Integrand size = 33, antiderivative size = 462 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {b g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \sqrt {b} f}+\frac {b g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^2 \sqrt {b} f}-\frac {g (g \cos (e+f x))^{3/2} \csc (e+f x)}{a f}-\frac {g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a f \sqrt {\cos (e+f x)}}-\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a b \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{a b \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]
-b*g^(5/2)*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^2/f+b*g^(5/2)*arctanh((g *cos(f*x+e))^(1/2)/g^(1/2))/a^2/f-g*(g*cos(f*x+e))^(3/2)*csc(f*x+e)/a/f+(- a^2+b^2)^(3/4)*g^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4 )/g^(1/2))/a^2/f/b^(1/2)-(-a^2+b^2)^(3/4)*g^(5/2)*arctanh(b^(1/2)*(g*cos(f *x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^2/f/b^(1/2)-(a^2-b^2)*g^3*(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/b/f/(b-(-a^2+b^2)^(1/2)) /(g*cos(f*x+e))^(1/2)-(a^2-b^2)*g^3*(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/b/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-g^2*(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/f/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 23.11 (sec) , antiderivative size = 1465, normalized size of antiderivative = 3.17 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \]
((g*Cos[e + f*x])^(5/2)*(-4*Cos[e + f*x]^(3/2)*Csc[e + f*x] - (5*b*Csc[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) + 3*(2*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)] - 2* 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)] + 4*a^2*ArcTan[Sqrt[Cos[e + f*x]]] - 4*b^2*ArcT an[Sqrt[Cos[e + f*x]]] + 2*a^2*Log[1 - Sqrt[Cos[e + f*x]]] - 2*b^2*Log[1 - Sqrt[Cos[e + f*x]]] - 2*a^2*Log[1 + Sqrt[Cos[e + f*x]]] + 2*b^2*Log[1 + S qrt[Cos[e + f*x]]] - 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]] + 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]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(12*a*(a^2 - b^2)*(b + a*Csc[e + f*x])) + (12*a*((a*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))/(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)*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[Sq rt[-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)))*(a + b*Sqrt[Sin[e + f*...
Time = 1.38 (sec) , antiderivative size = 462, 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 |
| \(\displaystyle \int \frac {(g \cos (e+f x))^{5/2}}{\sin (e+f x)^2 (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {b^2 (g \cos (e+f x))^{5/2}}{a^2 (a+b \sin (e+f x))}-\frac {b \csc (e+f x) (g \cos (e+f x))^{5/2}}{a^2}+\frac {\csc ^2(e+f x) (g \cos (e+f x))^{5/2}}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 \sqrt {b} f}-\frac {b g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}-\frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^2 \sqrt {b} f}+\frac {b g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^2 f}-\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a b f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{a b f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a f \sqrt {\cos (e+f x)}}-\frac {g \csc (e+f x) (g \cos (e+f x))^{3/2}}{a f}\) |
-((b*g^(5/2)*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f)) + ((-a^2 + b^2 )^(3/4)*g^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)* Sqrt[g])])/(a^2*Sqrt[b]*f) + (b*g^(5/2)*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[ g]])/(a^2*f) - ((-a^2 + b^2)^(3/4)*g^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*Sqrt[b]*f) - (g*(g*Cos[e + f*x ])^(3/2)*Csc[e + f*x])/(a*f) - (g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f* x)/2, 2])/(a*f*Sqrt[Cos[e + f*x]]) - ((a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*E llipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*b*(b - Sqrt[-a ^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - ((a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*E llipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a*b*(b + Sqrt[-a ^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])
3.14.87.3.1 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 = 24.84 (sec) , antiderivative size = 1484, normalized size of antiderivative = 3.21
(4*g^3*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)))-1/16/a^2*(a^2-b^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*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)))/b^2)-2*(g*(2*cos(1/2*f *x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a*(-1/4/a^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)^(1/2)*(cos(1/2*f*x+ 1/2*e)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*Ellip ticF(cos(1/2*f*x+1/2*e),2^(1/2))-cos(1/2*f*x+1/2*e)*(2*sin(1/2*f*x+1/2*e)^ 2-1)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/ 2))-2*sin(1/2*f*x+1/2*e)^4+sin(1/2*f*x+1/2*e)^2)+1/8/a^2/sin(1/2*f*x+1/2*e )^2/g/(2*sin(1/2*f*x+1/2*e)^2-1)*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+ 1/2*e)^2)^(1/2)*(2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2-EllipticF(co...
\[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^2\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]