\(\int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [189]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt [4]{a+b} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \] Output:

-1/2*(a+b)^(1/4)*(1+b^(1/2)*cos(d*x+c)^2/(a+b)^(1/2))*((a+b-2*b*cos(d*x+c) 
^2+b*cos(d*x+c)^4)/(a+b)/(1+b^(1/2)*cos(d*x+c)^2/(a+b)^(1/2))^2)^(1/2)*Inv 
erseJacobiAM(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)),1/2*(2+2*b^(1/2)/(a+ 
b)^(1/2))^(1/2))/b^(1/4)/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.79 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.72 \[ \int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {2 \sqrt {2} \cos ^3(c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {\left (1-\frac {i \sqrt {a}}{\sqrt {b}}\right ) \sec ^2(c+d x)} \left (\sqrt {a}+\left (\sqrt {a}+i \sqrt {b}\right ) \tan ^2(c+d x)\right ) \sqrt {-\frac {i \left (a+i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right )}{\sqrt {a} \sqrt {b}}}}{\left (\sqrt {a}+i \sqrt {b}\right ) d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))} \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}} \] Input:

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

Output:

(2*Sqrt[2]*Cos[c + d*x]^3*EllipticF[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + 
(I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[ 
a] - I*Sqrt[b])]*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]*(Sqrt[a] + 
 (Sqrt[a] + I*Sqrt[b])*Tan[c + d*x]^2)*Sqrt[((-I)*(a + I*Sqrt[a]*Sqrt[b] + 
 (a + b)*Tan[c + d*x]^2))/(Sqrt[a]*Sqrt[b])])/((Sqrt[a] + I*Sqrt[b])*d*Sqr 
t[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*Sqrt[1 + (I*Sqrt[ 
a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])])
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3694, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3694

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

\(\Big \downarrow \) 1416

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

Input:

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

Output:

-1/2*((a + b)^(1/4)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])*Sqrt[(a + b 
 - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4)/((a + b)*(1 + (Sqrt[b]*Cos[c + d 
*x]^2)/Sqrt[a + b])^2)]*EllipticF[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^ 
(1/4)], (1 + Sqrt[b]/Sqrt[a + b])/2])/(b^(1/4)*d*Sqrt[a + b - 2*b*Cos[c + 
d*x]^2 + b*Cos[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 3694

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.80 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {\sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \operatorname {EllipticF}\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )}{d \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \cos \left (d x +c \right )^{2}+b \cos \left (d x +c \right )^{4}}}\)	\(163\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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