Integrand size = 31, antiderivative size = 507 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a \left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {2}{a f g \sqrt {g \cos (e+f x)}}+\frac {2 b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \]
arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)-b^(5/2)*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)^(5/4)/f/g^(3/2)- arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)+b^(5/2)*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)^(5/4)/f/g^(3/2 )+2/a/f/g/(g*cos(f*x+e))^(1/2)+2*b*(b-a*sin(f*x+e))/a/(a^2-b^2)/f/g/(g*cos (f*x+e))^(1/2)+b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ellipti cPi(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)/ (a^2-b^2)/f/g/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+b^2*(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)/(a^2-b^2)/f/g/(b+(-a^2+b^2)^(1/2) )/(g*cos(f*x+e))^(1/2)+2*b*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e) *EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(g*cos(f*x+e))^(1/2)/(a^2-b^2)/f/g^ 2/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 25.67 (sec) , antiderivative size = 1587, normalized size of antiderivative = 3.13 \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \]
-1/2*(Cos[e + f*x]^(3/2)*((8*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Appe llF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]* Cos[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)* Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sq rt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + L og[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x] ] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4))))/(Sqrt[1 - Cos[e + f *x]^2]*(b + a*Csc[e + f*x])) - ((-2*a^2 + b^2)*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Csc[e + f*x]*(6*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3 /4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 6 *Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 12*(a^2 - b^2)*ArcTan[Sqrt[Cos[e + f*x]]] + 8*a*b*AppellF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^ 2 + b^2)]*Cos[e + f*x]^(3/2) + 6*a^2*Log[1 - Sqrt[Cos[e + f*x]]] - 6*b^2*L og[1 - Sqrt[Cos[e + f*x]]] - 6*a^2*Log[1 + Sqrt[Cos[e + f*x]]] + 6*b^2*Log [1 + Sqrt[Cos[e + f*x]]] - 3*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*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]] + 3*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*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]]))/(12*...
Time = 1.63 (sec) , antiderivative size = 507, 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.
| \(\displaystyle \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) (g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {\csc (e+f x)}{a (g \cos (e+f x))^{3/2}}-\frac {b}{a (g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac {2 b E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)}}+\frac {2 b (b-a \sin (e+f x))}{a f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{f g \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {\arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f g^{3/2}}+\frac {2}{a f g \sqrt {g \cos (e+f x)}}\) |
ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) - (b^(5/2)*ArcTan[(Sqrt [b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^( 5/4)*f*g^(3/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) + (b ^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g]) ])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)) + 2/(a*f*g*Sqrt[g*Cos[e + f*x]]) + (2* b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/((a^2 - b^2)*f*g^2*Sqrt[ Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b - Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Co s[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b ^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[ e + f*x]]) + (2*b*(b - a*Sin[e + f*x]))/(a*(a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]])
3.14.100.3.1 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 = 0.84 (sec) , antiderivative size = 436, normalized size of antiderivative = 0.86
| method | result | size |
| default | \(-\frac {4 g^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 \sqrt {-g}\, \ln \left (\frac {4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +2 \sqrt {-g}\, \ln \left (-\frac {2 \left (2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g -2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}}+4 \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}\, \sqrt {-g}\, \sqrt {g}-\ln \left (\frac {4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g \sqrt {-g}-\ln \left (-\frac {2 \left (2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) g \sqrt {-g}}{2 g^{\frac {5}{2}} a \sqrt {-g}\, \left (2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-1\right ) f}\) | \(436\) |
-1/2/g^(5/2)/a/(-g)^(1/2)/(2*sin(1/2*f*x+1/2*e)^2-1)*(4*g^(3/2)*ln(2/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))*sin(1/2* f*x+1/2*e)^2+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))*sin(1/2*f*x+1/2*e)^2* g+2*(-g)^(1/2)*ln(-2/(cos(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))*sin(1/2*f*x+1/2*e)^2*g-2*ln(2/co s(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^(3/ 2)+4*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-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*(-g)^(1/2)-ln(-2/(cos(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*(-g)^(1/2))/f
\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
integral(sqrt(g*cos(f*x + e))*csc(f*x + e)/(b*g^2*cos(f*x + e)^2*sin(f*x + e) + a*g^2*cos(f*x + e)^2), x)
\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
\[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]