Integrand size = 43, antiderivative size = 280 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \]
1/4*(39*A-20*B+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5 /2)/d-1/4*(A-B+C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(19* A-11*B+3*C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/32*(219*A-1 15*B+43*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a ^(5/2)/d*2^(1/2)-1/16*(63*A-35*B+11*C)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^( 1/2)+1/16*(31*A-15*B+7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/ 2)
Time = 10.33 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (\frac {\left ((-219 A+115 B-43 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+4 \sqrt {2} (39 A-20 B+8 C) \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {1}{4} (73 A-43 B+11 C+(89 A-55 B+15 C) \cos (c+d x)+2 (5 A-4 B) \cos (2 (c+d x))-2 A \cos (3 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{4 d (a (1+\sec (c+d x)))^{5/2}} \]
Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^(5/2),x]
(Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*((((-219*A + 115*B - 43*C)*ArcSin[T an[(c + d*x)/2]] + 4*Sqrt[2]*(39*A - 20*B + 8*C)*ArcTan[Tan[(c + d*x)/2]/S qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x] )]*Sqrt[1 + Sec[c + d*x]])/Sqrt[Sec[(c + d*x)/2]^2] + ((73*A - 43*B + 11*C + (89*A - 55*B + 15*C)*Cos[c + d*x] + 2*(5*A - 4*B)*Cos[2*(c + d*x)] - 2* A*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Sqrt[Sec[c + d*x]]*(Sin[(c + d*x)/2 ] - Sin[(3*(c + d*x))/2]))/4))/(4*d*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.92 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.06, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 4572, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4510, 25, 3042, 4408, 3042, 4261, 216, 4282, 216}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4572 |
\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (4 a (3 A-B+C)-a (7 A-7 B-C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (4 a (3 A-B+C)-a (7 A-7 B-C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {4 a (3 A-B+C)-a (7 A-7 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4508 |
\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (4 a^2 (31 A-15 B+7 C)-5 a^2 (19 A-11 B+3 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\cos ^2(c+d x) \left (4 a^2 (31 A-15 B+7 C)-5 a^2 (19 A-11 B+3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2 (31 A-15 B+7 C)-5 a^2 (19 A-11 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4510 |
\(\displaystyle \frac {\frac {\frac {\int -\frac {2 \cos (c+d x) \left (2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) \left (2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^3 (63 A-35 B+11 C)-3 a^3 (31 A-15 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4510 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A-20 B+8 C)-a^4 (63 A-35 B+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4408 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B+8 C) \int \sqrt {\sec (c+d x) a+a}dx-a^4 (219 A-115 B+43 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {4 a^3 (39 A-20 B+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {8 a^4 (39 A-20 B+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^4 (219 A-115 B+43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^4 (219 A-115 B+43 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (63 A-35 B+11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A-115 B+43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*((A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2) ) + (-1/2*(a*(19*A - 11*B + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[ c + d*x])^(3/2)) + ((2*a^2*(31*A - 15*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/ (d*Sqrt[a + a*Sec[c + d*x]]) - (-(((8*a^(7/2)*(39*A - 20*B + 8*C)*ArcTan[( Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a^(7/2)*(219 *A - 115*B + 43*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) + (2*a^3*(63*A - 35*B + 11*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*a^2))/(8*a^2)
3.6.33.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
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*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(2033\) vs. \(2(245)=490\).
Time = 3.04 (sec) , antiderivative size = 2034, normalized size of antiderivative = 7.26
int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, method=_RETURNVERBOSE)
1/32/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(156*A*arctanh(2 ^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2 ^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^4*csc(d*x+c) ^4+32*C*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+ c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x +c))^4*csc(d*x+c)^4+312*A*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1 )^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1 )^(3/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2+64*C*arctanh(2^(1/2)/((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*C*(1-cos(d*x+c))^ 3*csc(d*x+c)^3+85*A*(-cot(d*x+c)+csc(d*x+c))-21*B*(1-cos(d*x+c))^9*csc(d*x +c)^9-36*B*(1-cos(d*x+c))^7*csc(d*x+c)^7+74*B*(1-cos(d*x+c))^5*csc(d*x+c)^ 5+34*B*(1-cos(d*x+c))^3*csc(d*x+c)^3+156*A*((1-cos(d*x+c))^2*csc(d*x+c)^2- 1)^(3/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x +c)+csc(d*x+c)))*2^(1/2)+115*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*ln( csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-160*B*2^(1/ 2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*arctanh(2^(1/2)/((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-cos(d*x+c))^2*csc(d* x+c)^2-80*B*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*arctanh(2^(1/2 )/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-...
Time = 47.04 (sec) , antiderivative size = 835, normalized size of antiderivative = 2.98 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="fricas")
[-1/64*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^3 + 3*(219*A - 115*B + 43*C)*cos(d*x + c)^2 + 3*(219*A - 115*B + 43*C)*cos(d*x + c) + 219*A - 1 15*B + 43*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 8*((39*A - 20*B + 8*C)* cos(d*x + c)^3 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B + 8 *C)*cos(d*x + c) + 39*A - 20*B + 8*C)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2 *sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c ) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*(8*A*cos(d*x + c)^4 - 4*(5 *A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 - (63*A - 35*B + 11*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^3 + 3*(219 *A - 115*B + 43*C)*cos(d*x + c)^2 + 3*(219*A - 115*B + 43*C)*cos(d*x + c) + 219*A - 115*B + 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 8*((39*A - 20*B + 8*C) *cos(d*x + c)^3 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B + 8*C)*cos(d*x + c) + 39*A - 20*B + 8*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*(8*A*cos(d*x + c)^4 - 4*(5*A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B + 3*C)*cos(d*x +...
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a*(sec( c + d*x) + 1))**(5/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^2/(a*sec(d* x + c) + a)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 702 vs. \(2 (245) = 490\).
Time = 2.50 (sec) , antiderivative size = 702, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5 /2),x, algorithm="giac")
1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C* a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(29*A*a^5 - 21*B*a^5 + 13*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt( 2)*(219*A - 115*B + 43*C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan (1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*a^2*sgn(cos(d*x + c))) + 8*(39*A - 20*B + 8*C)*log(abs(309485009821345068724781056*(sqrt(-a)*tan(1/2*d*x + 1/ 2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 61897001964269013744956211 2*sqrt(2)*abs(a) - 928455029464035206174343168*a)/abs(30948500982134506872 4781056*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 618970019642690137449562112*sqrt(2)*abs(a) - 928455029464035206174 343168*a))/(sqrt(-a)*a*abs(a)*sgn(cos(d*x + c))) - 32*sqrt(2)*(41*(sqrt(-a )*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A - 12*(sq rt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B - 2 09*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 *A*a + 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a + 91*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 2*c)^2 + a))^2*A*a^2 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2 *d*x + 1/2*c)^2 + a))^2*B*a^2 - 11*A*a^3 + 4*B*a^3)/(((sqrt(-a)*tan(1/2*d* x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2* d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2*sqrt(-...
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]