Integrand size = 35, antiderivative size = 430 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (3 a^2 b^2 (5 A-8 C)+35 a^4 C-2 b^4 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 b^2 (9 A-20 C)+21 a^4 C-4 b^4 (3 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^5 \left (a^2-b^2\right ) d}-\frac {a^2 \left (5 A b^4-3 a^2 b^2 (A-3 C)-7 a^4 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^5 (a+b)^2 d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (5 A b^2+7 a^2 C-2 b^2 C\right ) \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {a \left (3 A b^2+7 a^2 C-4 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}} \] Output:
1/5*(3*a^2*b^2*(5*A-8*C)+35*a^4*C-2*b^4*(5*A+3*C))*cos(d*x+c)^(1/2)*Ellipt icE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/b^4/(a^2-b^2)/d-1/3*a*(a^ 2*b^2*(9*A-20*C)+21*a^4*C-4*b^4*(3*A+C))*cos(d*x+c)^(1/2)*InverseJacobiAM( 1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/b^5/(a^2-b^2)/d-a^2*(5*A*b^4-3*a^2 *b^2*(A-3*C)-7*a^4*C)*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/( a+b),2^(1/2))*sec(d*x+c)^(1/2)/(a-b)/b^5/(a+b)^2/d-(A*b^2+C*a^2)*sin(d*x+c )/b/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^(5/2)+1/5*(5*A*b^2+7*C*a^2-2*C *b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/sec(d*x+c)^(3/2)-1/3*a*(3*A*b^2+7*C*a^2-4 *C*b^2)*sin(d*x+c)/b^3/(a^2-b^2)/d/sec(d*x+c)^(1/2)
Time = 7.64 (sec) , antiderivative size = 762, normalized size of antiderivative = 1.77 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {2 \left (15 a^2 A b^2-30 A b^4+35 a^4 C-32 a^2 b^2 C-18 b^4 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 (60 a A b^3+56 a^3 b C+4 a b^3 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 (45 a^2 A b^2-30 A b^4+105 a^4 C-72 a^2 b^2 C-18 b^4 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 )}}{60 (a-b) b^3 (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {\left (10 a^2 A b^2+10 a^4 C-a^2 b^2 C+b^4 C\right ) \sin (c+d x)}{10 b^4 \left (a^2-b^2\right )}-\frac {a^3 A b^2 \sin (c+d x)+a^5 C \sin (c+d x)}{b^4 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}-\frac {2 a C \sin (2 (c+d x))}{3 b^3}+\frac {C \sin (3 (c+d x))}{10 b^2}\right )}{d} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2 )),x]
Output:
((2*(15*a^2*A*b^2 - 30*A*b^4 + 35*a^4*C - 32*a^2*b^2*C - 18*b^4*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), Ar cSin[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*(60*a* A*b^3 + 56*a^3*b*C + 4*a*b^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin[S qrt[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 - Cos[c + d*x]^2)) + ((45*a^2*A*b^2 - 30*A*b^4 + 105*a^4*C - 72*a^2*b^2*C - 18*b^4*C)*Cos[2*(c + d*x)]*(b + a*S ec[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]]*Sqrt[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)))/(60*(a - b)*b^3*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(-1/10*((10*a^2*A*b^2 + 10*a^4*C - a^2*b^2*C + b^4*C)*Sin[c + d*x ])/(b^4*(a^2 - b^2)) - (a^3*A*b^2*Sin[c + d*x] + a^5*C*Sin[c + d*x])/(b^4* (-a^2 + b^2)*(a + b*Cos[c + d*x])) - (2*a*C*Sin[2*(c + d*x)])/(3*b^3) + (C *Sin[3*(c + d*x)])/(10*b^2)))/d
Time = 2.85 (sec) , antiderivative size = 377, 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, 3527, 27, 3042, 3528, 27, 3042, 3528, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \cos (c+d x)^2}{\sec (c+d x)^{5/2} (a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (C \cos ^2(c+d x)+A\right )}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\left (\left (7 C a^2+5 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (A+C) \cos (c+d x)+5 \left (C a^2+A b^2\right )\right )}{2 (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\left (\left (7 C a^2+5 A b^2-2 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b (A+C) \cos (c+d x)+5 \left (C a^2+A b^2\right )\right )}{a+b \cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\left (-7 C a^2-5 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (C a^2+A b^2\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {2 \int -\frac {\sqrt {\cos (c+d x)} \left (-5 a \left (7 C a^2+3 A b^2-4 b^2 C\right ) \cos ^2(c+d x)-2 b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \cos (c+d x)+3 a \left (7 C a^2+5 A b^2-2 b^2 C\right )\right )}{2 (a+b \cos (c+d x))}dx}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-5 a \left (7 C a^2+3 A b^2-4 b^2 C\right ) \cos ^2(c+d x)-2 b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \cos (c+d x)+3 a \left (7 C a^2+5 A b^2-2 b^2 C\right )\right )}{a+b \cos (c+d x)}dx}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (-5 a \left (7 C a^2+3 A b^2-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (7 C a^2+5 A b^2-2 b^2 C\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {2 \int -\frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \cos (c+d x) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {5 \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2-2 b \left (15 A b^2+\left (14 a^2+b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 \left (35 C a^4+3 b^2 (5 A-8 C) a^2-2 b^4 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {5 \left (b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \cos (c+d x) a\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \cos (c+d x) a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \int \frac {b \left (7 C a^2+3 A b^2-4 b^2 C\right ) a^2+\left (21 C a^4+b^2 (9 A-20 C) a^2-4 b^4 (3 A+C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {5 \left (\frac {3 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \left (7 a^2 C+5 A b^2-2 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d}-\frac {-\frac {10 a \left (7 a^2 C+3 A b^2-4 b^2 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}-\frac {\frac {5 \left (\frac {2 a \left (21 a^4 C+a^2 b^2 (9 A-20 C)-4 b^4 (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {6 a^2 \left (-7 a^4 C-3 a^2 b^2 (A-3 C)+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}-\frac {6 \left (35 a^4 C+3 a^2 b^2 (5 A-8 C)-2 b^4 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}}{5 b}}{2 b \left (a^2-b^2\right )}\right )\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A*b^2 + a^2*C)*Cos[c + d*x]^(5/ 2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))) - ((-2*(5*A*b^2 + 7*a^2*C - 2*b^2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d) - (-1/3*((-6* (3*a^2*b^2*(5*A - 8*C) + 35*a^4*C - 2*b^4*(5*A + 3*C))*EllipticE[(c + d*x) /2, 2])/(b*d) + (5*((2*a*(a^2*b^2*(9*A - 20*C) + 21*a^4*C - 4*b^4*(3*A + C ))*EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a^2*(5*A*b^4 - 3*a^2*b^2*(A - 3*C ) - 7*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d)))/b) /b - (10*a*(3*A*b^2 + 7*a^2*C - 4*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/ (3*b*d))/(5*b))/(2*b*(a^2 - b^2)))
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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. \(1336\) vs. \(2(409)=818\).
Time = 7.73 (sec) , antiderivative size = 1337, normalized size of antiderivative = 3.11
Input:
int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x,method=_RETUR NVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*a^2/b^4*(3* A*b^2+5*C*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)*El lipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2*(2*A*a*b^2+A*b^3+4*C*a^3 +3*C*a^2*b+2*C*a*b^2+C*b^3)/b^5*(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))-4/5*C/b^2/(-2*sin(1/2*d*x+1/2*c)^4+s in(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-14*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+ 1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic E(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))+2/b^4*(A*b ^2+3*C*a^2+4*C*a*b+3*C*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)*(Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))- 4/3*C/b^3*(2*a+3*b)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x +1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(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)-2*a^3*...
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori thm="fricas")
Output:
integral((C*cos(d*x + c)^2 + A)/((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sec(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2/sec(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2 )), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(5/2 )), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^2) ,x)
Output:
int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^2) , x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \right ) c \] Input:
int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2/sec(d*x+c)^(5/2),x)
Output:
int(sqrt(sec(c + d*x))/(cos(c + d*x)**2*sec(c + d*x)**3*b**2 + 2*cos(c + d *x)*sec(c + d*x)**3*a*b + sec(c + d*x)**3*a**2),x)*a + int((sqrt(sec(c + d *x))*cos(c + d*x)**2)/(cos(c + d*x)**2*sec(c + d*x)**3*b**2 + 2*cos(c + d* x)*sec(c + d*x)**3*a*b + sec(c + d*x)**3*a**2),x)*c