Integrand size = 35, antiderivative size = 270 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (3 A b^4-a^4 C-a^2 b^2 (5 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
(3*A*b^2-a^2*(2*A-C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)/d+(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)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2) )/a/b/(a^2-b^2)/d+(3*A*b^4-a^4*C-a^2*b^2*(5*A+C))*(cos(1/2*d*x+1/2*c)^2)^( 1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a ^2/(a-b)/b/(a+b)^2/d-(3*A*b^2-a^2*(2*A-C))*sin(d*x+c)/a^2/(a^2-b^2)/d/cos( d*x+c)^(1/2)+(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))/cos(d *x+c)^(1/2)
Time = 4.67 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.13 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {-\frac {-\frac {2 \left (-9 A b^3+a^2 b (10 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (8 a A b^2-4 a^3 (A-C)\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}-\frac {2 \left (-3 A b^2+a^2 (2 A-C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (\frac {\left (A b^3+a^2 b C\right ) \sin (c+d x)}{\left (-a^2+b^2\right ) (a+b \cos (c+d x))}+2 A \tan (c+d x)\right )}{4 a^2 d} \]
(-(((-2*(-9*A*b^3 + a^2*b*(10*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/ 2, 2])/(a + b) + ((8*a*A*b^2 - 4*a^3*(A - C))*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b - (2*(-3*A* b^2 + a^2*(2*A - 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)*Elli pticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Si n[c + d*x]^2]))/((-a + b)*(a + b))) + 4*Sqrt[Cos[c + d*x]]*(((A*b^3 + a^2* b*C)*Sin[c + d*x])/((-a^2 + b^2)*(a + b*Cos[c + d*x])) + 2*A*Tan[c + d*x]) )/(4*a^2*d)
Time = 1.88 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3538, 25, 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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \frac {\int -\frac {-2 \left (A-\frac {C}{2}\right ) a^2+2 b (A+C) \cos (c+d x) a+3 A b^2-\left (C a^2+A b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A-C) a^2\right )+2 b (A+C) \cos (c+d x) a+3 A b^2-\left (C a^2+A b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A-C) a^2\right )+2 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+\left (-C a^2-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \int -\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \cos ^2(c+d x)+2 a \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)+b \left (3 A b^2-a^2 (4 A+C)\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \cos ^2(c+d x)+2 a \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)+b \left (3 A b^2-a^2 (4 A+C)\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a \left (2 A b^2-a^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (3 A b^2-a^2 (4 A+C)\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {\left (3 A b^2-a^2 (4 A+C)\right ) b^2+a \left (C a^2+A b^2\right ) \cos (c+d x) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {\left (3 A b^2-a^2 (4 A+C)\right ) b^2+a \left (C a^2+A b^2\right ) \cos (c+d x) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (3 A b^2-a^2 (2 A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {\left (3 A b^2-a^2 (4 A+C)\right ) b^2+a \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {\left (3 A b^2-a^2 (4 A+C)\right ) b^2+a \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\left (a^4 (-C)-a^2 b^2 (5 A+C)+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a \left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^4 (-C)-a^2 b^2 (5 A+C)+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\left (a^4 (-C)-a^2 b^2 (5 A+C)+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a \left (a^2 C+A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}+\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {2 \left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {\frac {2 a \left (a^2 C+A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 \left (a^4 (-C)-a^2 b^2 (5 A+C)+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b* Cos[c + d*x])) - (-(((2*(3*A*b^2 - a^2*(2*A - C))*EllipticE[(c + d*x)/2, 2 ])/d + ((2*a*(A*b^2 + a^2*C)*EllipticF[(c + d*x)/2, 2])/d + (2*(3*A*b^4 - a^4*C - a^2*b^2*(5*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (2*(3*A*b^2 - a^2*(2*A - C))*Sin[c + d*x])/(a*d*Sqrt[Cos[ c + d*x]]))/(2*a*(a^2 - b^2))
3.8.18.3.1 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. \(871\) vs. \(2(344)=688\).
Time = 12.41 (sec) , antiderivative size = 872, normalized size of antiderivative = 3.23
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^2/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)/a^2/(-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))+2*( -A*b^2-C*a^2)/a/b*(-1/a*b^2/(a^2-b^2)*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)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/( a+b)*(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*si n(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))-1/2/(a^2-b^2)*b/a*(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)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/( -2*a*b+2*b^2)*b*(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))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^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...
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]