Integrand size = 35, antiderivative size = 140 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} \sqrt {c} f}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} \sqrt {c-d} f} \]
-2*arctanh(cos(f*x+e)*a^(1/2)*c^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+ e))^(1/2))/f/a^(1/2)/c^(1/2)+arctanh(1/2*cos(f*x+e)*a^(1/2)*(c-d)^(1/2)*2^ (1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))*2^(1/2)/f/a^(1/2)/(c- d)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(885\) vs. \(2(140)=280\).
Time = 14.91 (sec) , antiderivative size = 885, normalized size of antiderivative = 6.32 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\csc (e+f x) \left (\frac {\log \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c}}-\frac {\sqrt {2} \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c-d}}+\frac {\log \left (d+\sqrt {2} \sqrt {c} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c}}-\frac {\log \left (c+\sqrt {2} \sqrt {c} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+d \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c}}+\frac {\sqrt {2} \log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c-d}}\right )}{f \sqrt {a (1+\sin (e+f x))} \sqrt {c+d \sin (e+f x)} \left (\frac {\csc \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right )}{2 \sqrt {c}}-\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {2} \sqrt {c-d} \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {\frac {1}{2} c \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c} d \cos (e+f x) \sqrt {\frac {1}{1+\cos (e+f x)}}}{\sqrt {2} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {c} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} \sin (e+f x) \sqrt {c+d \sin (e+f x)}}{\sqrt {2}}}{\sqrt {c} \left (d+\sqrt {2} \sqrt {c} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+c \tan \left (\frac {1}{2} (e+f x)\right )\right )}-\frac {\frac {1}{2} d \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c} d \cos (e+f x) \sqrt {\frac {1}{1+\cos (e+f x)}}}{\sqrt {2} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {c} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} \sin (e+f x) \sqrt {c+d \sin (e+f x)}}{\sqrt {2}}}{\sqrt {c} \left (c+\sqrt {2} \sqrt {c} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+d \tan \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {\sqrt {2} \left (\frac {1}{2} (-c+d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} d \cos (e+f x) \sqrt {\frac {1}{1+\cos (e+f x)}}}{\sqrt {c+d \sin (e+f x)}}+\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} \sin (e+f x) \sqrt {c+d \sin (e+f x)}\right )}{\sqrt {c-d} \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )}\right )} \]
(Csc[e + f*x]*(Log[Tan[(e + f*x)/2]]/Sqrt[c] - (Sqrt[2]*Log[1 + Tan[(e + f *x)/2]])/Sqrt[c - d] + Log[d + Sqrt[2]*Sqrt[c]*Sqrt[(1 + Cos[e + f*x])^(-1 )]*Sqrt[c + d*Sin[e + f*x]] + c*Tan[(e + f*x)/2]]/Sqrt[c] - Log[c + Sqrt[2 ]*Sqrt[c]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + d*Tan[( e + f*x)/2]]/Sqrt[c] + (Sqrt[2]*Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]])/Sqrt[ c - d]))/(f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c + d*Sin[e + f*x]]*((Csc[(e + f*x)/2]*Sec[(e + f*x)/2])/(2*Sqrt[c]) - Sec[(e + f*x)/2]^2/(Sqrt[2]*Sqrt[ c - d]*(1 + Tan[(e + f*x)/2])) + ((c*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c]*d*Co s[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/(Sqrt[2]*Sqrt[c + d*Sin[e + f*x] ]) + (Sqrt[c]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[ e + f*x]])/Sqrt[2])/(Sqrt[c]*(d + Sqrt[2]*Sqrt[c]*Sqrt[(1 + Cos[e + f*x])^ (-1)]*Sqrt[c + d*Sin[e + f*x]] + c*Tan[(e + f*x)/2])) - ((d*Sec[(e + f*x)/ 2]^2)/2 + (Sqrt[c]*d*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/(Sqrt[2]* Sqrt[c + d*Sin[e + f*x]]) + (Sqrt[c]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/Sqrt[2])/(Sqrt[c]*(c + Sqrt[2]*Sqrt[c]*S qrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + d*Tan[(e + f*x)/2] )) + (Sqrt[2]*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*d*Cos[e + f* x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqrt[c - d]*( (1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]))/...
Time = 0.89 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3426, 3042, 3261, 221, 3422, 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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3426 |
| \(\displaystyle \frac {\int \frac {\csc (e+f x) \sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}-\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x) \sqrt {c+d \sin (e+f x)}}dx}{a}-\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {2 a \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}+\frac {\int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x) \sqrt {c+d \sin (e+f x)}}dx}{a}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a}}{\sin (e+f x) \sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}}\) |
| \(\Big \downarrow \) 3422 |
| \(\displaystyle \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}}-\frac {2 \int \frac {1}{1-\frac {a c \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {\cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} \sqrt {c} f}\) |
(-2*ArcTanh[(Sqrt[a]*Sqrt[c]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[ c + d*Sin[e + f*x]])])/(Sqrt[a]*Sqrt[c]*f) + (Sqrt[2]*ArcTanh[(Sqrt[a]*Sqr t[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[a]*Sqrt[c - d]*f)
3.1.38.3.1 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[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(sin[(e_.) + (f_.)*(x_)]*Sqr t[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subs t[Int[1/(1 - a*c*x^2), x], x, Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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[b*c + a*d, 0]
Int[1/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*S qrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-b/a Int[1/ (Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] + Simp[1/a In t[Sqrt[a + b*Sin[e + f*x]]/(Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && (NeQ[a^2 - b^2, 0] || NeQ[c^2 - d^2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(346\) vs. \(2(113)=226\).
Time = 1.65 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.48
| method | result | size |
| default | \(\frac {\sqrt {c +d \sin \left (f x +e \right )}\, \sqrt {2}\, \left (\ln \left (-\frac {-\sqrt {c}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+c \cot \left (f x +e \right )-c \csc \left (f x +e \right )-d}{\sqrt {c}}\right ) \sqrt {2 c -2 d}-\ln \left (-\frac {2 \left (\sqrt {c}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )+c \sin \left (f x +e \right )-\cos \left (f x +e \right ) d +d \right )}{\cos \left (f x +e \right )-1}\right ) \sqrt {2 c -2 d}+2 \ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 \cos \left (f x +e \right ) d -2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {c}\right ) \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right )}{2 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {c}\, \sqrt {2 c -2 d}}\) | \(347\) |
1/2/f*(c+d*sin(f*x+e))^(1/2)*2^(1/2)*(ln(-(-c^(1/2)*2^(1/2)*((c+d*sin(f*x+ e))/(1+cos(f*x+e)))^(1/2)+c*cot(f*x+e)-c*csc(f*x+e)-d)/c^(1/2))*(2*c-2*d)^ (1/2)-ln(-2*(c^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(1+cos(f*x+e)))^(1/2)*sin(f *x+e)+c*sin(f*x+e)-cos(f*x+e)*d+d)/(cos(f*x+e)-1))*(2*c-2*d)^(1/2)+2*ln(2* ((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(1+cos(f*x+e)))^(1/2)*sin(f*x+e )+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-cos(f*x+e)*d-c+d)/(-cos(f*x+e)+1+ sin(f*x+e)))*c^(1/2))*(cos(f*x+e)+sin(f*x+e)+1)/(1+cos(f*x+e))/(a*(1+sin(f *x+e)))^(1/2)/((c+d*sin(f*x+e))/(1+cos(f*x+e)))^(1/2)/c^(1/2)/(2*c-2*d)^(1 /2)
Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (113) = 226\).
Time = 0.71 (sec) , antiderivative size = 3005, normalized size of antiderivative = 21.46 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/4*(sqrt(2)*a*c*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 - (13*c^2 - 22*c*d - 3*d^2)*cos(f*x + e)^2 + 4*sqrt(2)*((c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 - 4*c^2 + 8*c*d - 4*d^2 - (3*c^2 - 4*c*d + d^2)*cos(f*x + e) + (4*c ^2 - 8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqr t(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/sqrt(a*c - a*d) - 4*c^2 - 8 *c*d - 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 + 2*(7*c^2 - 18*c*d + 7*d^2 )*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f* x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4))/sqrt(a* c - a*d) + sqrt(a*c)*log(((a*c^4 - 28*a*c^3*d + 70*a*c^2*d^2 - 28*a*c*d^3 + a*d^4)*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a* d^4 - (31*a*c^4 - 196*a*c^3*d + 154*a*c^2*d^2 - 4*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 2*(81*a*c^4 - 252*a*c^3*d + 150*a*c^2*d^2 - 28*a*c*d^3 + a*d^4)* cos(f*x + e)^3 + 2*(79*a*c^4 - 100*a*c^3*d + 74*a*c^2*d^2 - 4*a*c*d^3 - a* d^4)*cos(f*x + e)^2 - 8*((c^3 - 7*c^2*d + 7*c*d^2 - d^3)*cos(f*x + e)^4 - 2*(5*c^3 - 14*c^2*d + 5*c*d^2)*cos(f*x + e)^3 + 51*c^3 - 59*c^2*d + 17*c*d ^2 - d^3 - 2*(18*c^3 - 33*c^2*d + 12*c*d^2 - d^3)*cos(f*x + e)^2 + 2*(13*c ^3 - 14*c^2*d + 5*c*d^2)*cos(f*x + e) + ((c^3 - 7*c^2*d + 7*c*d^2 - d^3)*c os(f*x + e)^3 - 51*c^3 + 59*c^2*d - 17*c*d^2 + d^3 + (11*c^3 - 35*c^2*d + 17*c*d^2 - d^3)*cos(f*x + e)^2 - (25*c^3 - 31*c^2*d + 7*c*d^2 - d^3)*co...
\[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}}\, dx \]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \]