Integrand size = 43, antiderivative size = 311 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^{5/2} (1015 A+1132 B+1304 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{512 d}+\frac {a^3 (1015 A+1132 B+1304 C) \tan (c+d x)}{512 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (1015 A+1132 B+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (545 A+628 B+680 C) \sec ^2(c+d x) \tan (c+d x)}{960 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (115 A+156 B+120 C) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{480 d}+\frac {a (5 A+12 B) (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{60 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d} \]
1/512*a^(5/2)*(1015*A+1132*B+1304*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d *x+c))^(1/2))/d+1/60*a*(5*A+12*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^4*tan( d*x+c)/d+1/6*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^5*tan(d*x+c)/d+1/512*a^3* (1015*A+1132*B+1304*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/768*a^3*(1015 *A+1132*B+1304*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/960*a^3 *(545*A+628*B+680*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/48 0*a^2*(115*A+156*B+120*C)*sec(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 4.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.78 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (120 \sqrt {2} (1015 A+1132 B+1304 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^6(c+d x)+(137060 A+112464 B+93600 C+(321370 A+303048 B+283920 C) \cos (c+d x)+16 (8555 A+8444 B+7480 C) \cos (2 (c+d x))+108605 A \cos (3 (c+d x))+121124 B \cos (3 (c+d x))+127240 C \cos (3 (c+d x))+20300 A \cos (4 (c+d x))+22640 B \cos (4 (c+d x))+26080 C \cos (4 (c+d x))+15225 A \cos (5 (c+d x))+16980 B \cos (5 (c+d x))+19560 C \cos (5 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{122880 d} \]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^7,x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^6*(120*Sqrt[ 2]*(1015*A + 1132*B + 1304*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d* x]^6 + (137060*A + 112464*B + 93600*C + (321370*A + 303048*B + 283920*C)*C os[c + d*x] + 16*(8555*A + 8444*B + 7480*C)*Cos[2*(c + d*x)] + 108605*A*Co s[3*(c + d*x)] + 121124*B*Cos[3*(c + d*x)] + 127240*C*Cos[3*(c + d*x)] + 2 0300*A*Cos[4*(c + d*x)] + 22640*B*Cos[4*(c + d*x)] + 26080*C*Cos[4*(c + d* x)] + 15225*A*Cos[5*(c + d*x)] + 16980*B*Cos[5*(c + d*x)] + 19560*C*Cos[5* (c + d*x)])*Sin[(c + d*x)/2]))/(122880*d)
Time = 1.91 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3522, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3251, 3042, 3252, 219}
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.
| \(\displaystyle \int \sec ^7(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^7}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (a (5 A+12 B)+a (5 A+12 C) \cos (c+d x)) \sec ^6(c+d x)dx}{6 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (a (5 A+12 B)+a (5 A+12 C) \cos (c+d x)) \sec ^6(c+d x)dx}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+12 B)+a (5 A+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left ((115 A+156 B+120 C) a^2+15 (5 A+4 B+8 C) \cos (c+d x) a^2\right ) \sec ^5(c+d x)dx+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{10} \int (\cos (c+d x) a+a)^{3/2} \left ((115 A+156 B+120 C) a^2+15 (5 A+4 B+8 C) \cos (c+d x) a^2\right ) \sec ^5(c+d x)dx+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((115 A+156 B+120 C) a^2+15 (5 A+4 B+8 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{4} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 (545 A+628 B+680 C) a^3+5 (235 A+252 B+312 C) \cos (c+d x) a^3\right ) \sec ^4(c+d x)dx+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \int \sqrt {\cos (c+d x) a+a} \left (3 (545 A+628 B+680 C) a^3+5 (235 A+252 B+312 C) \cos (c+d x) a^3\right ) \sec ^4(c+d x)dx+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (545 A+628 B+680 C) a^3+5 (235 A+252 B+312 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \int \sqrt {\cos (c+d x) a+a} \sec ^3(c+d x)dx+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )+\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 (5 A+12 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}+\frac {1}{10} \left (\frac {a^3 (115 A+156 B+120 C) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {a^4 (545 A+628 B+680 C) \tan (c+d x) \sec ^2(c+d x)}{d \sqrt {a \cos (c+d x)+a}}+\frac {5}{2} a^3 (1015 A+1132 B+1304 C) \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )\right )}{12 a}+\frac {A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}\) |
(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + ((a^2*( 5*A + 12*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a^3*(115*A + 156*B + 120*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Ta n[c + d*x])/(4*d) + ((a^4*(545*A + 628*B + 680*C)*Sec[c + d*x]^2*Tan[c + d *x])/(d*Sqrt[a + a*Cos[c + d*x]]) + (5*a^3*(1015*A + 1132*B + 1304*C)*((a* Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((Sqrt[a]*A rcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Tan[c + d* x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4))/2)/8)/10)/(12*a)
3.4.100.3.1 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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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^2*C - B*c*d + 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/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(3357\) vs. \(2(279)=558\).
Time = 6.14 (sec) , antiderivative size = 3358, normalized size of antiderivative = 10.80
\[\text {output too large to display}\]
1/240*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(960*a*(10 15*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^ (1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+1015*A*ln(4/(2*cos(1/2* d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1 /2*c)^2)^(1/2)*a^(1/2)+2*a))+1132*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*( 2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2 )-2*a))+1132*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+ 1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+1304*C*ln(-4/( 2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin (1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+1304*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2 ^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/ 2)*a^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^12-960*(1015*A*a^(1/2)*2^(1/2)*(a*sin (1/2*d*x+1/2*c)^2)^(1/2)+1132*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 1/2)+1304*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3045*A*ln(-4/(2 *cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin( 1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+3045*A*ln(4/(2*cos(1/2*d*x+1/2*c)+ 2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1 /2)*a^(1/2)+2*a))*a+3396*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a *cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a +3396*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*...
Time = 0.40 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.93 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {15 \, {\left ({\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{7} + {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{6}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (15 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 10 \, {\left (1015 \, A + 1132 \, B + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (1015 \, A + 1132 \, B + 920 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 48 \, {\left (145 \, A + 116 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 128 \, {\left (35 \, A + 12 \, B\right )} a^{2} \cos \left (d x + c\right ) + 1280 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{30720 \, {\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^7,x, algorithm="fricas")
1/30720*(15*((1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c)^7 + (1015*A + 113 2*B + 1304*C)*a^2*cos(d*x + c)^6)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos( d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(15*(1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c)^5 + 10*(1015*A + 1132*B + 1304*C)*a^2*cos(d*x + c)^4 + 8*(1015*A + 1132*B + 920*C)*a^2*cos(d*x + c)^3 + 48*(145*A + 116*B + 40*C)*a^2*cos(d*x + c)^2 + 128*(35*A + 12*B)*a^2*cos(d*x + c) + 1280*A*a ^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^7,x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (279) = 558\).
Time = 2.59 (sec) , antiderivative size = 606, normalized size of antiderivative = 1.95 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^7,x, algorithm="giac")
-1/30720*sqrt(2)*(15*sqrt(2)*(1015*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 1132* B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 1304*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*lo g(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(487200*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) ^11 + 543360*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^11 + 625 920*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^11 - 1380400*A*a^ 2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 - 1539520*B*a^2*sgn(cos (1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 - 1773440*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 1607760*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) *sin(1/2*d*x + 1/2*c)^7 + 1793088*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2* d*x + 1/2*c)^7 + 2040960*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2 *c)^7 - 977880*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 10 81824*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 1191360*C*a ^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 318970*A*a^2*sgn(cos (1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 340040*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 351760*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*s in(1/2*d*x + 1/2*c)^3 - 46215*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 44460*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 41 880*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 1)^6)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^7} \,d x \]