Integrand size = 33, antiderivative size = 448 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f \sqrt {g}}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f \sqrt {g}}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {\sqrt {g \cos (e+f x)} \csc (e+f x)}{a f g}+\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{a f \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b^2+b \sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}} \]
b*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(1/2)-b^(5/2)*arctan(b^(1/2 )*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/(-a^2+b^2)^(3/4)/f/g^ (1/2)+b*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(1/2)-b^(5/2)*arctan h(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/(-a^2+b^2)^(3 /4)/f/g^(1/2)+(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(si n(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/a/f/(g*cos(f*x+e))^(1/2)+b^2*(c os(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/f/(a^2-b^2+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/f/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-cs c(f*x+e)*(g*cos(f*x+e))^(1/2)/a/f/g
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 26.86 (sec) , antiderivative size = 2093, normalized size of antiderivative = 4.67 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Result too large to show} \]
-(Cot[e + f*x]/(a*f*Sqrt[g*Cos[e + f*x]])) - (Sqrt[Cos[e + f*x]]*((4*a*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(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]])/(S qrt[1 - 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)*A ppellF1[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)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt [b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[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]] - 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]]))/(-a^2 + b^2)^(3/4)))/(S qrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - (b*(-1 + Cos[e + f*x]^2)*( a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*Csc[e + f*x]*((-10*Sqrt[2 ]*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2 )^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) + (10*Sqrt[2]*(2*a^2 - b^2)*ArcTan [1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*( a^2 - b^2)^(3/4)) - (20*ArcTan[Sqrt[Cos[e + f*x]]])/a - (16*b*AppellF1[5/4 , 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e...
Time = 1.24 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
| \(\displaystyle \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^2 \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {b^2}{a^2 \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}-\frac {b \csc (e+f x)}{a^2 \sqrt {g \cos (e+f x)}}+\frac {\csc ^2(e+f x)}{a \sqrt {g \cos (e+f x)}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f \sqrt {g}}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f \sqrt {g}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f \left (b \sqrt {b^2-a^2}+a^2-b^2\right ) \sqrt {g \cos (e+f x)}}+\frac {b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {\csc (e+f x) \sqrt {g \cos (e+f x)}}{a f g}+\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{a f \sqrt {g \cos (e+f x)}}\) |
(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*Sqrt[g]) - (b^(5/2)*ArcTan [(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f *Sqrt[g]) - (b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^ (1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) - (Sqrt[g*Cos[e + f*x] ]*Csc[e + f*x])/(a*f*g) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/( a*f*Sqrt[g*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*(a^2 - b^2 + b*Sqrt[-a^2 + b^2])*f* Sqrt[g*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*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt [g*Cos[e + f*x]])
3.14.94.3.1 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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 5.27 (sec) , antiderivative size = 1278, normalized size of antiderivative = 2.85
(4*b*(-1/4/a^2/(-g)^(1/2)*ln((-2*g+2*(-g)^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2- g)^(1/2))/cos(1/2*f*x+1/2*e))+1/8/a^2/g^(1/2)*ln((-4*g*cos(1/2*f*x+1/2*e)+ 2*g^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-2*g)/(cos(1/2*f*x+1/2*e)+1)) +1/8/a^2/g^(1/2)*ln((4*g*cos(1/2*f*x+1/2*e)+2*g^(1/2)*(-2*g*sin(1/2*f*x+1/ 2*e)^2+g)^(1/2)-2*g)/(-1+cos(1/2*f*x+1/2*e)))+b^2/a^2*(g^2*(a^2-b^2)/b^2)^ (1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2 *g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/(2*g*c os(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2- g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*g*cos(1/ 2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/ 4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2*(a^2-b^2)/b^ 2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4)))/(16*a^2-16*b^2)/g)+1/8*(g*(2*cos(1/2 *f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/a/cos(1/2*f*x+1/2*e)/(-2*g*si n(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(3/2)*(8*cos(1/2*f*x+1/2*e)*(2* sin(1/2*f*x+1/2*e)^2-1)^(3/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticF(cos(1 /2*f*x+1/2*e),2^(1/2))*sin(1/2*f*x+1/2*e)^2*g-16*g*sin(1/2*f*x+1/2*e)^6-co s(1/2*f*x+1/2*e)*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(3/2)/ a^2*sum(1/_alpha/(2*_alpha^2-1)*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2 )*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticPi (cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*_alpha^3*b^2-8...
Timed out. \[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\csc ^{2}{\left (e + f x \right )}}{\sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
Exception generated. \[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^2\,\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]