\(\int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [493]

Optimal result

Integrand size = 16, antiderivative size = 162 \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}} \] Output:

1/2*cos(d*x+c)^2*InverseJacobiAM(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)), 
1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)*(( 
a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^ 
2)^(1/2)/a^(1/4)/(a+b)^(1/4)/d/(a+b*sin(d*x+c)^4)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {2 i \cos ^2(c+d x) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {2+\left (2-\frac {2 i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))}} \] Input:

Integrate[1/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

((-2*I)*Cos[c + d*x]^2*EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*T 
an[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])]*Sqrt[1 + (1 + ( 
I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[2 + (2 - ((2*I)*Sqrt[b])/Sqrt[a]) 
*Tan[c + d*x]^2])/(Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*b*Co 
s[2*(c + d*x)] + b*Cos[4*(c + d*x)]])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 3689, 1416}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3689

\(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \int \frac {1}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{d \sqrt {a+b \sin ^4(c+d x)}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{a} d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}}\)

Input:

Int[1/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(Cos[c + d*x]^2*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], 
(1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[( 
a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Ta 
n[c + d*x]^2)^2])/(2*a^(1/4)*(a + b)^(1/4)*d*Sqrt[a + b*Sin[c + d*x]^4])
 

Defintions of rubi rules used

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 3042

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

rule 3689

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff*(a + b*Sin[e + f*x]^4)^p*((Sec[e + f 
*x]^2)^(2*p)/(f*(a + 2*a*Tan[e + f*x]^2 + (a + b)*Tan[e + f*x]^4)^p))   Sub 
st[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, p}, x] && IntegerQ[p - 1/2 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(395\) vs. \(2(136)=272\).

Time = 2.34 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.44

method	result	size
default	\(-\frac {\sqrt {\left (4 a +\cos \left (2 d x +2 c \right )^{2} b +b -2 b \cos \left (2 d x +2 c \right )\right ) \sin \left (2 d x +2 c \right )^{2}}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \left (\cos \left (2 d x +2 c \right )+1\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )}{\left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {4 a +\cos \left (2 d x +2 c \right )^{2} b +b -2 b \cos \left (2 d x +2 c \right )}\, d}\)	\(396\)

Input:

int(1/(a+b*sin(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-((4*a+cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+2*c))*sin(2*d*x+2*c)^2)^(1/2)*(- 
a*b)^(1/2)*((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+ 
2*c)+1))^(1/2)*(cos(2*d*x+2*c)+1)^2*((-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)/ 
(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2)*((b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b 
)/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2)*EllipticF(((-b+(-a*b)^(1/2))*(-1+ 
cos(2*d*x+2*c))/(-a*b)^(1/2)/(cos(2*d*x+2*c)+1))^(1/2),((b+(-a*b)^(1/2))/( 
-b+(-a*b)^(1/2)))^(1/2))/(-b+(-a*b)^(1/2))/(1/b*(-1+cos(2*d*x+2*c))*(cos(2 
*d*x+2*c)+1)*(-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)*(b*cos(2*d*x+2*c)+2*(-a* 
b)^(1/2)-b))^(1/2)/sin(2*d*x+2*c)/(4*a+cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+ 
2*c))^(1/2)/d
 

Fricas [F]

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

integrate(1/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 

Output:

integral(1/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b), x)
 

Sympy [F]

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

integrate(1/(a+b*sin(d*x+c)**4)**(1/2),x)
 

Output:

Integral(1/sqrt(a + b*sin(c + d*x)**4), x)
 

Maxima [F]

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

integrate(1/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/sqrt(b*sin(d*x + c)^4 + a), x)
 

Giac [F]

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

integrate(1/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

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

int(1/(a + b*sin(c + d*x)^4)^(1/2),x)
 

Output:

int(1/(a + b*sin(c + d*x)^4)^(1/2), x)
 

Reduce [F]

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

int(1/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

int(sqrt(sin(c + d*x)**4*b + a)/(sin(c + d*x)**4*b + a),x)