Integrand size = 45, antiderivative size = 301 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^{5/2} (400 A+326 B+283 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{128 d}+\frac {a^3 (1040 A+950 B+787 C) \sin (c+d x)}{960 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (400 A+326 B+283 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {a (2 B+C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/8*a*(2*B+C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1/5*C*( a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1/128*a^(5/2)*(400*A+3 26*B+283*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^ (1/2)*sec(d*x+c)^(1/2)/d+1/960*a^3*(1040*A+950*B+787*C)*sin(d*x+c)/d/cos(d *x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/128*a^3*(400*A+326*B+283*C)*sin(d*x+c )/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/240*a^2*(80*A+110*B+79*C)*si n(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)
Leaf count is larger than twice the leaf count of optimal. \(608\) vs. \(2(301)=602\).
Time = 12.80 (sec) , antiderivative size = 608, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 \sec ^5\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )}} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )} \left (\frac {25 A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 \sqrt {2}}+\frac {163 B \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{512 \sqrt {2}}+\frac {283 C \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{1024 \sqrt {2}}+\frac {C \sin \left (\frac {1}{2} (c+d x)\right )}{40 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^5}+\frac {B \sin \left (\frac {1}{2} (c+d x)\right )}{32 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {29 C \sin \left (\frac {1}{2} (c+d x)\right )}{320 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{24 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {23 B \sin \left (\frac {1}{2} (c+d x)\right )}{192 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {283 C \sin \left (\frac {1}{2} (c+d x)\right )}{1920 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {17 A \sin \left (\frac {1}{2} (c+d x)\right )}{96 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {163 B \sin \left (\frac {1}{2} (c+d x)\right )}{768 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {283 C \sin \left (\frac {1}{2} (c+d x)\right )}{1536 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {25 A \sin \left (\frac {1}{2} (c+d x)\right )}{64 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {163 B \sin \left (\frac {1}{2} (c+d x)\right )}{512 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {283 C \sin \left (\frac {1}{2} (c+d x)\right )}{1024 \left (1-2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {9}{2}}(c+d x)} \]
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x] ^2))/Cos[c + d*x]^(3/2),x]
(4*Sec[(c + d*x)/2]^5*(a*(1 + Sec[c + d*x]))^(5/2)*(A + B*Sec[c + d*x] + C *Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]*((25*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]])/(64*Sqrt[2]) + (163* B*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]])/(512*Sqrt[2]) + (283*C*ArcTanh[Sqrt[2 ]*Sin[(c + d*x)/2]])/(1024*Sqrt[2]) + (C*Sin[(c + d*x)/2])/(40*(1 - 2*Sin[ (c + d*x)/2]^2)^5) + (B*Sin[(c + d*x)/2])/(32*(1 - 2*Sin[(c + d*x)/2]^2)^4 ) + (29*C*Sin[(c + d*x)/2])/(320*(1 - 2*Sin[(c + d*x)/2]^2)^4) + (A*Sin[(c + d*x)/2])/(24*(1 - 2*Sin[(c + d*x)/2]^2)^3) + (23*B*Sin[(c + d*x)/2])/(1 92*(1 - 2*Sin[(c + d*x)/2]^2)^3) + (283*C*Sin[(c + d*x)/2])/(1920*(1 - 2*S in[(c + d*x)/2]^2)^3) + (17*A*Sin[(c + d*x)/2])/(96*(1 - 2*Sin[(c + d*x)/2 ]^2)^2) + (163*B*Sin[(c + d*x)/2])/(768*(1 - 2*Sin[(c + d*x)/2]^2)^2) + (2 83*C*Sin[(c + d*x)/2])/(1536*(1 - 2*Sin[(c + d*x)/2]^2)^2) + (25*A*Sin[(c + d*x)/2])/(64*(1 - 2*Sin[(c + d*x)/2]^2)) + (163*B*Sin[(c + d*x)/2])/(512 *(1 - 2*Sin[(c + d*x)/2]^2)) + (283*C*Sin[(c + d*x)/2])/(1024*(1 - 2*Sin[( c + d*x)/2]^2))))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec [c + d*x]^(9/2))
Time = 1.96 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\cos (c+d x)^{3/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} (a (10 A+3 C)+5 a (2 B+C) \sec (c+d x))dx}{5 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2} (a (10 A+3 C)+5 a (2 B+C) \sec (c+d x))dx}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (10 A+3 C)+5 a (2 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{4} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((80 A+30 B+39 C) a^2+(80 A+110 B+79 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (240 A+170 B+157 C) a^3+(1040 A+950 B+787 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {5 a^2 (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}+\frac {1}{8} \left (\frac {a^3 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^4 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {15}{4} a^3 (400 A+326 B+283 C) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )}{10 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d}\right )\) |
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/C os[c + d*x]^(3/2),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + ((5*a^2*(2*B + C)*Sec[c + d*x]^(5/2)*( a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^3*(80*A + 110*B + 79*C )*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^4* (1040*A + 950*B + 787*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a* Sec[c + d*x]]) + (15*a^3*(400*A + 326*B + 283*C)*((Sqrt[a]*ArcSinh[(Sqrt[a ]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4)/6)/8)/(10*a))
3.13.70.3.1 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[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[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[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*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] && GtQ[m, 1/2] && !LtQ[n, -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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(731\) vs. \(2(259)=518\).
Time = 1.13 (sec) , antiderivative size = 732, normalized size of antiderivative = 2.43
| method | result | size |
| default | \(\frac {a^{2} \left (12000 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-6000 A \cos \left (d x +c \right )^{5} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-6000 A \cos \left (d x +c \right )^{5} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+9780 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-4890 B \cos \left (d x +c \right )^{5} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4890 B \cos \left (d x +c \right )^{5} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+8490 C \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-4245 C \cos \left (d x +c \right )^{5} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4245 C \cos \left (d x +c \right )^{5} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+5440 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+6520 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+5660 C \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+1280 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+3680 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+4528 C \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+960 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+2784 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+768 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{3840 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(732\) |
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
1/3840*a^2/d*(12000*A*cos(d*x+c)^4*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-60 00*A*cos(d*x+c)^5*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/ (1+cos(d*x+c)))^(1/2))-6000*A*cos(d*x+c)^5*arctan(1/2*(-cos(d*x+c)+sin(d*x +c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+9780*B*sin(d*x+c)*cos(d*x +c)^4*(-1/(1+cos(d*x+c)))^(1/2)-4890*B*cos(d*x+c)^5*arctan(1/2*(cos(d*x+c) +sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-4890*B*cos(d*x+c) ^5*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)) )^(1/2))+8490*C*cos(d*x+c)^4*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-4245*C*c os(d*x+c)^5*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos (d*x+c)))^(1/2))-4245*C*cos(d*x+c)^5*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1) /(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+5440*A*cos(d*x+c)^3*sin(d*x+c)* (-1/(1+cos(d*x+c)))^(1/2)+6520*B*sin(d*x+c)*cos(d*x+c)^3*(-1/(1+cos(d*x+c) ))^(1/2)+5660*C*cos(d*x+c)^3*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+1280*A*s in(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)+3680*B*cos(d*x+c)^2*sin(d *x+c)*(-1/(1+cos(d*x+c)))^(1/2)+4528*C*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos( d*x+c)))^(1/2)+960*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+2784* C*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+768*C*sin(d*x+c)*(-1/(1+ cos(d*x+c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x +c)))^(1/2)/cos(d*x+c)^(9/2)
Time = 0.54 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.92 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {4 \, {\left (15 \, {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (272 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 230 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 29 \, C\right )} a^{2} \cos \left (d x + c\right ) + 384 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 15 \, {\left ({\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}, \frac {2 \, {\left (15 \, {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (272 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 230 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 29 \, C\right )} a^{2} \cos \left (d x + c\right ) + 384 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 15 \, {\left ({\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{3840 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}\right ] \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="fricas")
[1/7680*(4*(15*(400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4 + 10*(272*A + 32 6*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^2*cos(d*x + c )^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((400*A + 326*B + 283*C)*a^2*cos(d*x + c)^6 + (400*A + 326*B + 283*C)*a^2*cos(d*x + c)^5)*sq rt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c )^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^6 + d*cos(d *x + c)^5), 1/3840*(2*(15*(400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4 + 10* (272*A + 326*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^2* cos(d*x + c)^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*C*a^2)*sqrt((a*co s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((400*A + 326*B + 283*C)*a^2*cos(d*x + c)^6 + (400*A + 326*B + 283*C)*a^2*cos(d*x + c)^5)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) )*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a )))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 12710 vs. \(2 (259) = 518\).
Time = 1.49 (sec) , antiderivative size = 12710, normalized size of antiderivative = 42.23 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="maxima")
1/7680*(80*(300*sqrt(2)*a^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c)))*sin(6*d*x + 6*c) - 28*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 28 *sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 28*(sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - sqrt(2)*a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) - 300*(sqrt(2)*a^2*sin (6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/ 2*d*x + 3/2*c))) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos (3/2*d*x + 3/2*c))))*cos(11/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*(7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2* d*x + 3/2*c) - 114*sqrt(2)*a^2*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 /2*d*x + 3/2*c))) + 114*sqrt(2)*a^2*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2* c), cos(3/2*d*x + 3/2*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c))) - 456*(sqrt(2)*a^2*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(4/3 *arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin (3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 456*(sqrt(2)*a^2*sin(6*d*x + 6 *c) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/ 2*c))))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12* (7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) )))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*(...
\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="giac")
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]
int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^(3/2),x)