Integrand size = 33, antiderivative size = 651 \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \left (-a^2+b^2\right )^{7/4} f g^{5/2}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a \left (a^2-b^2\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a \left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}} \]
b*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*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)^(7/4)/f/g^ (5/2)+b*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f/g^(5/2)-b^(9/2)*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)^(7 /4)/f/g^(5/2)-2/3*b/a^2/f/g/(g*cos(f*x+e))^(3/2)-csc(f*x+e)/a/f/g/(g*cos(f *x+e))^(3/2)+5/3*sin(f*x+e)/a/f/g/(g*cos(f*x+e))^(3/2)-2/3*b^2*(b-a*sin(f* x+e))/a^2/(a^2-b^2)/f/g/(g*cos(f*x+e))^(3/2)+5/3*(cos(1/2*f*x+1/2*e)^2)^(1 /2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1 /2)/a/f/g^2/(g*cos(f*x+e))^(1/2)+2/3*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos( 1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/a/(a ^2-b^2)/f/g^2/(g*cos(f*x+e))^(1/2)-b^4*(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/(a^2-b^2)/f/g^2/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos( f*x+e))^(1/2)-b^4*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Elliptic Pi(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 /(a^2-b^2)/f/g^2/(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 = 28.96 (sec) , antiderivative size = 2183, normalized size of antiderivative = 3.35 \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Result too large to show} \]
(Cos[e + f*x]^(5/2)*((-2*(10*a^3 - 18*a*b^2)*(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]])/(Sqrt[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)*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)*(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)*S qrt[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)))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - ((-5*a^2*b + 3*b^3)*(-1 + Cos[e + f*x]^2)*(a + b*Sqr t[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 + (Sqr t[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 + f*x]^(...
Time = 1.76 (sec) , antiderivative size = 651, 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.
| \(\displaystyle \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^2 (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {b^2}{a^2 (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}-\frac {b \csc (e+f x)}{a^2 (g \cos (e+f x))^{5/2}}+\frac {\csc ^2(e+f x)}{a (g \cos (e+f x))^{5/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}-\frac {b^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f g^{5/2}}+\frac {2 b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b^2 (b-a \sin (e+f x))}{3 a^2 f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 b}{3 a^2 f g (g \cos (e+f x))^{3/2}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 a f g^2 \sqrt {g \cos (e+f x)}}+\frac {5 \sin (e+f x)}{3 a f g (g \cos (e+f x))^{3/2}}-\frac {\csc (e+f x)}{a f g (g \cos (e+f x))^{3/2}}\) |
(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*g^(5/2)) - (b^(9/2)*ArcTan [(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(7/4)*f*g^(5/2)) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f *g^(5/2)) - (b^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^ (1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(7/4)*f*g^(5/2)) - (2*b)/(3*a^2*f*g*(g* Cos[e + f*x])^(3/2)) - Csc[e + f*x]/(a*f*g*(g*Cos[e + f*x])^(3/2)) + (5*Sq rt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*a*f*g^2*Sqrt[g*Cos[e + f*x] ]) + (2*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*a*(a^2 - b^2) *f*g^2*Sqrt[g*Cos[e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*(a^2 - b*(b - Sqrt[- a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) - (b^4*Sqrt[Cos[e + f*x]]*Ellipti cPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*(a^2 - b^2)*(a^2 - b *(b + Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (5*Sin[e + f*x])/(3 *a*f*g*(g*Cos[e + f*x])^(3/2)) - (2*b^2*(b - a*Sin[e + f*x]))/(3*a^2*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))
3.15.7.3.1 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 = 8.32 (sec) , antiderivative size = 1662, normalized size of antiderivative = 2.55
(-1/g^2*b*(1/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))-2/(2+2^(1/2))^2/(-2+2^(1/2))^2/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))-2/(2+2^(1/2))^2/(-2+2^(1/2))^2/a^2/g^(1/2)*l n((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)))+4*b^4/(a-b)/(a+b)/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*c os(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*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*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-1/3*2^(1/2)/(2+2^(1/2) )/(-2+2^(1/2))/(a^2-b^2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(2^(1/2)+cos( 1/2*f*x+1/2*e))/g/(2*cos(1/2*f*x+1/2*e)*2^(1/2)-2*sin(1/2*f*x+1/2*e)^2+3)- 1/3*2^(1/2)/(2+2^(1/2))/(-2+2^(1/2))/(a^2-b^2)*(-2*g*sin(1/2*f*x+1/2*e)^2+ g)^(1/2)*(-2^(1/2)+cos(1/2*f*x+1/2*e))/g/(2*cos(1/2*f*x+1/2*e)*2^(1/2)+2*s in(1/2*f*x+1/2*e)^2-3))+1/24*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2 *e)^2)^(1/2)/a/g^2/cos(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-b^2)*(40*cos(1/2*f*x+1/2*e)*(2*sin(1/2*f*x+1/...
\[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
integral(sqrt(g*cos(f*x + e))*csc(f*x + e)^2/(b*g^3*cos(f*x + e)^3*sin(f*x + e) + a*g^3*cos(f*x + e)^3), x)
Timed out. \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\csc ^2(e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^2\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]