Integrand size = 37, antiderivative size = 494 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (16 A b^4-2 a^2 b^2 (4 A-5 C)-a^4 (3 A+5 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{5 a^5 \sqrt {a+b} d}-\frac {2 \left (12 a A b^2+16 A b^3+2 a^2 b (2 A+5 C)+a^3 (3 A+5 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{5 a^4 \sqrt {a+b} d}+\frac {2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)} \]
2*(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c)) ^(1/2)-2/5*(6*A*b^2-a^2*(A-5*C))*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^2/(a^ 2-b^2)/d/cos(d*x+c)^(5/2)+2/5*b*(8*A*b^2-a^2*(3*A-5*C))*sin(d*x+c)*(a+b*co s(d*x+c))^(1/2)/a^3/(a^2-b^2)/d/cos(d*x+c)^(3/2)-2/5*(16*A*b^4-2*a^2*b^2*( 4*A-5*C)-a^4*(3*A+5*C))*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^ (1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2 )*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^5/d/(a+b)^(1/2)-2/5*(12*a*A*b^2+16*A*b^ 3+2*a^2*b*(2*A+5*C)+a^3*(3*A+5*C))*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^( 1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/ (a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d/(a+b)^(1/2)
Result contains complex when optimal does not.
Time = 7.36 (sec) , antiderivative size = 1418, normalized size of antiderivative = 2.87 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \]
((-4*a*(4*a^4*A*b + 12*a^2*A*b^3 - 16*A*b^5 + 10*a^4*b*C - 10*a^2*b^3*C)*S qrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Cs c[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Cs c[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2) /a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d *x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(3*a^5*A + 8*a^3*A*b^2 - 16*a*A*b^4 + 5*a^5*C - 10*a^3*b^2*C)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqr t[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x ])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2 ]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b )*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x) /2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]* EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a ]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqr t[a + b*Cos[c + d*x]])) + 2*(3*a^4*A*b + 8*a^2*A*b^3 - 16*A*b^5 + 5*a^4*b* C - 10*a^2*b^3*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[ I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d *x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*S ec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a...
Time = 2.32 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {2 \int -\frac {-\left ((A-5 C) a^2\right )+b (A+C) \cos (c+d x) a+6 A b^2-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-\left ((A-5 C) a^2\right )+b (A+C) \cos (c+d x) a+6 A b^2-4 \left (C a^2+A b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {-\left ((A-5 C) a^2\right )+b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+6 A b^2-4 \left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \int -\frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \cos ^2(c+d x)+a \left ((3 A+5 C) a^2+2 A b^2\right ) \cos (c+d x)+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}+\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \cos ^2(c+d x)+a \left ((3 A+5 C) a^2+2 A b^2\right ) \cos (c+d x)+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 b \left (6 A b^2-a^2 (A-5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((3 A+5 C) a^2+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 b \left (8 A b^2-a^2 (3 A-5 C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 \left (-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \cos (c+d x) a+16 A b^4\right )}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \cos (c+d x) a+16 A b^4}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-\left ((3 A+5 C) a^4\right )-2 b^2 (4 A-5 C) a^2+b \left ((A+5 C) a^2+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+16 A b^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (6 A b^2-a^2 (A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 b \left (8 A b^2-a^2 (3 A-5 C)\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (-\left (a^4 (3 A+5 C)\right )-2 a^2 b^2 (4 A-5 C)+16 A b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (a^3 (3 A+5 C)+2 a^2 b (2 A+5 C)+12 a A b^2+16 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a}}{5 a}}{a \left (a^2-b^2\right )}\) |
(2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*Sqrt[ a + b*Cos[c + d*x]]) - ((2*(6*A*b^2 - a^2*(A - 5*C))*Sqrt[a + b*Cos[c + d* x]]*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (-(((2*(a - b)*Sqrt[a + b]* (16*A*b^4 - 2*a^2*b^2*(4*A - 5*C) - a^4*(3*A + 5*C))*Cot[c + d*x]*Elliptic E[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d *x]))/(a - b)])/(a^2*d) + (2*(a - b)*Sqrt[a + b]*(12*a*A*b^2 + 16*A*b^3 + 2*a^2*b*(2*A + 5*C) + a^3*(3*A + 5*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[ a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))] *Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)] )/(a*d))/a) + (2*b*(8*A*b^2 - a^2*(3*A - 5*C))*Sqrt[a + b*Cos[c + d*x]]*Si n[c + d*x])/(a*d*Cos[c + d*x]^(3/2)))/(5*a))/(a*(a^2 - b^2))
3.8.60.3.1 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[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(4603\) vs. \(2(458)=916\).
Time = 33.16 (sec) , antiderivative size = 4604, normalized size of antiderivative = 9.32
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4604\) |
| default | \(\text {Expression too large to display}\) | \(5476\) |
-2/5*A/d*(a^3*b^2*sin(d*x+c)-a^5*sin(d*x+c)-8*a*b^4*cos(d*x+c)^3*sin(d*x+c )+a^3*b^2*cos(d*x+c)*sin(d*x+c)-2*a^2*b^3*cos(d*x+c)*sin(d*x+c)-2*a^2*b^3* cos(d*x+c)^2*sin(d*x+c)+8*a*b^4*cos(d*x+c)^2*sin(d*x+c)+16*b^5*cos(d*x+c)^ 3*sin(d*x+c)-4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos( d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2) *a^2*b^3*cos(d*x+c)^4-16*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1 /2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a +b))^(1/2)*a*b^4*cos(d*x+c)^4+16*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/( a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+ cos(d*x+c)))^(1/2)*a^3*b^2*cos(d*x+c)^3-8*EllipticF(cot(d*x+c)-csc(d*x+c), (-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d *x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b^3*cos(d*x+c)^3-32*EllipticF(cot(d*x+c)-c sc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1 /2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^4*cos(d*x+c)^3-3*EllipticE(cot(d *x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a +b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^4*b*cos(d*x+c)^2-2*Elliptic F(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x +c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^4*b*cos(d*x+c)^3-6*E llipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+ cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^4*b*cos(d*...
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c) )/(b^2*cos(d*x + c)^6 + 2*a*b*cos(d*x + c)^5 + a^2*cos(d*x + c)^4), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]