Integrand size = 43, antiderivative size = 306 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\left (3 A b^2-a b B-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (3 A b^4+3 a^3 b B-a b^3 B-a^4 C-a^2 b^2 (5 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (3 A b^2-a b B-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \]
(3*A*b^2-B*a*b-a^2*(2*A-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^2/(a^2-b^2)/d+(A*b^2-a*(B*b-C*a) )*(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+3*B*a^3*b-B*a*b^3-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-B*a*b-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-a*(B*b-C*a))*sin(d *x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))/cos(d*x+c)^(1/2)
Time = 5.13 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 \left (2 a A \left (a^2-b^2\right )+b \left (-3 A b^2+a b B+a^2 (2 A-C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (9 A b^3+4 a^3 B-3 a b^2 B-a^2 b (10 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 a \left (-2 A b^2+a b B+a^2 (A-C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}-\frac {2 \left (-3 A b^2+a b B+a^2 (2 A-C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 a^2 d} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2),x]
((4*(2*a*A*(a^2 - b^2) + b*(-3*A*b^2 + a*b*B + a^2*(2*A - C))*Cos[c + d*x] )*Sin[c + d*x])/((a^2 - b^2)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])) - (( 2*(9*A*b^3 + 4*a^3*B - 3*a*b^2*B - a^2*b*(10*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*a*(-2*A*b^2 + a*b*B + a^2*(A - C))*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) - (2*(-3*A*b^2 + a*b*B + 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)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x ]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(4 *a^2*d)
Time = 2.18 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.93, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 3534, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\int -\frac {-\left ((2 A-C) a^2\right )-b B a+2 (A b+C b-a B) \cos (c+d x) a+3 A b^2-\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A-C) a^2\right )-b B a+2 (A b+C b-a B) \cos (c+d x) a+3 A b^2-\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A-C) a^2\right )-b B a+2 (A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+\left (a (b B-a C)-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \int -\frac {2 B a^3-b (4 A+C) a^2-b^2 B a+2 \left (-\left ((A-C) a^2\right )-b B a+2 A b^2\right ) \cos (c+d x) a+3 A b^3+b \left (-\left ((2 A-C) a^2\right )-b B a+3 A b^2\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 B a^3-b (4 A+C) a^2-b^2 B a+2 \left (-\left ((A-C) a^2\right )-b B a+2 A b^2\right ) \cos (c+d x) a+3 A b^3+b \left (-\left ((2 A-C) a^2\right )-b B a+3 A b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 B a^3-b (4 A+C) a^2-b^2 B a+2 \left (-\left ((A-C) a^2\right )-b B a+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^3+b \left (-\left ((2 A-C) a^2\right )-b B a+3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {b \left (2 B a^3-b (4 A+C) a^2-b^2 B a+3 A b^3\right )+a b \left (A b^2-a (b B-a C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {b \left (2 B a^3-b (4 A+C) a^2-b^2 B a+3 A b^3\right )+a b \left (A b^2-a (b B-a C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {b \left (2 B a^3-b (4 A+C) a^2-b^2 B a+3 A b^3\right )+a b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {b \left (2 B a^3-b (4 A+C) a^2-b^2 B a+3 A b^3\right )+a b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (A b^2-a (b B-a C)\right )}{d}}{b}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \sin (c+d x) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{d}+\frac {\frac {2 \left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (A b^2-a (b B-a C)\right )}{d}}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
((A*b^2 - a*(b*B - a*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*b*B - a^2*(2*A - C))*Elliptic E[(c + d*x)/2, 2])/d + ((2*a*(A*b^2 - a*(b*B - a*C))*EllipticF[(c + d*x)/2 , 2])/d + (2*(3*A*b^4 + 3*a^3*b*B - a*b^3*B - a^4*C - a^2*b^2*(5*A + C))*E llipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (2*(3*A*b^2 - a*b*B - a^2*(2*A - C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(2*a*(a^ 2 - b^2))
3.12.7.3.1 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_)] + (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. \(875\) vs. \(2(380)=760\).
Time = 4.14 (sec) , antiderivative size = 876, normalized size of antiderivative = 2.86
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^2,x, method=_RETURNVERBOSE)
-(-(-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+B*a*b-C*a^2)/a/b*(-b^2/a/(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*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2 /(a+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)*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2))-1/2*b/a/(a^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) *EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a^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)*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(c os(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*(si n(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...
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^2,x, algorithm="fricas")
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^2,x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^2* cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]