\(\int \frac {1}{a+a \sin ^4(x)} \, dx\) [65]

Optimal result

Integrand size = 10, antiderivative size = 290 \[ \int \frac {1}{a+a \sin ^4(x)} \, dx=\frac {\sqrt {1+\sqrt {2}} x}{2 a}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)+\left (2-\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}-\left (2-\sqrt {2}\right ) \cos ^2(x)-2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )}{4 a}-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \sqrt {-1+\sqrt {2}} \cos ^2(x)-\left (2-\sqrt {2}\right ) \cos (x) \sin (x)}{2+\sqrt {1+\sqrt {2}}-\left (2-\sqrt {2}\right ) \cos ^2(x)+2 \sqrt {-1+\sqrt {2}} \cos (x) \sin (x)}\right )}{4 a}+\frac {\sqrt {-1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} \tan (x)}{1+\sqrt {2} \tan ^2(x)}\right )}{4 a} \] Output:

1/2*(1+2^(1/2))^(1/2)*x/a+1/4*(1+2^(1/2))^(1/2)*arctan(((2^(1/2)-1)^(1/2)- 
2*(2^(1/2)-1)^(1/2)*cos(x)^2+(2-2^(1/2))*cos(x)*sin(x))/(2+(1+2^(1/2))^(1/ 
2)-(2-2^(1/2))*cos(x)^2-2*(2^(1/2)-1)^(1/2)*cos(x)*sin(x)))/a-1/4*(1+2^(1/ 
2))^(1/2)*arctan(((2^(1/2)-1)^(1/2)-2*(2^(1/2)-1)^(1/2)*cos(x)^2-(2-2^(1/2 
))*cos(x)*sin(x))/(2+(1+2^(1/2))^(1/2)-(2-2^(1/2))*cos(x)^2+2*(2^(1/2)-1)^ 
(1/2)*cos(x)*sin(x)))/a+1/4*(2^(1/2)-1)^(1/2)*arctanh((-2+2*2^(1/2))^(1/2) 
*tan(x)/(1+2^(1/2)*tan(x)^2))/a
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.17 \[ \int \frac {1}{a+a \sin ^4(x)} \, dx=\frac {\frac {2 \arctan \left (\sqrt {1-i} \tan (x)\right )}{\sqrt {1-i}}+\frac {2 \arctan \left (\sqrt {1+i} \tan (x)\right )}{\sqrt {1+i}}}{4 a} \] Input:

Integrate[(a + a*Sin[x]^4)^(-1),x]
 

Output:

((2*ArcTan[Sqrt[1 - I]*Tan[x]])/Sqrt[1 - I] + (2*ArcTan[Sqrt[1 + I]*Tan[x] 
])/Sqrt[1 + I])/(4*a)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.71, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3688, 1483, 1142, 27, 1083, 217, 1103}

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.

\(\displaystyle \int \frac {1}{a \sin ^4(x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{a \sin (x)^4+a}dx\)

\(\Big \downarrow \) 3688

\(\displaystyle \int \frac {\tan ^2(x)+1}{2 a \tan ^4(x)+2 a \tan ^2(x)+a}d\tan (x)\)

\(\Big \downarrow \) 1483

\(\displaystyle \frac {\int \frac {2 \sqrt {-1+\sqrt {2}}-\left (2-\sqrt {2}\right ) \tan (x)}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\int \frac {\left (2-\sqrt {2}\right ) \tan (x)+2 \sqrt {-1+\sqrt {2}}}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)-\frac {1}{4} \left (2-\sqrt {2}\right ) \int -\frac {2 \left (\sqrt {-1+\sqrt {2}}-2 \tan (x)\right )}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)+\frac {1}{4} \left (2-\sqrt {2}\right ) \int \frac {2 \left (2 \tan (x)+\sqrt {-1+\sqrt {2}}\right )}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)+\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {\sqrt {-1+\sqrt {2}}-2 \tan (x)}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)+\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {2 \tan (x)+\sqrt {-1+\sqrt {2}}}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {\sqrt {-1+\sqrt {2}}-2 \tan (x)}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\left (4 \tan (x)-2 \sqrt {-1+\sqrt {2}}\right )^2-4 \left (1+\sqrt {2}\right )}d\left (4 \tan (x)-2 \sqrt {-1+\sqrt {2}}\right )}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {2 \tan (x)+\sqrt {-1+\sqrt {2}}}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\left (4 \tan (x)+2 \sqrt {-1+\sqrt {2}}\right )^2-4 \left (1+\sqrt {2}\right )}d\left (4 \tan (x)+2 \sqrt {-1+\sqrt {2}}\right )}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {\sqrt {-1+\sqrt {2}}-2 \tan (x)}{2 \tan ^2(x)-2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)+\frac {\arctan \left (\frac {4 \tan (x)-2 \sqrt {\sqrt {2}-1}}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {2}}}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\frac {1}{2} \left (2-\sqrt {2}\right ) \int \frac {2 \tan (x)+\sqrt {-1+\sqrt {2}}}{2 \tan ^2(x)+2 \sqrt {-1+\sqrt {2}} \tan (x)+\sqrt {2}}d\tan (x)+\frac {\arctan \left (\frac {4 \tan (x)+2 \sqrt {\sqrt {2}-1}}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {2}}}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {\arctan \left (\frac {4 \tan (x)-2 \sqrt {\sqrt {2}-1}}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {2}}-\frac {1}{4} \left (2-\sqrt {2}\right ) \log \left (2 \tan ^2(x)-2 \sqrt {\sqrt {2}-1} \tan (x)+\sqrt {2}\right )}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}+\frac {\frac {\arctan \left (\frac {4 \tan (x)+2 \sqrt {\sqrt {2}-1}}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {2}}+\frac {1}{4} \left (2-\sqrt {2}\right ) \log \left (\sqrt {2} \tan ^2(x)+\sqrt {2 \left (\sqrt {2}-1\right )} \tan (x)+1\right )}{2 \sqrt {2 \left (\sqrt {2}-1\right )} a}\)

Input:

Int[(a + a*Sin[x]^4)^(-1),x]
 

Output:

(ArcTan[(-2*Sqrt[-1 + Sqrt[2]] + 4*Tan[x])/(2*Sqrt[1 + Sqrt[2]])]/Sqrt[2] 
- ((2 - Sqrt[2])*Log[Sqrt[2] - 2*Sqrt[-1 + Sqrt[2]]*Tan[x] + 2*Tan[x]^2])/ 
4)/(2*Sqrt[2*(-1 + Sqrt[2])]*a) + (ArcTan[(2*Sqrt[-1 + Sqrt[2]] + 4*Tan[x] 
)/(2*Sqrt[1 + Sqrt[2]])]/Sqrt[2] + ((2 - Sqrt[2])*Log[1 + Sqrt[2*(-1 + Sqr 
t[2])]*Tan[x] + Sqrt[2]*Tan[x]^2])/4)/(2*Sqrt[2*(-1 + Sqrt[2])]*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 1083

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

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1483

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]}, Simp[1/(2*c*q*r)   In 
t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[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] && N 
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3688

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = 
 FreeFactors[Tan[e + f*x], x]}, Simp[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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.48

method	result	size
risch	\(\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2-2 i}-\sqrt {-2-2 i}-1-2 i\right )}{8 a}-\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2-2 i}+\sqrt {-2-2 i}-1-2 i\right )}{8 a}+\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2+2 i}+\sqrt {-2+2 i}-1+2 i\right )}{8 a}-\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2+2 i}-\sqrt {-2+2 i}-1+2 i\right )}{8 a}\)	\(138\)
default	\(\frac {\frac {\sqrt {2}\, \left (-\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (-\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \tan \left (x \right )^{2}+\sqrt {2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{4}+\frac {\sqrt {2}\, \left (\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (\sqrt {2}+\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \tan \left (x \right )^{2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{4}}{a}\)	\(194\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.44 \[ \int \frac {1}{a+a \sin ^4(x)} \, dx=-\frac {1}{8} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} + 1}{a^{2}}} \log \left (\frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (a^{3} \sqrt {-\frac {1}{a^{4}}} \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {-\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} + 1}{a^{2}}} + \frac {1}{4} \, \cos \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} \cos \left (x\right )^{2} - a^{2}\right )} \sqrt {-\frac {1}{a^{4}}} - \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} + 1}{a^{2}}} \log \left (-\frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (a^{3} \sqrt {-\frac {1}{a^{4}}} \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {-\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} + 1}{a^{2}}} + \frac {1}{4} \, \cos \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} \cos \left (x\right )^{2} - a^{2}\right )} \sqrt {-\frac {1}{a^{4}}} - \frac {1}{4}\right ) - \frac {1}{8} \, \sqrt {\frac {1}{2}} \sqrt {\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} - 1}{a^{2}}} \log \left (\frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (a^{3} \sqrt {-\frac {1}{a^{4}}} \cos \left (x\right ) \sin \left (x\right ) - a \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} - 1}{a^{2}}} - \frac {1}{4} \, \cos \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} \cos \left (x\right )^{2} - a^{2}\right )} \sqrt {-\frac {1}{a^{4}}} + \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {\frac {1}{2}} \sqrt {\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} - 1}{a^{2}}} \log \left (-\frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (a^{3} \sqrt {-\frac {1}{a^{4}}} \cos \left (x\right ) \sin \left (x\right ) - a \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {a^{2} \sqrt {-\frac {1}{a^{4}}} - 1}{a^{2}}} - \frac {1}{4} \, \cos \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} \cos \left (x\right )^{2} - a^{2}\right )} \sqrt {-\frac {1}{a^{4}}} + \frac {1}{4}\right ) \] Input:

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

Output:

-1/8*sqrt(1/2)*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a^2)*log(1/2*sqrt(1/2)*(a^3*sq 
rt(-1/a^4)*cos(x)*sin(x) + a*cos(x)*sin(x))*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a 
^2) + 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) - 1/4) + 1/8* 
sqrt(1/2)*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a^2)*log(-1/2*sqrt(1/2)*(a^3*sqrt(- 
1/a^4)*cos(x)*sin(x) + a*cos(x)*sin(x))*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a^2) 
+ 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) - 1/4) - 1/8*sqrt 
(1/2)*sqrt((a^2*sqrt(-1/a^4) - 1)/a^2)*log(1/2*sqrt(1/2)*(a^3*sqrt(-1/a^4) 
*cos(x)*sin(x) - a*cos(x)*sin(x))*sqrt((a^2*sqrt(-1/a^4) - 1)/a^2) - 1/4*c 
os(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) + 1/4) + 1/8*sqrt(1/2)*s 
qrt((a^2*sqrt(-1/a^4) - 1)/a^2)*log(-1/2*sqrt(1/2)*(a^3*sqrt(-1/a^4)*cos(x 
)*sin(x) - a*cos(x)*sin(x))*sqrt((a^2*sqrt(-1/a^4) - 1)/a^2) - 1/4*cos(x)^ 
2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) + 1/4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.91 \[ \int \frac {1}{a+a \sin ^4(x)} \, dx=\frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} + 2} + a \sqrt {2 \, \sqrt {2} - 2} {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )}}{8 \, a^{3}} + \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} + 2} + a \sqrt {2 \, \sqrt {2} - 2} {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )}}{8 \, a^{3}} - \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} - 2} - a \sqrt {2 \, \sqrt {2} + 2} {\left | a \right |}\right )} \log \left (\tan \left (x\right )^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right )}{16 \, a^{3}} + \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} - 2} - a \sqrt {2 \, \sqrt {2} + 2} {\left | a \right |}\right )} \log \left (\tan \left (x\right )^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right )}{16 \, a^{3}} \] Input:

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

Output:

1/8*(a^2*sqrt(2*sqrt(2) + 2) + a*sqrt(2*sqrt(2) - 2)*abs(a))*(pi*floor(x/p 
i + 1/2) + arctan(2*(1/2)^(3/4)*((1/2)^(1/4)*sqrt(-sqrt(2) + 2) + 2*tan(x) 
)/sqrt(sqrt(2) + 2)))/a^3 + 1/8*(a^2*sqrt(2*sqrt(2) + 2) + a*sqrt(2*sqrt(2 
) - 2)*abs(a))*(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)))/a^3 - 1/16*(a^2*sqrt(2* 
sqrt(2) - 2) - a*sqrt(2*sqrt(2) + 2)*abs(a))*log(tan(x)^2 + (1/2)^(1/4)*sq 
rt(-sqrt(2) + 2)*tan(x) + sqrt(1/2))/a^3 + 1/16*(a^2*sqrt(2*sqrt(2) - 2) - 
 a*sqrt(2*sqrt(2) + 2)*abs(a))*log(tan(x)^2 - (1/2)^(1/4)*sqrt(-sqrt(2) + 
2)*tan(x) + sqrt(1/2))/a^3
 

Mupad [B] (verification not implemented)

Time = 37.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.20 \[ \int \frac {1}{a+a \sin ^4(x)} \, dx=\mathrm {atan}\left (a\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {-\frac {1}{32}-\frac {1}{32}{}\mathrm {i}}{a^2}}\,\left (4+4{}\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{32}-\frac {1}{32}{}\mathrm {i}}{a^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (a\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {-\frac {1}{32}+\frac {1}{32}{}\mathrm {i}}{a^2}}\,\left (4-4{}\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{32}+\frac {1}{32}{}\mathrm {i}}{a^2}}\,2{}\mathrm {i} \] Input:

int(1/(a + a*sin(x)^4),x)
 

Output:

atan(a*tan(x)*((- 1/32 - 1i/32)/a^2)^(1/2)*(4 + 4i))*((- 1/32 - 1i/32)/a^2 
)^(1/2)*2i - atan(a*tan(x)*((- 1/32 + 1i/32)/a^2)^(1/2)*(4 - 4i))*((- 1/32 
 + 1i/32)/a^2)^(1/2)*2i
 

Reduce [F]

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

int(1/(a+a*sin(x)^4),x)
 

Output:

int(1/(sin(x)**4 + 1),x)/a