Integrand size = 35, antiderivative size = 206 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {2 \left (5 A b^2+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 A b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {2 b \left (A b^2+a^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a+b) d}+\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt {\cos (c+d x)}} \]
-2/5*(5*A*b^2+a^2*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2* c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d-2/3*A*b*(cos(1/2*d*x+1/2*c) ^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-2 *b*(A*b^2+C*a^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticP i(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^3/(a+b)/d+2/5*A*sin(d*x+c)/a/d/c os(d*x+c)^(5/2)-2/3*A*b*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)+2/5*(5*A*b^2+a^2 *(3*A+5*C))*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)
Time = 4.75 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.43 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {\frac {2 \left (45 A b^3+a^2 b (19 A+45 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (40 a A b^2+6 a^3 (3 A+5 C)\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {6 \left (5 A b^2+a^2 (3 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}-\frac {2 \left (-10 a A b \sin (c+d x)+3 \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (2 (c+d x))+6 a^2 A \tan (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}}{30 a^3 d} \]
-1/30*((2*(45*A*b^3 + a^2*b*(19*A + 45*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ((40*a*A*b^2 + 6*a^3*(3*A + 5*C))*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (6*(5*A*b^2 + a^2*(3*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x] ]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/( a*b*Sqrt[Sin[c + d*x]^2]) - (2*(-10*a*A*b*Sin[c + d*x] + 3*(5*A*b^2 + a^2* (3*A + 5*C))*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x]))/Cos[c + d*x]^(3/2)) /(a^3*d)
Time = 1.98 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \frac {2 \int -\frac {-3 A b \cos ^2(c+d x)-a (3 A+5 C) \cos (c+d x)+5 A b}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{5 a}+\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-3 A b \cos ^2(c+d x)-a (3 A+5 C) \cos (c+d x)+5 A b}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-3 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 a}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {-5 A b^2 \cos ^2(c+d x)+4 a A b \cos (c+d x)+3 \left ((3 A+5 C) a^2+5 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}+\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 A b^2 \cos ^2(c+d x)+4 a A b \cos (c+d x)+3 \left ((3 A+5 C) a^2+5 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 a A b \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2+5 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \cos (c+d x)+5 b \left ((A+3 C) a^2+3 A b^2\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \cos (c+d x)+5 b \left ((A+3 C) a^2+3 A b^2\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (3 (3 A+5 C) a^2+20 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b \left ((A+3 C) a^2+3 A b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (a A \cos (c+d x) b^3+\left ((A+3 C) a^2+3 A b^2\right ) b^2\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx+\frac {5 \int \frac {a A \cos (c+d x) b^3+\left ((A+3 C) a^2+3 A b^2\right ) b^2}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 \int \frac {a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left ((A+3 C) a^2+3 A b^2\right ) b^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \int \frac {a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left ((A+3 C) a^2+3 A b^2\right ) b^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+a A b^2 \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a A b^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a A b^2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}+\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 A b \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {5 \left (\frac {6 b^2 \left (a^2 C+A b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a A b^2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}}{a}}{3 a}}{5 a}\) |
(2*A*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - ((10*A*b*Sin[c + d*x])/(3* a*d*Cos[c + d*x]^(3/2)) - (-(((6*(5*A*b^2 + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (5*((2*a*A*b^2*EllipticF[(c + d*x)/2, 2])/d + (6*b^2*(A* b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d)))/b)/a ) + (6*(5*A*b^2 + a^2*(3*A + 5*C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]])) /(3*a))/(5*a)
3.8.11.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(758\) vs. \(2(268)=536\).
Time = 14.99 (sec) , antiderivative size = 759, normalized size of antiderivative = 3.68
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*A/a/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d *x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 *c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c) ^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2*A/a^2*b*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1 /2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1 /3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c), 2^(1/2)))+2*(A*b^2+C*a^2)/a^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2 -1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/ 2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*(A*b^2+C*a^2)*b^2/a ^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1) ^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos (1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+...
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]