\(\int \frac {1}{1-\sin ^4(x)} \, dx\) [236]
Optimal result
Integrand size = 10, antiderivative size = 25 \[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{2 \sqrt {2}}+\frac {\tan (x)}{2}
\]
[Out]
1/4*arctan(2^(1/2)*tan(x))*2^(1/2)+1/2*tan(x)
Rubi [A] (verified)
Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80, number of
steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3288, 396, 209}
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {\arctan \left (\frac {\sin (x) \cos (x)}{\sin ^2(x)+\sqrt {2}+1}\right )}{2 \sqrt {2}}+\frac {x}{2 \sqrt {2}}+\frac {\tan (x)}{2}
\]
[In]
Int[(1 - Sin[x]^4)^(-1),x]
[Out]
x/(2*Sqrt[2]) + ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)]/(2*Sqrt[2]) + Tan[x]/2
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 396
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
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} \, dx,x,\tan (x)\right ) \\ & = \frac {\tan (x)}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\tan (x)\right ) \\ & = \frac {x}{2 \sqrt {2}}+\frac {\arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{2 \sqrt {2}}+\frac {\tan (x)}{2} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {1}{4} \left (\sqrt {2} \arctan \left (\sqrt {2} \tan (x)\right )+2 \tan (x)\right )
\]
[In]
Integrate[(1 - Sin[x]^4)^(-1),x]
[Out]
(Sqrt[2]*ArcTan[Sqrt[2]*Tan[x]] + 2*Tan[x])/4
Maple [A] (verified)
Time = 1.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
| | |
method | result | size |
| | |
default |
\(\frac {\arctan \left (\sqrt {2}\, \tan \left (x \right )\right ) \sqrt {2}}{4}+\frac {\tan \left (x \right )}{2}\) |
\(18\) |
risch |
\(\frac {i}{{\mathrm e}^{2 i x}+1}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{8}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{8}\) |
\(52\) |
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[In]
int(1/(1-sin(x)^4),x,method=_RETURNVERBOSE)
[Out]
1/4*arctan(2^(1/2)*tan(x))*2^(1/2)+1/2*tan(x)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (17) = 34\).
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right ) - 4 \, \sin \left (x\right )}{8 \, \cos \left (x\right )}
\]
[In]
integrate(1/(1-sin(x)^4),x, algorithm="fricas")
[Out]
-1/8*(sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*sin(x)))*cos(x) - 4*sin(x))/cos(x)
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 724 vs. \(2 (20) = 40\).
Time = 21.73 (sec) , antiderivative size = 724, normalized size of antiderivative = 28.96
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\text {Too large to display}
\]
[In]
integrate(1/(1-sin(x)**4),x)
[Out]
54608393*sqrt(2)*sqrt(3 - 2*sqrt(2))*(atan(tan(x/2)/sqrt(3 - 2*sqrt(2))) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)
**2/(63977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) + 77227930*sqrt(3 - 2*
sqrt(2))*(atan(tan(x/2)/sqrt(3 - 2*sqrt(2))) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**2/(63977712*sqrt(2)*tan(x/
2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) - 77227930*sqrt(3 - 2*sqrt(2))*(atan(tan(x/2)/sqrt
(3 - 2*sqrt(2))) + pi*floor((x/2 - pi/2)/pi))/(63977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148
- 63977712*sqrt(2)) - 54608393*sqrt(2)*sqrt(3 - 2*sqrt(2))*(atan(tan(x/2)/sqrt(3 - 2*sqrt(2))) + pi*floor((x/2
- pi/2)/pi))/(63977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) + 9369319*sq
rt(2)*sqrt(2*sqrt(2) + 3)*(atan(tan(x/2)/sqrt(2*sqrt(2) + 3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**2/(639777
12*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) + 13250218*sqrt(2*sqrt(2) + 3)*(a
tan(tan(x/2)/sqrt(2*sqrt(2) + 3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**2/(63977712*sqrt(2)*tan(x/2)**2 + 904
78148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) - 13250218*sqrt(2*sqrt(2) + 3)*(atan(tan(x/2)/sqrt(2*sqrt(2)
+ 3)) + pi*floor((x/2 - pi/2)/pi))/(63977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*
sqrt(2)) - 9369319*sqrt(2)*sqrt(2*sqrt(2) + 3)*(atan(tan(x/2)/sqrt(2*sqrt(2) + 3)) + pi*floor((x/2 - pi/2)/pi)
)/(63977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) - 90478148*tan(x/2)/(639
77712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2)) - 63977712*sqrt(2)*tan(x/2)/(6
3977712*sqrt(2)*tan(x/2)**2 + 90478148*tan(x/2)**2 - 90478148 - 63977712*sqrt(2))
Maxima [A] (verification not implemented)
none
Time = 0.40 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) + \frac {1}{2} \, \tan \left (x\right )
\]
[In]
integrate(1/(1-sin(x)^4),x, algorithm="maxima")
[Out]
1/4*sqrt(2)*arctan(sqrt(2)*tan(x)) + 1/2*tan(x)
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).
Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {1}{4} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - 2 \, \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - 2 \, \cos \left (2 \, x\right ) + 2}\right )\right )} + \frac {1}{2} \, \tan \left (x\right )
\]
[In]
integrate(1/(1-sin(x)^4),x, algorithm="giac")
[Out]
1/4*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - 2*sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - 2*cos(2*x) + 2))) + 1/2
*tan(x)
Mupad [B] (verification not implemented)
Time = 13.77 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
\[
\int \frac {1}{1-\sin ^4(x)} \, dx=\frac {\mathrm {tan}\left (x\right )}{2}+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{4}
\]
[In]
int(-1/(sin(x)^4 - 1),x)
[Out]
tan(x)/2 + (2^(1/2)*atan(2^(1/2)*tan(x)))/4