Integrand size = 31, antiderivative size = 369 \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]
-arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)+b^(3/2)*arctan(b^(1/2)*( g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(3/4)/f/g^(1/2) -arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)+b^(3/2)*arctanh(b^(1/2) *(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(3/4)/f/g^(1/ 2)-b*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f* x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/f/(a^2-b*(b-(- a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-b*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1 /2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/ 2))*cos(f*x+e)^(1/2)/f/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 15.71 (sec) , antiderivative size = 663, normalized size of antiderivative = 1.80 \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {\cos (e+f x)} \left (-\frac {-2 \sqrt {2} b^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt {2} b^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 \left (a^2-b^2\right )^{3/4} \arctan \left (\sqrt {\cos (e+f x)}\right )-2 \left (a^2-b^2\right )^{3/4} \log \left (1-\sqrt {\cos (e+f x)}\right )+2 \left (a^2-b^2\right )^{3/4} \log \left (1+\sqrt {\cos (e+f x)}\right )-\sqrt {2} b^{3/2} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\sqrt {2} b^{3/2} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )}{8 a \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (a^2-b^2+b^2 \cos ^2(e+f x)\right ) \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \sqrt {\sin ^2(e+f x)}}\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{f \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \]
(2*Sqrt[Cos[e + f*x]]*(-1/8*(-2*Sqrt[2]*b^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b ]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*Sqrt[2]*b^(3/2)*ArcTan[1 + (S qrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4*(a^2 - b^2)^(3/4 )*ArcTan[Sqrt[Cos[e + f*x]]] - 2*(a^2 - b^2)^(3/4)*Log[1 - Sqrt[Cos[e + f* x]]] + 2*(a^2 - b^2)^(3/4)*Log[1 + Sqrt[Cos[e + f*x]]] - Sqrt[2]*b^(3/2)*L og[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] + Sqrt[2]*b^(3/2)*Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]* (a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]])/(a*(a^2 - b^2)^(3/ 4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos [e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/((a^2 - b^2 + b^2*Cos[e + f *x]^2)*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[ e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f* x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^ 2)*Sqrt[Sin[e + f*x]^2]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(f*Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x]))
Time = 1.18 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3042, 3377, 2009}
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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {\csc (e+f x)}{a \sqrt {g \cos (e+f x)}}-\frac {b}{a \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f \sqrt {g}}\) |
-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g])) + (b^(3/2)*ArcTan[(S qrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2 )^(3/4)*f*Sqrt[g]) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g]) + (b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[ g])])/(a*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[ (2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b*(b + Sqrt[-a^2 + b^2]))*f *Sqrt[g*Cos[e + f*x]])
3.14.93.3.1 Defintions of rubi rules used
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 2, 0])
Time = 0.77 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.50
| method | result | size |
| default | \(-\frac {\ln \left (\frac {4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sqrt {-g}+\ln \left (-\frac {2 \left (2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}-2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sqrt {g}}{2 a \sqrt {-g}\, \sqrt {g}\, f}\) | \(184\) |
-1/2/a/(-g)^(1/2)/g^(1/2)*(ln(2/(-1+cos(1/2*f*x+1/2*e))*(2*g*cos(1/2*f*x+1 /2*e)+g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*(-g)^(1/2)+ln(-2/(co s(1/2*f*x+1/2*e)+1)*(2*g*cos(1/2*f*x+1/2*e)-g^(1/2)*(-2*g*sin(1/2*f*x+1/2* e)^2+g)^(1/2)+g))*(-g)^(1/2)-2*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*s in(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*g^(1/2))/f
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
\[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]