Integrand size = 35, antiderivative size = 501 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\left (15 A b^2-a^2 (7 A-8 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt {a+b} d}-\frac {\left (5 a A b+15 A b^2-2 a^2 (A-4 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt {a+b} d}-\frac {\sqrt {a+b} \left (15 A b^2+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^4 d}-\frac {5 A b \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]
1/4*(15*A*b^2-a^2*(7*A-8*C))*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/( a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec( d*x+c))/(a-b))^(1/2)/a^3/d/(a+b)^(1/2)-1/4*(5*a*A*b+15*A*b^2-2*a^2*(A-4*C) )*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1 /2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/d/ (a+b)^(1/2)-1/4*(15*A*b^2+4*a^2*(A+2*C))*cot(d*x+c)*EllipticPi((a+b*sec(d* x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec (d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d-5/4*A*b*sin(d* x+c)/a^2/d/(a+b*sec(d*x+c))^(1/2)+1/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec (d*x+c))^(1/2)+1/4*b^2*(15*A*b^2-a^2*(7*A-8*C))*tan(d*x+c)/a^3/(a^2-b^2)/d /(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1870\) vs. \(2(501)=1002\).
Time = 18.56 (sec) , antiderivative size = 1870, normalized size of antiderivative = 3.73 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \]
(((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((4*b*(A*b^2 + a^2*C)*Sin[c + d*x] )/(a^3*(-a^2 + b^2)) + (4*(A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/( a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(2*a^2)))/(d* (a + b*Sec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/ 2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-7*a^3*A*b*Tan[(c + d*x)/2] - 7*a^2*A*b^2*Tan[(c + d*x)/2] + 15*a*A*b^3*Tan[(c + d*x)/2] + 15*A*b^4*Tan[(c + d*x)/2] + 8*a^3*b*C*Tan[ (c + d*x)/2] + 8*a^2*b^2*C*Tan[(c + d*x)/2] + 14*a^3*A*b*Tan[(c + d*x)/2]^ 3 - 30*a*A*b^3*Tan[(c + d*x)/2]^3 - 16*a^3*b*C*Tan[(c + d*x)/2]^3 - 7*a^3* A*b*Tan[(c + d*x)/2]^5 + 7*a^2*A*b^2*Tan[(c + d*x)/2]^5 + 15*a*A*b^3*Tan[( c + d*x)/2]^5 - 15*A*b^4*Tan[(c + d*x)/2]^5 + 8*a^3*b*C*Tan[(c + d*x)/2]^5 - 8*a^2*b^2*C*Tan[(c + d*x)/2]^5 + 8*a^4*A*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*T an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 22*a^2*A*b^2*Elliptic Pi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2 ]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 30*A*b^4*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x )/2]^2)/(a + b)] + 16*a^4*C*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/...
Time = 2.35 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4593, 27, 3042, 4592, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4593 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (A+2 C) \sec (c+d x)+5 A b\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{2 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (A+2 C) \sec (c+d x)+5 A b\right )}{(a+b \sec (c+d x))^{3/2}}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{4 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2+6 A b \sec (c+d x) a+15 A b^2-5 A b^2 \sec ^2(c+d x)}{2 (a+b \sec (c+d x))^{3/2}}dx}{a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2+6 A b \sec (c+d x) a+15 A b^2-5 A b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2+6 A b \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^2-5 A b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {-b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \sec ^2(c+d x)-2 a b \left (5 A b^2-a^2 (A-4 C)\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2+15 A b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {-b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \sec ^2(c+d x)-2 a b \left (5 A b^2-a^2 (A-4 C)\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2+15 A b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {-b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b \left (5 A b^2-a^2 (A-4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2+15 A b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\left (a^2-b^2\right ) \left (4 (A+2 C) a^2+15 A b^2\right )+\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right )-2 a b \left (5 A b^2-a^2 (A-4 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\left (a^2-b^2\right ) \left (4 (A+2 C) a^2+15 A b^2\right )+\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right )-2 a b \left (5 A b^2-a^2 (A-4 C)\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b (a-b) \left (-2 a^2 (A-4 C)+5 a A b+15 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (a^2-b^2\right ) \left (4 a^2 (A+2 C)+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b (a-b) \left (-2 a^2 (A-4 C)+5 a A b+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^2-b^2\right ) \left (4 a^2 (A+2 C)+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-b (a-b) \left (-2 a^2 (A-4 C)+5 a A b+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 (A+2 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\left (b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 (A-4 C)+5 a A b+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 (A+2 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {5 A b \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 (A-4 C)+5 a A b+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (15 A b^2-a^2 (7 A-8 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 (A+2 C)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (15 A b^2-a^2 (7 A-8 C)\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
(A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + b*Sec[c + d*x]]) - ((5*A*b*S in[c + d*x])/(a*d*Sqrt[a + b*Sec[c + d*x]]) - (((2*(a - b)*Sqrt[a + b]*(15 *A*b^2 - a^2*(7*A - 8*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)] *Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*(a - b)*Sqrt[a + b]*(5*a* A*b + 15*A*b^2 - 2*a^2*(A - 4*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b *Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/ (a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*Sqrt[a + b]*(a^2 - b^2)*(15*A*b^2 + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[(a + b)/a, Arc Sin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S ec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a* (a^2 - b^2)) + (2*b^2*(15*A*b^2 - a^2*(7*A - 8*C))*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(2*a))/(4*a)
3.8.47.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^( m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x ] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[( -A)*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && NeQ[a^2 - b^2 , 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4798\) vs. \(2(456)=912\).
Time = 5.82 (sec) , antiderivative size = 4799, normalized size of antiderivative = 9.58
1/4/d/a^3/(a+b)/(a-b)*(2*A*a^4*cos(d*x+c)^3*sin(d*x+c)+2*A*a^4*cos(d*x+c)^ 2*sin(d*x+c)+15*A*b^4*cos(d*x+c)*sin(d*x+c)+16*C*EllipticPi(cot(d*x+c)-csc (d*x+c),-1,((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ (1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)^2+7*A*(1/(a+b)* (b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E llipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-15*A*(1/(a+b)* (b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E llipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-8*C*(1/(a+b)*(b+ a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli pticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b-8*C*(1/(a+b)*(b+a*c os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti cE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-44*A*EllipticPi(cot( d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d *x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)+14*A* EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x +c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x +c)-4*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d *x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b *cos(d*x+c)+8*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c )/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(...
\[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + A*cos(d*x + c)^2)*sqrt(b*sec(d *x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]