\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [417]

Optimal result

Integrand size = 43, antiderivative size = 284 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(47 A-38 B+24 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 a^{3/2} d}+\frac {(17 A-13 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(21 A-14 B+12 C) \tan (c+d x)}{8 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A-12 B+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-3 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}} \] Output:

-1/8*(47*A-38*B+24*C)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a 
^(3/2)/d+1/4*(17*A-13*B+9*C)*arctanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*c 
os(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d+1/8*(21*A-14*B+12*C)*tan(d*x+c)/a/d/(a 
+a*cos(d*x+c))^(1/2)-1/12*(13*A-12*B+6*C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*c 
os(d*x+c))^(1/2)-1/2*(A-B+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(3 
/2)+1/6*(5*A-3*B+3*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 4.81 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.88 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sec ^3(c+d x) \left (12 (17 A-13 B+9 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x)-3 \sqrt {2} (47 A-38 B+24 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \cos ^3(c+d x)+\frac {1}{4} \cos ^3\left (\frac {1}{2} (c+d x)\right ) (106 A-36 B+48 C+3 (55 A-26 B+36 C) \cos (c+d x)+(74 A-36 B+48 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))-42 B \cos (3 (c+d x))+36 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{12 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a* 
Cos[c + d*x])^(3/2),x]
 

Output:

-1/12*(Sec[c + d*x]^3*(12*(17*A - 13*B + 9*C)*ArcTanh[Sin[(c + d*x)/2]]*Co 
s[(c + d*x)/2]^5*Cos[c + d*x]^3 - 3*Sqrt[2]*(47*A - 38*B + 24*C)*ArcTanh[S 
qrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5*Cos[c + d*x]^3 + (Cos[(c + d*x 
)/2]^3*(106*A - 36*B + 48*C + 3*(55*A - 26*B + 36*C)*Cos[c + d*x] + (74*A 
- 36*B + 48*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 42*B*Cos[3*(c + 
d*x)] + 36*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/4))/(d*(a*(1 + Cos[c + d* 
x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))
 

Rubi [A] (verified)

Time = 1.97 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.442, Rules used = {3042, 3520, 27, 3042, 3463, 25, 3042, 3463, 27, 3042, 3463, 27, 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.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 3520

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {\int -\frac {\left (2 a^2 (13 A-12 B+6 C)-5 a^2 (5 A-3 B+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (2 a^2 (13 A-12 B+6 C)-5 a^2 (5 A-3 B+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {2 a^2 (13 A-12 B+6 C)-5 a^2 (5 A-3 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {\int -\frac {3 \left (a^3 (21 A-14 B+12 C)-a^3 (13 A-12 B+6 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}+\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {\left (a^3 (21 A-14 B+12 C)-a^3 (13 A-12 B+6 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \int \frac {a^3 (21 A-14 B+12 C)-a^3 (13 A-12 B+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {\left (a^4 (47 A-38 B+24 C)-a^4 (21 A-14 B+12 C) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\left (a^4 (47 A-38 B+24 C)-a^4 (21 A-14 B+12 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^4 (47 A-38 B+24 C)-a^4 (21 A-14 B+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (47 A-38 B+24 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-4 a^4 (17 A-13 B+9 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (47 A-38 B+24 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-4 a^4 (17 A-13 B+9 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (47 A-38 B+24 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {8 a^4 (17 A-13 B+9 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a^3 (47 A-38 B+24 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {4 \sqrt {2} a^{7/2} (17 A-13 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a^4 (47 A-38 B+24 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 \sqrt {2} a^{7/2} (17 A-13 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (5 A-3 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {a^2 (13 A-12 B+6 C) \tan (c+d x) \sec (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 \left (\frac {a^3 (21 A-14 B+12 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^{7/2} (47 A-38 B+24 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 \sqrt {2} a^{7/2} (17 A-13 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\right )}{2 a}}{3 a}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c 
+ d*x])^(3/2),x]
 

Output:

-1/2*((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/ 
2)) + ((2*a*(5*A - 3*B + 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a 
*Cos[c + d*x]]) - ((a^2*(13*A - 12*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(d* 
Sqrt[a + a*Cos[c + d*x]]) - (3*(-1/2*((2*a^(7/2)*(47*A - 38*B + 24*C)*ArcT 
anh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (4*Sqrt[2]*a^(7/ 
2)*(17*A - 13*B + 9*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a* 
Cos[c + d*x]])])/d)/a + (a^3*(21*A - 14*B + 12*C)*Tan[c + d*x])/(d*Sqrt[a 
+ a*Cos[c + d*x]])))/(2*a))/(3*a))/(4*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 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 3520

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*A - b*B + 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))   Int[(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)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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. \(2412\) vs. \(2(249)=498\).

Time = 0.88 (sec) , antiderivative size = 2413, normalized size of antiderivative = 8.50

method	result	size
parts	\(\text {Expression too large to display}\)	\(2413\)
default	\(\text {Expression too large to display}\)	\(2993\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, 
method=_RETURNVERBOSE)
 

Output:

1/6*A*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(1632*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-11 
28*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-11 
28*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-244 
8*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+504*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+1692*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+1692*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+1224*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-608*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)^4-846*ln(-4/(2*c 
os(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-846*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)*(si 
n(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-204*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(...
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.42 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {12 \, \sqrt {2} {\left ({\left (17 \, A - 13 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (17 \, A - 13 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (17 \, A - 13 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 3 \, {\left ({\left (47 \, A - 38 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (47 \, A - 38 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (47 \, A - 38 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (21 \, A - 14 \, B + 12 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (37 \, A - 18 \, B + 24 \, C\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3 
/2),x, algorithm="fricas")
 

Output:

1/96*(12*sqrt(2)*((17*A - 13*B + 9*C)*cos(d*x + c)^5 + 2*(17*A - 13*B + 9* 
C)*cos(d*x + c)^4 + (17*A - 13*B + 9*C)*cos(d*x + c)^3)*sqrt(a)*log(-(a*co 
s(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)) + 3*((47*A - 
 38*B + 24*C)*cos(d*x + c)^5 + 2*(47*A - 38*B + 24*C)*cos(d*x + c)^4 + (47 
*A - 38*B + 24*C)*cos(d*x + c)^3)*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*(3*(21*A - 14*B + 12*C 
)*cos(d*x + c)^3 + (37*A - 18*B + 24*C)*cos(d*x + c)^2 - 6*(A - 2*B)*cos(d 
*x + c) + 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^ 
5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))* 
*(3/2),x)
 

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**4/(a*(cos( 
c + d*x) + 1))**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3 
/2),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3 
/2),x, algorithm="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. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + 
 d*x))^(3/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + 
 d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**4)/(cos(c 
 + d*x)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(cos(c + d*x) + 1)*cos(c 
+ d*x)**2*sec(c + d*x)**4)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*c + i 
nt((sqrt(cos(c + d*x) + 1)*sec(c + d*x)**4)/(cos(c + d*x)**2 + 2*cos(c + d 
*x) + 1),x)*a))/a**2