\(\int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [190]

Optimal result

Integrand size = 23, antiderivative size = 458 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\right )}{2 \sqrt {a} d}+\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 a d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (\sqrt {a+b}+\sqrt {b} \cos ^2(c+d x)\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{\left (\sqrt {a+b}+\sqrt {b} \cos ^2(c+d x)\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{4 a \sqrt [4]{b} \sqrt [4]{a+b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \] Output:

-1/2*arctanh(a^(1/2)*cos(d*x+c)/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2 
))/a^(1/2)/d+1/2*b^(1/4)*(a+b)^(1/4)*(b^(1/2)-(a+b)^(1/2))*(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)*InverseJacobiAM(2*arctan(b^(1/4)*co 
s(d*x+c)/(a+b)^(1/4)),1/2*(2+2*b^(1/2)/(a+b)^(1/2))^(1/2))/a/d/(a+b-2*b*co 
s(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)-1/4*(b^(1/2)-(a+b)^(1/2))^2*((a+b)^(1/2)+ 
b^(1/2)*cos(d*x+c)^2)*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/((a+b)^(1/2)+ 
b^(1/2)*cos(d*x+c)^2)^2)^(1/2)*EllipticPi(sin(2*arctan(b^(1/4)*cos(d*x+c)/ 
(a+b)^(1/4))),1/4*(b^(1/2)+(a+b)^(1/2))^2/b^(1/2)/(a+b)^(1/2),1/2*(2+2*b^( 
1/2)/(a+b)^(1/2))^(1/2))/a/b^(1/4)/(a+b)^(1/4)/d/(a+b-2*b*cos(d*x+c)^2+b*c 
os(d*x+c)^4)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.69 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.07 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^3(c+d x) \sqrt {\left (1-\frac {i \sqrt {a}}{\sqrt {b}}\right ) \sec ^2(c+d x)} \left (\operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \left (a-i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right ) \sqrt {-\frac {i \left (a+i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right )}{\sqrt {a} \sqrt {b}}}+i \left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {b} \operatorname {EllipticPi}\left (-\frac {2 i \sqrt {b}}{\sqrt {a}-i \sqrt {b}},\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\frac {(a+b) \left (a \sec ^4(c+d x)+b \tan ^4(c+d x)\right )}{a b}}\right )}{(a+b) d \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\cos ^4(c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}} \] Input:

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

Output:

(Cos[c + d*x]^3*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]*(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])]*(a - I*Sqrt[a]*Sqrt 
[b] + (a + 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])] + I*(Sqrt[a] + I*Sqrt[b])*Sqrt[b]*Elli 
pticPi[((-2*I)*Sqrt[b])/(Sqrt[a] - I*Sqrt[b]), 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] + (I*(a + b)*Tan[ 
c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[((a + b)*(a*Sec[c + d*x]^4 + b*Tan[c + 
 d*x]^4))/(a*b)]))/((a + b)*d*Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Ta 
n[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x] 
^2 + (a + b)*Tan[c + d*x]^4)])
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3694, 1540, 1416, 2222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3694

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

\(\Big \downarrow \) 1540

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 2222

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

Input:

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

Output:

-((-1/2*(b^(1/4)*(a + b)^(1/4)*(Sqrt[b] - Sqrt[a + b])*(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*Arc 
Tan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4)], (1 + Sqrt[b]/Sqrt[a + b])/2])/( 
a*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4]) + ((a + b)*(1 - Sqr 
t[b]/Sqrt[a + b])*(((1 + Sqrt[b]/Sqrt[a + b])*ArcTanh[(Sqrt[a]*Cos[c + d*x 
])/Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4]])/(2*Sqrt[a]) + ((a 
 + b)^(1/4)*(1 - Sqrt[b]/Sqrt[a + b])*(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)]*EllipticPi[(Sqrt[b] + Sqrt[a + b 
])^2/(4*Sqrt[b]*Sqrt[a + b]), 2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4 
)], (1 + Sqrt[b]/Sqrt[a + b])/2])/(4*b^(1/4)*Sqrt[a + b - 2*b*Cos[c + d*x] 
^2 + b*Cos[c + d*x]^4])))/a)/d)
 

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 1540

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_S 
ymbol] :> With[{q = Rt[c/a, 2]}, Simp[(c*d + a*e*q)/(c*d^2 - a*e^2)   Int[1 
/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[(a*e*(e + d*q))/(c*d^2 - a*e^2)   I 
nt[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), 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] && 
NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]
 

rule 2222

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A 
rcTanh[Rt[b - c*(d/e) - a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ 
b - c*(d/e) - a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + 
 b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*Ell 
ipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] 
/; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && 
 EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && NegQ[-b + c*(d/e) + a*(e/d)]
 

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 [F]

\[\int \frac {\csc \left (d x +c \right )}{\sqrt {a +b \sin \left (d x +c \right )^{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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