left (\mathrm {tan}\left (x\right )\right )\right )\,\left (\pi \,\left (2\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}-2\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}\right )\,1{}\mathrm {i}+\pi \,\left (2\,\sqrt {-\frac {\sqrt {2}}{64}-\frac {1}{64}}+2\,\sqrt {\frac {\sqrt {2}}{64}-\frac {1}{64}}\right )\,1{}\mathrm {i}\right )}{\pi } \]

[In]

int(1/(sin(x)^4 + 1),x)

[Out]

atanh(tan(x)/(8*(- 2^(1/2)/64 - 1/64)^(1/2)) - tan(x)/(8*(2^(1/2)/64 - 1/64)^(1/2)) + (2^(1/2)*tan(x))/(16*(-
2^(1/2)/64 - 1/64)^(1/2)) + (2^(1/2)*tan(x))/(16*(2^(1/2)/64 - 1/64)^(1/2)))*(2*(- 2^(1/2)/64 - 1/64)^(1/2) -
2*(2^(1/2)/64 - 1/64)^(1/2)) + atanh(tan(x)/(8*(- 2^(1/2)/64 - 1/64)^(1/2)) + tan(x)/(8*(2^(1/2)/64 - 1/64)^(1
/2)) + (2^(1/2)*tan(x))/(16*(- 2^(1/2)/64 - 1/64)^(1/2)) - (2^(1/2)*tan(x))/(16*(2^(1/2)/64 - 1/64)^(1/2)))*(2
*(- 2^(1/2)/64 - 1/64)^(1/2) + 2*(2^(1/2)/64 - 1/64)^(1/2)) - ((x - atan(tan(x)))*(pi*(2*(- 2^(1/2)/64 - 1/64)
^(1/2) - 2*(2^(1/2)/64 - 1/64)^(1/2))*1i + pi*(2*(- 2^(1/2)/64 - 1/64)^(1/2) + 2*(2^(1/2)/64 - 1/64)^(1/2))*1i
))/pi