\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\) [1101]

Optimal result

Integrand size = 43, antiderivative size = 234 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {2 b \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a+b) d}+\frac {2 A \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 5.36 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.42 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=-\frac {\frac {2 \left (45 A b^3-10 a^3 B-45 a b^2 B+a^2 b (19 A+45 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {4 a \left (20 A b^2-20 a b B+3 a^2 (3 A+5 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}+\frac {6 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}-\frac {2 \left (10 a (-A b+a B) \sin (c+d x)+3 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (2 (c+d x))+6 a^2 A \tan (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}}{30 a^3 d} \] Input:

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

Output:

-1/30*((2*(45*A*b^3 - 10*a^3*B - 45*a*b^2*B + a^2*b*(19*A + 45*C))*Ellipti 
cPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (4*a*(20*A*b^2 - 20*a*b*B + 
3*a^2*(3*A + 5*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b) 
/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) + (6*(5*A*b^2 - 5*a*b*B + a^2*(3*A 
 + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*E 
llipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a 
), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2 
]) - (2*(10*a*(-(A*b) + a*B)*Sin[c + d*x] + 3*(5*A*b^2 - 5*a*b*B + a^2*(3* 
A + 5*C))*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x]))/Cos[c + d*x]^(3/2))/(a 
^3*d)
 

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

(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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(774\) vs. \(2(223)=446\).

Time = 6.28 (sec) , antiderivative size = 775, normalized size of antiderivative = 3.31

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)/cos(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*(2*sin(1/2*d 
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticE(cos(1/2*d*x+1/2*c),2^(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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5} b +\cos \left (d x +c \right )^{4} a}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b +\cos \left (d x +c \right )^{3} a}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} b +\cos \left (d x +c \right )^{2} a}d x \right ) c \] Input:

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**5*b + cos(c + d*x)**4*a),x)*a + int( 
sqrt(cos(c + d*x))/(cos(c + d*x)**4*b + cos(c + d*x)**3*a),x)*b + int(sqrt 
(cos(c + d*x))/(cos(c + d*x)**3*b + cos(c + d*x)**2*a),x)*c