Integrand size = 35, antiderivative size = 477 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (15 A b^4+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (5 A b^4-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^2 b \left (a^2-b^2\right )^2 d}-\frac {\left (15 A b^6+3 a^6 C-a^2 b^4 (38 A+C)+5 a^4 b^2 (7 A+2 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac {\left (15 A b^4+a^4 (8 A-5 C)-a^2 b^2 (29 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (5 A b^4-3 a^4 C-a^2 b^2 (11 A+3 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
1/4*(15*A*b^4+a^4*(8*A-5*C)-a^2*b^2*(29*A+C))*sin(d*x+c)*sec(d*x+c)^(1/2)/ a^3/(a^2-b^2)^2/d+1/2*(A*b^2+C*a^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2 )/d/(a+b*cos(d*x+c))^2-1/4*(5*A*b^4-3*a^4*C-a^2*b^2*(11*A+3*C))*sin(d*x+c) *sec(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))-1/4*(15*A*b^4+a^4*(8* A-5*C)-a^2*b^2*(29*A+C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E llipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3 /(a^2-b^2)^2/d-1/4*(5*A*b^4-3*a^4*C-a^2*b^2*(11*A+3*C))*(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))*cos(d* x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/b/(a^2-b^2)^2/d-1/4*(15*A*b^6+3*a^6*C-a^2* b^4*(38*A+C)+5*a^4*b^2*(7*A+2*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))*cos(d*x+c)^(1/2)* sec(d*x+c)^(1/2)/a^3/(a-b)^2/b/(a+b)^3/d
Time = 7.16 (sec) , antiderivative size = 840, normalized size of antiderivative = 1.76 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {2 \left (56 a^4 A b-95 a^2 A b^3+45 A b^5+9 a^4 b C-3 a^2 b^3 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (16 a^5 A-80 a^3 A b^2+40 a A b^4-16 a^5 C-8 a^3 b^2 C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (8 a^4 A b-29 a^2 A b^3+15 A b^5-5 a^4 b C-a^2 b^3 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 a^3 (a-b)^2 (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4-5 a^4 C-a^2 b^2 C\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2}+\frac {A b^2 \sin (c+d x)+a^2 C \sin (c+d x)}{2 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {11 a^2 A b^2 \sin (c+d x)-5 A b^4 \sin (c+d x)+3 a^4 C \sin (c+d x)+3 a^2 b^2 C \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d} \]
-1/16*((2*(56*a^4*A*b - 95*a^2*A*b^3 + 45*A*b^5 + 9*a^4*b*C - 3*a^2*b^3*C) *Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-( a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ( 2*(16*a^5*A - 80*a^3*A*b^2 + 40*a*A*b^4 - 16*a^5*C - 8*a^3*b^2*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d *x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - C os[c + d*x]^2)) + ((8*a^4*A*b - 29*a^2*A*b^3 + 15*A*b^5 - 5*a^4*b*C - a^2* b^3*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^ 2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqr t[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]] , -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b ), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x ]^2] + 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c + d*x ])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(a^3*(a - b)^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x]]*(((8*a^4*A - 29*a^2*A*b^2 + 15*A *b^4 - 5*a^4*C - a^2*b^2*C)*Sin[c + d*x])/(4*a^3*(a^2 - b^2)^2) + (A*b^2*S in[c + d*x] + a^2*C*Sin[c + d*x])/(2*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (11*a^2*A*b^2*Sin[c + d*x] - 5*A*b^4*Sin[c + d*x] + 3*a^4*C*Sin[c + ...
Time = 3.14 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.88, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4709, 3042, 3535, 27, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{3/2} \left (A+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int -\frac {-\left ((4 A-C) a^2\right )+4 b (A+C) \cos (c+d x) a+5 A b^2-3 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\int \frac {-\left ((4 A-C) a^2\right )+4 b (A+C) \cos (c+d x) a+5 A b^2-3 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}dx}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\int \frac {-\left ((4 A-C) a^2\right )+4 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+5 A b^2-3 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\int -\frac {(8 A-5 C) a^4-b^2 (29 A+C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x) a+15 A b^4-\left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {(8 A-5 C) a^4-b^2 (29 A+C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x) a+15 A b^4-\left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {(8 A-5 C) a^4-b^2 (29 A+C) a^2+4 b \left (A b^2-a^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^4+\left (3 C a^4+b^2 (11 A+3 C) a^2-5 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \int -\frac {b \left ((8 A-5 C) a^4-b^2 (29 A+C) a^2+15 A b^4\right ) \cos ^2(c+d x)+4 a \left (2 (A-C) a^4-b^2 (10 A+C) a^2+5 A b^4\right ) \cos (c+d x)+b \left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \left ((8 A-5 C) a^4-b^2 (29 A+C) a^2+15 A b^4\right ) \cos ^2(c+d x)+4 a \left (2 (A-C) a^4-b^2 (10 A+C) a^2+5 A b^4\right ) \cos (c+d x)+b \left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {b \left ((8 A-5 C) a^4-b^2 (29 A+C) a^2+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 a \left (2 (A-C) a^4-b^2 (10 A+C) a^2+5 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {\left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right ) b^2+a \left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \cos (c+d x) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {\left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right ) b^2+a \left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \cos (c+d x) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {\left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right ) b^2+a \left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {\left ((24 A+7 C) a^4-b^2 (33 A+C) a^2+15 A b^4\right ) b^2+a \left (-3 C a^4-b^2 (11 A+3 C) a^2+5 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a \left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\left (3 a^6 C+5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+15 A b^6\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a \left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^6 C+5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+15 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\left (3 a^6 C+5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+15 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a \left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}+\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\frac {\left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {2 \left (a^4 (8 A-5 C)-a^2 b^2 (29 A+C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {\frac {2 a \left (-3 a^4 C-a^2 b^2 (11 A+3 C)+5 A b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 \left (3 a^6 C+5 a^4 b^2 (7 A+2 C)-a^2 b^4 (38 A+C)+15 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a *(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2) - (((5*A*b^4 - 3 *a^4*C - a^2*b^2*(11*A + 3*C))*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])) - (-(((2*(15*A*b^4 + a^4*(8*A - 5*C) - a^2*b^ 2*(29*A + C))*EllipticE[(c + d*x)/2, 2])/d + ((2*a*(5*A*b^4 - 3*a^4*C - a^ 2*b^2*(11*A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (2*(15*A*b^6 + 3*a^6*C - a^2*b^4*(38*A + C) + 5*a^4*b^2*(7*A + 2*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (2*(15*A*b^4 + a^4*(8*A - 5*C) - a^2*b ^2*(29*A + C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]))/(2*a*(a^2 - b^2)))/ (4*a*(a^2 - b^2)))
3.15.5.3.1 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]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1995\) vs. \(2(525)=1050\).
Time = 8.33 (sec) , antiderivative size = 1996, normalized size of antiderivative = 4.18
-(-(-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^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/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))+2*(-A*b^2-C* a^2)/a/b*(-1/2*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)^2-3/4*b^2*(3*a ^2-b^2)/a^2/(a^2-b^2)^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)-7/8/(a+b)/(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))+1/4/(a+b)/(a^2-b^2)/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)*Ellip ticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^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+s in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b /(a^2-b^2)^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/...
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]