\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [125]

Optimal result

Integrand size = 35, antiderivative size = 262 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(39 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A+43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(63 A+11 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A+3 C) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A+7 C) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \] Output:

1/4*(39*A+8*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/ 
d-1/32*(219*A+43*C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c) 
)^(1/2))*2^(1/2)/a^(5/2)/d-1/16*(63*A+11*C)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+ 
c))^(1/2)-1/4*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(1 
9*A+3*C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/16*(31*A+7*C)* 
sec(d*x+c)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 6.21 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.40 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {4 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (-\frac {1}{8} (219 A+43 C) \text {arctanh}\left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )+\frac {3 A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{\sqrt {2}}-6 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )+4 \sqrt {2} (6 A+C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )-\frac {A+C}{16 \left (1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}-\frac {27 A+11 C}{16 \left (1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {A+C}{16 \left (1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {27 A+11 C}{16 \left (1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {2 A \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}-\frac {9 A \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{d (a (1+\cos (c+d x)))^{5/2} (2 A+C+C \cos (2 c+2 d x))} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/ 
2),x]
 

Output:

(4*Cos[c/2 + (d*x)/2]^5*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-1/8*((219* 
A + 43*C)*ArcTanh[Sin[c/2 + (d*x)/2]]) + (3*A*ArcTanh[Sqrt[2]*Sin[c/2 + (d 
*x)/2]])/Sqrt[2] - 6*Sqrt[2]*A*ArcTanh[Sqrt[2]*Sin[c/2 + (d*x)/2]] + 4*Sqr 
t[2]*(6*A + C)*ArcTanh[Sqrt[2]*Sin[c/2 + (d*x)/2]] - (A + C)/(16*(1 - Sin[ 
c/2 + (d*x)/2])^2) - (27*A + 11*C)/(16*(1 - Sin[c/2 + (d*x)/2])) + (A + C) 
/(16*(1 + Sin[c/2 + (d*x)/2])^2) + (27*A + 11*C)/(16*(1 + Sin[c/2 + (d*x)/ 
2])) + (2*A*Sin[c/2 + (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)^2 - (9*A*Sin[ 
c/2 + (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)))/(d*(a*(1 + Cos[c + d*x]))^( 
5/2)*(2*A + C + C*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 3521, 27, 3042, 3457, 27, 3042, 3463, 27, 3042, 3463, 25, 3042, 3464, 3042, 3128, 219, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {(4 a (3 A+C)-a (7 A-C) \cos (c+d x)) \sec ^3(c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(4 a (3 A+C)-a (7 A-C) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {4 a (3 A+C)-a (7 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\left (4 a^2 (31 A+7 C)-5 a^2 (19 A+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (4 a^2 (31 A+7 C)-5 a^2 (19 A+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {4 a^2 (31 A+7 C)-5 a^2 (19 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {\frac {\int -\frac {2 \left (2 a^3 (63 A+11 C)-3 a^3 (31 A+7 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (2 a^3 (63 A+11 C)-3 a^3 (31 A+7 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {2 a^3 (63 A+11 C)-3 a^3 (31 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {\left (4 a^4 (39 A+8 C)-a^4 (63 A+11 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (4 a^4 (39 A+8 C)-a^4 (63 A+11 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {4 a^4 (39 A+8 C)-a^4 (63 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-a^4 (219 A+43 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^4 (219 A+43 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^4 (219 A+43 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {4 a^3 (39 A+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\sqrt {2} a^{7/2} (219 A+43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {8 a^4 (39 A+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A+43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a^2 (31 A+7 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (63 A+11 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {8 a^{7/2} (39 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{7/2} (219 A+43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}}{a}}{4 a^2}-\frac {a (19 A+3 C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]
 

Output:

-1/4*((A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) + 
(-1/2*(a*(19*A + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^( 
3/2)) + ((2*a^2*(31*A + 7*C)*Sec[c + d*x]*Tan[c + d*x])/(d*Sqrt[a + a*Cos[ 
c + d*x]]) - (-(((8*a^(7/2)*(39*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sq 
rt[a + a*Cos[c + d*x]]])/d - (Sqrt[2]*a^(7/2)*(219*A + 43*C)*ArcTanh[(Sqrt 
[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/a) + (2*a^3*(63* 
A + 11*C)*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/a)/(4*a^2))/(8*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3252

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]
 

rule 3457

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 3463

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 0])
 

rule 3464

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A 
*b - a*B)/(b*c - a*d)   Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c 
- A*d)/(b*c - a*d)   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]
 

rule 3521

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*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
 + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) 
- b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* 
c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] && LtQ[m, -2^(-1)]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1164\) vs. \(2(227)=454\).

Time = 0.50 (sec) , antiderivative size = 1165, normalized size of antiderivative = 4.45

method	result	size
parts	\(\text {Expression too large to display}\)	\(1165\)
default	\(\text {Expression too large to display}\)	\(1610\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/8*A*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(876*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^8*a-62 
4*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/ 
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^8*a-624 
*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2) 
*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-876*2 
^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2 
*c))*cos(1/2*d*x+1/2*c)^6*a+252*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^( 
1/2)*cos(1/2*d*x+1/2*c)^6+624*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1 
/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a) 
)*cos(1/2*d*x+1/2*c)^6*a+624*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2 
)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))* 
cos(1/2*d*x+1/2*c)^6*a+219*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a 
)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-188*2^(1/2)*(sin(1 
/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-156*ln(-4/(2*cos(1/2 
*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/ 
2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-156*ln(4/(2*cos(1/2*d 
*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2* 
d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+19*a^(1/2)*2^(1/2)*(sin 
(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(1/2*d*x+1/2*c)^2+2*a^(1/2)*2^(1/2)*(sin(...
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.61 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (219 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (219 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (219 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (219 \, A + 43 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (39 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (39 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (39 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (63 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 20 \, A \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algori 
thm="fricas")
 

Output:

1/64*(sqrt(2)*((219*A + 43*C)*cos(d*x + c)^5 + 3*(219*A + 43*C)*cos(d*x + 
c)^4 + 3*(219*A + 43*C)*cos(d*x + c)^3 + (219*A + 43*C)*cos(d*x + c)^2)*sq 
rt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)* 
sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 
1)) + 4*((39*A + 8*C)*cos(d*x + c)^5 + 3*(39*A + 8*C)*cos(d*x + c)^4 + 3*( 
39*A + 8*C)*cos(d*x + c)^3 + (39*A + 8*C)*cos(d*x + c)^2)*sqrt(a)*log((a*c 
os(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(c 
os(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 
4*((63*A + 11*C)*cos(d*x + c)^3 + 5*(19*A + 3*C)*cos(d*x + c)^2 + 20*A*cos 
(d*x + c) - 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c 
)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c) 
^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algori 
thm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**3)/(co 
s(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c + int((sqrt(c 
os(c + d*x) + 1)*sec(c + d*x)**3)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3 
*cos(c + d*x) + 1),x)*a))/a**3