Integrand size = 23, antiderivative size = 469 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {-a} \cos (c+d x)}{\sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\right )}{2 \sqrt {-a} d}+\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 a d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right )^2 \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{4 a \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]
-1/2*arctan(cos(d*x+c)*(-a)^(1/2)/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1 /2))/d/(-a)^(1/2)+1/2*b^(1/4)*(a+b)^(1/4)*(cos(2*arctan(b^(1/4)*cos(d*x+c) /(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))*Elli pticF(sin(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))*(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))*(b^(1/2)-(a+b)^(1/2))* ((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c)^2*b^(1/2)/(a+b) ^(1/2))^2)^(1/2)/a/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)-1/4*(a+b) ^(1/4)*(cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arcta n(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))*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))*(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))*(b^(1/2) -(a+b)^(1/2))^2*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c) ^2*b^(1/2)/(a+b)^(1/2))^2)^(1/2)/a/b^(1/4)/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d *x+c)^4)^(1/2)
Result contains complex when optimal does not.
Time = 14.57 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.04 \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^3(c+d x) \sqrt {\left (1-\frac {i \sqrt {a}}{\sqrt {b}}\right ) \sec ^2(c+d x)} \left (\operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \left (a-i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right ) \sqrt {-\frac {i \left (a+i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right )}{\sqrt {a} \sqrt {b}}}+i \left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {b} \operatorname {EllipticPi}\left (-\frac {2 i \sqrt {b}}{\sqrt {a}-i \sqrt {b}},\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\frac {(a+b) \left (a \sec ^4(c+d x)+b \tan ^4(c+d x)\right )}{a b}}\right )}{(a+b) d \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}} \sqrt {\cos ^4(c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}} \]
(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)])
Time = 0.68 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.06, 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 |
\(\displaystyle -\frac {\frac {(a+b) \left (1-\frac {\sqrt {b}}{\sqrt {a+b}}\right ) \left (\frac {\sqrt [4]{a+b} \left (1-\frac {\sqrt {b}}{\sqrt {a+b}}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {b}+\sqrt {a+b}\right )^2}{4 \sqrt {b} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{4 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac {\left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{2 \sqrt {a}}\right )}{a}-\frac {\sqrt [4]{b} \sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 a \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}}{d}\) |
-((-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)
3.3.44.3.1 Defintions of rubi rules used
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]
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]
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)]
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]
\[\int \frac {\csc \left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}d x\]
\[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\csc \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
\[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\csc {\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\csc \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
\[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\csc \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {1}{\sin \left (c+d\,x\right )\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]