\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) [890]

Optimal result

Integrand size = 42, antiderivative size = 345 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b \left (7 a^2 b B-5 b^3 B-5 a^3 C+3 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \] Output:

(4*B*a^2*b-5*B*b^3-2*C*a^3+3*C*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) 
)/a^3/(a^2-b^2)/d+1/3*(2*B*a^2-5*B*b^2+3*C*a*b)*InverseJacobiAM(1/2*d*x+1/ 
2*c,2^(1/2))/a^2/(a^2-b^2)/d+b*(7*B*a^2*b-5*B*b^3-5*C*a^3+3*C*a*b^2)*Ellip 
ticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^3/(a-b)/(a+b)^2/d+1/3*(2*B*a 
^2-5*B*b^2+3*C*a*b)*sin(d*x+c)/a^2/(a^2-b^2)/d/cos(d*x+c)^(3/2)-(4*B*a^2*b 
-5*B*b^3-2*C*a^3+3*C*a*b^2)*sin(d*x+c)/a^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)+b* 
(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 7.24 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.24 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \left (4 a^4 B+44 a^2 b^2 B-45 b^4 B-30 a^3 b C+27 a b^3 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (28 a^3 b B-40 a b^3 B-12 a^4 C+24 a^2 b^2 C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {2 \left (12 a^2 b^2 B-15 b^4 B-6 a^3 b C+9 a b^3 C\right ) \cos (2 (c+d x)) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{12 a^3 (a-b) (a+b) d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (-2 b B \sin (c+d x)+a C \sin (c+d x))}{a^3}+\frac {b^4 B \sin (c+d x)-a b^3 C \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 B \sec (c+d x) \tan (c+d x)}{3 a^2}\right )}{d} \] Input:

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

Output:

((2*(4*a^4*B + 44*a^2*b^2*B - 45*b^4*B - 30*a^3*b*C + 27*a*b^3*C)*Elliptic 
Pi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ((28*a^3*b*B - 40*a*b^3*B - 1 
2*a^4*C + 24*a^2*b^2*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2* 
b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (2*(12*a^2*b^2*B - 15*b^4*B - 6 
*a^3*b*C + 9*a*b^3*C)*Cos[2*(c + d*x)]*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c 
 + d*x]]], -1] + 2*a*(a + b)*EllipticF[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^2*Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(12*a^3*(a 
 - b)*(a + b)*d) + (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(-2*b*B*Sin[c + d* 
x] + a*C*Sin[c + d*x]))/a^3 + (b^4*B*Sin[c + d*x] - a*b^3*C*Sin[c + d*x])/ 
(a^3*(a^2 - b^2)*(a + b*Cos[c + d*x])) + (2*B*Sec[c + d*x]*Tan[c + d*x])/( 
3*a^2)))/d
 

Rubi [A] (verified)

Time = 2.76 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.95, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3508, 3042, 3479, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a+2 \left (-3 C a^3+7 b B a^2+6 b^2 C a-10 b^3 B\right ) \cos (c+d x) a+3 b \left (-2 C a^3+4 b B a^2+3 b^2 C a-5 b^3 B\right ) \cos ^2(c+d x)-15 b^4 B}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a+2 \left (-3 C a^3+7 b B a^2+6 b^2 C a-10 b^3 B\right ) \cos (c+d x) a+3 b \left (-2 C a^3+4 b B a^2+3 b^2 C a-5 b^3 B\right ) \cos ^2(c+d x)-15 b^4 B}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a+2 \left (-3 C a^3+7 b B a^2+6 b^2 C a-10 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 b \left (-2 C a^3+4 b B a^2+3 b^2 C a-5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-15 b^4 B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {a \left (2 B a^2+3 b C a-5 b^2 B\right ) \cos (c+d x) b^2+\left (2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a-15 b^4 B\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {a \left (2 B a^2+3 b C a-5 b^2 B\right ) \cos (c+d x) b^2+\left (2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a-15 b^4 B\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {a \left (2 B a^2+3 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a-15 b^4 B\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {a \left (2 B a^2+3 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (2 B a^4-12 b C a^3+16 b^2 B a^2+9 b^3 C a-15 b^4 B\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a b \left (2 a^2 B+3 a b C-5 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 b^2 \left (-5 a^3 C+7 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a b \left (2 a^2 B+3 a b C-5 b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 b^2 \left (-5 a^3 C+7 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {3 b^2 \left (-5 a^3 C+7 a^2 b B+3 a b^2 C-5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a b \left (2 a^2 B+3 a b C-5 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}+\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\frac {2 \left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 \left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {\frac {2 a b \left (2 a^2 B+3 a b C-5 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 b^2 \left (-5 a^3 C+7 a^2 b B+3 a b^2 C-5 b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\)

Input:

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

Output:

(b*(b*B - a*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Co 
s[c + d*x])) + ((2*(2*a^2*B - 5*b^2*B + 3*a*b*C)*Sin[c + d*x])/(3*a*d*Cos[ 
c + d*x]^(3/2)) - (-(((6*(4*a^2*b*B - 5*b^3*B - 2*a^3*C + 3*a*b^2*C)*Ellip 
ticE[(c + d*x)/2, 2])/d + ((2*a*b*(2*a^2*B - 5*b^2*B + 3*a*b*C)*EllipticF[ 
(c + d*x)/2, 2])/d + (6*b^2*(7*a^2*b*B - 5*b^3*B - 5*a^3*C + 3*a*b^2*C)*El 
lipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (6*(4*a^2*b* 
B - 5*b^3*B - 2*a^3*C + 3*a*b^2*C)*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]])) 
/(3*a))/(2*a*(a^2 - b^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 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 3479

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 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. \(1004\) vs. \(2(340)=680\).

Time = 6.56 (sec) , antiderivative size = 1005, normalized size of antiderivative = 2.91

method	result	size
default	\(\text {Expression too large to display}\)	\(1005\)

Input:

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

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/a^2*(-1/6*c 
os(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)/(co 
s(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(2*B*b-C*a)/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*sin(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*b^2*(2*B*b-C*a)/a^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2* 
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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*(B*b-C*a)*b/a^2 
*(-1/a*b^2/(a^2-b^2)*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)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/(a+b)/a*(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*(1-2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b 
/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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)*EllipticF(cos(1/2*d*x+ 
1/2*c),2^(1/2))+1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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) 
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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