\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [1482]

Optimal result

Integrand size = 43, antiderivative size = 294 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^3 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d} \] Output:

-2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*C))*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d* 
x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/a^3/d-2/3*(A*b-B*a)*cos(d*x+c)^(1/2)*In 
verseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^2/d-2*b*(A*b^2-a*( 
B*b-C*a))*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2) 
)*sec(d*x+c)^(1/2)/a^3/(a+b)/d+2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*C))*sec(d*x 
+c)^(1/2)*sin(d*x+c)/a^3/d-2/3*(A*b-B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d 
+2/5*A*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d
 

Mathematica [A] (warning: unable to verify)

Time = 8.62 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.26 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {4 a \sec (c+d x) \left (3 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)+a (-5 A b+5 a B+3 a A \sec (c+d x)) \tan (c+d x)\right )-\frac {4 \cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (-3 a \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+\left (15 A b^3+15 a b^2 (A-B)+5 a^2 b (A-3 B+3 C)+a^3 (9 A-5 B+15 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+3 \left (a \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \tan ^2(c+d x)-5 b \left (A b^2+a (-b B+a C)\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{a+b \cos (c+d x)}}{30 a^4 d \sqrt {\sec (c+d x)}} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a 
+ b*Cos[c + d*x]),x]
 

Output:

(4*a*Sec[c + d*x]*(3*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x] + 
a*(-5*A*b + 5*a*B + 3*a*A*Sec[c + d*x])*Tan[c + d*x]) - (4*Cos[c + d*x]*Co 
t[c + d*x]*(b + a*Sec[c + d*x])*(-3*a*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C) 
)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c 
 + d*x]^2] + (15*A*b^3 + 15*a*b^2*(A - B) + 5*a^2*b*(A - 3*B + 3*C) + a^3* 
(9*A - 5*B + 15*C))*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + 
 d*x]]*Sqrt[-Tan[c + d*x]^2] + 3*(a*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))* 
Tan[c + d*x]^2 - 5*b*(A*b^2 + a*(-(b*B) + a*C))*EllipticPi[-(a/b), ArcSin[ 
Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/(a + 
b*Cos[c + d*x]))/(30*a^4*d*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 2.47 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.93, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4709, 3042, 3534, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}

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.

\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{7/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{a+b \cos (c+d x)}dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {-5 b (A b-a B) \cos ^2(c+d x)+a (4 A b+5 a B) \cos (c+d x)+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}+\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 b (A b-a B) \cos ^2(c+d x)+a (4 A b+5 a B) \cos (c+d x)+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \cos (c+d x)+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \cos ^2(c+d x)+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \cos (c+d x)+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 b \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (3 (3 A+5 C) a^2-20 b B a+20 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (a (A b-a B) \cos (c+d x) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx+\frac {5 \int \frac {a (A b-a B) \cos (c+d x) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 \int \frac {a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \int \frac {a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+a b (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a b (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}+\frac {5 \left (\frac {6 b^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{b}}{a}}{3 a}}{5 a}\right )\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + b*Co 
s[c + d*x]),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(5*a*d*Cos[c + d 
*x]^(5/2)) - ((10*(A*b - a*B)*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - ( 
-(((6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + 
 (5*((2*a*b*(A*b - a*B)*EllipticF[(c + d*x)/2, 2])/d + (6*b^2*(A*b^2 - a*( 
b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d)))/b)/a) 
 + (6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(a*d*Sqrt[Cos[c 
+ d*x]]))/(3*a))/(5*a))
 

Defintions of rubi rules used

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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3534

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])))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(774\) vs. \(2(271)=542\).

Time = 210.34 (sec) , antiderivative size = 775, normalized size of antiderivative = 2.64

method	result	size
default	\(\text {Expression too large to display}\)	\(775\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x,me 
thod=_RETURNVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*A/a/(8*sin 
(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 
2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 
*c)+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c) 
^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) 
^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^ 
4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2*(A*b-B*a)/a^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2 
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/ 
2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(- 
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2)))+2*(A*b^2-B*a*b+C*a^2)/a^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2* 
d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 
1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*(A*b^2- 
B*a*b+C*a^2)*b^2/a^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1 
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1 
/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) 
),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2)/(a+b*cos(d*x+ 
c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) 
),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(b*co 
s(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c) 
),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(b*co 
s(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
b*cos(c + d*x)),x)
 

Output:

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
b*cos(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x)
 

Output:

int((sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3)/(cos(c + d*x)*b + a) 
,x)*b + int((sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3)/(cos(c + 
d*x)*b + a),x)*c + int((sqrt(sec(c + d*x))*sec(c + d*x)**3)/(cos(c + d*x)* 
b + a),x)*a