Integrand size = 35, antiderivative size = 431 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \left (3 a^2 A+A b^2-4 a b B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}+\frac {2 (a (3 A+B)-b (A+3 B)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) (a+b)^{3/2} d \sqrt {\sec (c+d x)}}-\frac {2 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {2 \left (3 a^2 A+A b^2-4 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \]
-2/3*(A*b-B*a)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(1 /2)+2/3*(3*A*a^2+A*b^2-4*B*a*b)*sin(d*x+c)*sec(d*x+c)^(1/2)/(a^2-b^2)^2/d/ (a+b*cos(d*x+c))^(1/2)-2/3*(3*A*a^2+A*b^2-4*B*a*b)*csc(d*x+c)*EllipticE((a +b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*co s(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/ 2)/a^2/(a-b)/(a+b)^(3/2)/d/sec(d*x+c)^(1/2)+2/3*(a*(3*A+B)-b*(A+3*B))*csc( d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a- b)/(a-b))^(1/2))*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec (d*x+c))/(a-b))^(1/2)/a/(a-b)/(a+b)^(3/2)/d/sec(d*x+c)^(1/2)
Time = 14.53 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {2 \left (3 a^2 A+A b^2-4 a b B\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2}+\frac {2 \left (-a A b \sin (c+d x)+a^2 B \sin (c+d x)\right )}{3 b \left (-a^2+b^2\right ) (a+b \cos (c+d x))^2}+\frac {2 \left (2 a^2 A b \sin (c+d x)+2 A b^3 \sin (c+d x)+a^3 B \sin (c+d x)-5 a b^2 B \sin (c+d x)\right )}{3 b \left (-a^2+b^2\right )^2 (a+b \cos (c+d x))}\right )}{d}-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-2 (a+b) \left (3 a^2 A+A b^2-4 a b B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+2 a (a+b) (3 a A+A b-a B-3 b B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-\left (3 a^2 A+A b^2-4 a b B\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \]
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(3*a^2*A + A*b^2 - 4*a*b *B)*Sin[c + d*x])/(3*a*(a^2 - b^2)^2) + (2*(-(a*A*b*Sin[c + d*x]) + a^2*B* Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (2*(2*a^2*A*b*S in[c + d*x] + 2*A*b^3*Sin[c + d*x] + a^3*B*Sin[c + d*x] - 5*a*b^2*B*Sin[c + d*x]))/(3*b*(-a^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d - (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(a + b)*(3*a^2*A + A*b^2 - 4*a*b*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a + b)*(3*a*A + A*b - a*B - 3*b*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt [(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[( c + d*x)/2]], (-a + b)/(a + b)] - (3*a^2*A + A*b^2 - 4*a*b*B)*Cos[c + d*x] *(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*a*(a^2 - b^ 2)^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2])
Time = 1.79 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3440, 3042, 3478, 27, 3042, 3472, 3042, 3477, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3440 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3478 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {2 \int \frac {A b-a B-3 (a A-b B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {A b-a B-3 (a A-b B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {A b-a B-3 (a A-b B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3472 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {3 A a^2-4 b B a+A b^2+\left (-B a^2+4 A b a-3 b^2 B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {3 A a^2-4 b B a+A b^2+\left (-B a^2+4 A b a-3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\left (3 a^2 A-4 a b B+A b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) (a (3 A+B)-b (A+3 B)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\left (3 a^2 A-4 a b B+A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (a (3 A+B)-b (A+3 B)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\left (3 a^2 A-4 a b B+A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (a (3 A+B)-b (A+3 B)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 A-4 a b B+A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 (a-b) \sqrt {a+b} (a (3 A+B)-b (A+3 B)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*(A*b - a*B)*Sqrt[Cos[c + d*x]]* Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (((2*(a - b)* Sqrt[a + b]*(3*a^2*A + A*b^2 - 4*a*b*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt [a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b)) ]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b) ])/(a^2*d) - (2*(a - b)*Sqrt[a + b]*(a*(3*A + B) - b*(A + 3*B))*Cot[c + d* x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x ]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))/(a^2 - b^2) - (2*(3*a^2*A + A*b^2 - 4*a* b*B)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d* x]]))/(3*(a^2 - b^2)))
3.7.28.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(d_.)*sin[(e_.) + (f_.)*( x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)), x_Symbol] :> Simp[2*(A *b - a*B)*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[d*Sin[ e + f*x]])), x] + Simp[d/(a^2 - b^2) Int[(A*b - a*B + (a*A - b*B)*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] /; FreeQ[{ a, b, d, e, f, A, B}, x] && NeQ[a^2 - b^2, 0]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(B*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f* x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[c*(a*A - b*B)*( m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*(m + 2))* Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; Free Q[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(4566\) vs. \(2(391)=782\).
Time = 17.01 (sec) , antiderivative size = 4567, normalized size of antiderivative = 10.60
method | result | size |
default | \(\text {Expression too large to display}\) | \(4567\) |
parts | \(\text {Expression too large to display}\) | \(4847\) |
2/3/d*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2- csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(1 /2)*(-2*B*a^4*(csc(d*x+c)-cot(d*x+c))-3*A*a^4*(csc(d*x+c)-cot(d*x+c))+A*b^ 4*(csc(d*x+c)-cot(d*x+c))+6*A*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((c sc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b)) ^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b+4*A*(-c sc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-cs c(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d *x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^2+2*A*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1) ^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a +b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b ^3+5*B*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d* x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b))^(1/2)*EllipticF(cot(d* x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b+7*B*(-csc(d*x+c)^2*(1-cos(d*x+ c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x +c))^2+a+b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1 /2))*a^2*b^2+3*B*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a *(1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/(a+b))^(1/2)*Ellipt icF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^3-4*B*(-csc(d*x+c)^2*( 1-cos(d*x+c))^2+1)^(1/2)*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c)^2...
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
integral((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/((b^3*cos(d*x + c)^ 3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sqrt(sec(d*x + c) )), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]