Integrand size = 33, antiderivative size = 165 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} c f}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} c (c-d) \sqrt {c+d} f} \]
-2*arctanh(cos(f*x+e)*a^(1/2)/(a+a*sin(f*x+e))^(1/2))/c/f/a^(1/2)+arctanh( 1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/(c-d)/f/a^( 1/2)-2*d^(3/2)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x +e))^(1/2))/c/(c-d)/f/a^(1/2)/(c+d)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 3.37 (sec) , antiderivative size = 654, normalized size of antiderivative = 3.96 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\left (-2 \sqrt {c+d} \left ((2+2 i) (-1)^{3/4} c \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+(c-d) \left (\log \left (1+\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1-\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )-d^{3/2} \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]+d^{3/2} \text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 c (c-d) \sqrt {c+d} f \sqrt {a (1+\sin (e+f x))}} \]
((-2*Sqrt[c + d]*((2 + 2*I)*(-1)^(3/4)*c*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(- 1 + Tan[(e + f*x)/4])] + (c - d)*(Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x) /2]] - Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])) - d^(3/2)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4 ]]) + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3* d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^ 2 - c*#1^3) & ] + d^(3/2)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^ 4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Ta n[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*Sqrt[d]*Sqrt[c + d] *Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + S qrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + S in[(e + f*x)/2]))/(2*c*(c - d)*Sqrt[c + d]*f*Sqrt[a*(1 + Sin[e + f*x])])
Time = 1.00 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3419, 3042, 3252, 221, 3464, 3042, 3128, 219, 3252, 219}
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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3419 |
| \(\displaystyle \frac {d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{a c (c-d)}+\frac {\int \frac {\csc (e+f x) (a (c-d)-a d \sin (e+f x))}{\sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{a c (c-d)}+\frac {\int \frac {a (c-d)-a d \sin (e+f x)}{\sin (e+f x) \sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\int \frac {a (c-d)-a d \sin (e+f x)}{\sin (e+f x) \sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}-\frac {2 d^2 \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{c f (c-d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\int \frac {a (c-d)-a d \sin (e+f x)}{\sin (e+f x) \sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {(c-d) \int \csc (e+f x) \sqrt {\sin (e+f x) a+a}dx-a c \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(c-d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx-a c \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {2 a c \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}+(c-d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(c-d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x)}dx+\frac {\sqrt {2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {\sqrt {2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}-\frac {2 a (c-d) \int \frac {1}{a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sqrt {2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}-\frac {2 \sqrt {a} (c-d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{f}}{a c (c-d)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}\) |
((-2*Sqrt[a]*(c - d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x ]]])/f + (Sqrt[2]*Sqrt[a]*c*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f)/(a*c*(c - d)) - (2*d^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[ d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*c*(c - d)*Sqrt[c + d]*f)
3.1.24.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[1/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*( (c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[d^2/(c*(b*c - a*d )) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] + Simp[1/(c* (b*c - a*d)) Int[(b*c - a*d - b*d*Sin[e + f*x])/(Sin[e + f*x]*Sqrt[a + b* Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Time = 0.59 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.25
| method | result | size |
| default | \(\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-2 d^{2} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a^{\frac {5}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \sqrt {a \left (c +d \right ) d}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {a}}\right ) a^{2} \sqrt {a \left (c +d \right ) d}\, c +2 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {a}}\right ) a^{2} \sqrt {a \left (c +d \right ) d}\, d \right )}{\left (c -d \right ) c \sqrt {a \left (c +d \right ) d}\, a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(206\) |
(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-2*d^2*arctanh((a-a*sin(f*x+e))^ (1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)+2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^ (1/2)*2^(1/2)/a^(1/2))*a^2*c*(a*(c+d)*d)^(1/2)-2*arctanh((a-a*sin(f*x+e))^ (1/2)/a^(1/2))*a^2*(a*(c+d)*d)^(1/2)*c+2*arctanh((a-a*sin(f*x+e))^(1/2)/a^ (1/2))*a^2*(a*(c+d)*d)^(1/2)*d)/(c-d)/c/(a*(c+d)*d)^(1/2)/a^(5/2)/cos(f*x+ e)/(a+a*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (136) = 272\).
Time = 1.51 (sec) , antiderivative size = 1044, normalized size of antiderivative = 6.33 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]
[-1/2*(a*d*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*c os(f*x + e)^2 - c^2 - 2*c*d - d^2 + 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) + (c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/ (a*c + a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos (f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^ 2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + sqrt(2)*sqrt(a)*c*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f* x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - sqrt(a)*(c - d)*log((a*cos (f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)* sin(f*x + e) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a* cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a) /(cos(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - co s(f*x + e) - 1)))/((a*c^2 - a*c*d)*f), -1/2*(2*a*d*sqrt(-d/(a*c + a*d))*ar ctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + sqrt(2)*sqrt(a)*c*log(-(cos(f*x + e)^2 - (cos( f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*...
\[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int \frac {1}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right ) \sin {\left (e + f x \right )}}\, dx \]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )} \,d x } \]
Time = 0.40 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.15 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} d^{2} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c^{2} - c d\right )} \sqrt {-c d - d^{2}}} + \frac {\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}\right )}{c} + \frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c - d} - \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c - d}\right )}}{2 \, \sqrt {a} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
-1/2*sqrt(2)*(2*sqrt(2)*d^2*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e )/sqrt(-c*d - d^2))/((c^2 - c*d)*sqrt(-c*d - d^2)) + sqrt(2)*log(abs(-2*sq rt(2) + 4*sin(-1/4*pi + 1/2*f*x + 1/2*e))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c + log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(c - d) - l og(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(c - d))/(sqrt(a)*f*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e)))
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (c+d\,\sin \left (e+f\,x\right )\right )} \,d x \]