\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\) [894]

Optimal result

Integrand size = 42, antiderivative size = 337 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 b B+3 b^3 B+a^3 C-7 a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^4 b B+10 a^2 b^3 B-b^5 B+a^5 C-10 a^3 b^2 C-3 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}-\frac {(b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

1/4*(5*B*a^2*b+B*b^3-C*a^3-5*C*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) 
)/a/b/(a^2-b^2)^2/d+1/4*(3*B*a^2*b+3*B*b^3+C*a^3-7*C*a*b^2)*InverseJacobiA 
M(1/2*d*x+1/2*c,2^(1/2))/b^2/(a^2-b^2)^2/d-1/4*(3*B*a^4*b+10*B*a^2*b^3-B*b 
^5+C*a^5-10*C*a^3*b^2-3*C*a*b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2 
^(1/2))/a/(a-b)^2/b^2/(a+b)^3/d-1/2*(B*b-C*a)*cos(d*x+c)^(1/2)*sin(d*x+c)/ 
(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/4*(5*B*a^2*b+B*b^3-C*a^3-5*C*a*b^2)*cos(d 
*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 4.81 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.08 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 \sqrt {\cos (c+d x)} \left (a \left (-7 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right )+b \left (-5 a^2 b B-b^3 B+a^3 C+5 a b^2 C\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {2 \left (-9 a^2 b B+3 b^3 B+5 a^3 C+a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {16 a \left (2 a^2 B+b^2 B-3 a b C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}-\frac {2 \left (-5 a^2 b B-b^3 B+a^3 C+5 a b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 a d} \] Input:

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*C 
os[c + d*x])^3),x]
 

Output:

((4*Sqrt[Cos[c + d*x]]*(a*(-7*a^2*b*B + b^3*B + 3*a^3*C + 3*a*b^2*C) + b*( 
-5*a^2*b*B - b^3*B + a^3*C + 5*a*b^2*C)*Cos[c + d*x])*Sin[c + d*x])/((a^2 
- b^2)^2*(a + b*Cos[c + d*x])^2) + ((2*(-9*a^2*b*B + 3*b^3*B + 5*a^3*C + a 
*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*a*(2*a^2* 
B + b^2*B - 3*a*b*C)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2* 
b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) - (2*(-5*a^2*b*B - b^3*B + a^3*C 
 + 5*a*b^2*C)*(-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[Sin[c + 
 d*x]^2]))/((a - b)^2*(a + b)^2))/(16*a*d)
 

Rubi [A] (verified)

Time = 2.19 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.99, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3508, 3042, 3478, 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.

\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3478

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

\(\displaystyle -\frac {\frac {\frac {\frac {\left (a^5 C+3 a^4 b B-10 a^3 b^2 C+10 a^2 b^3 B-3 a b^4 C-b^5 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {a \left (a^3 C+3 a^2 b B-7 a b^2 C+3 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}}{b}-\frac {2 \left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}+\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\frac {\left (a^5 C+3 a^4 b B-10 a^3 b^2 C+10 a^2 b^3 B-3 a b^4 C-b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {a \left (a^3 C+3 a^2 b B-7 a b^2 C+3 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 \left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}+\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {\frac {\frac {\left (a^5 C+3 a^4 b B-10 a^3 b^2 C+10 a^2 b^3 B-3 a b^4 C-b^5 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 a \left (a^3 C+3 a^2 b B-7 a b^2 C+3 b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}+\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\frac {2 \left (a^5 C+3 a^4 b B-10 a^3 b^2 C+10 a^2 b^3 B-3 a b^4 C-b^5 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}-\frac {2 a \left (a^3 C+3 a^2 b B-7 a b^2 C+3 b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 \left (a^2-b^2\right )}\)

Input:

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + 
 d*x])^3),x]
 

Output:

-1/2*((b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*C 
os[c + d*x])^2) - (((-2*(5*a^2*b*B + b^3*B - a^3*C - 5*a*b^2*C)*EllipticE[ 
(c + d*x)/2, 2])/(b*d) + ((-2*a*(3*a^2*b*B + 3*b^3*B + a^3*C - 7*a*b^2*C)* 
EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(3*a^4*b*B + 10*a^2*b^3*B - b^5*B + 
a^5*C - 10*a^3*b^2*C - 3*a*b^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2 
])/(b*(a + b)*d))/b)/(2*a*(a^2 - b^2)) + ((5*a^2*b*B + b^3*B - a^3*C - 5*a 
*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d 
*x])))/(4*(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 3478

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[(B*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f* 
x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a 
+ b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[c*(a*A - b*B)*( 
m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*(m + 2))* 
Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
 NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 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. \(1849\) vs. \(2(328)=656\).

Time = 6.92 (sec) , antiderivative size = 1850, normalized size of antiderivative = 5.49

method	result	size
default	\(\text {Expression too large to display}\)	\(1850\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,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)*(-4*C/b/(-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*s 
in(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-2*C*a)/b^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)*Elli 
pticF(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*(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))+1/a/(a^2-b^2)/(-2*a* 
b+2*b^2)*b^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)*EllipticPi(cos(1/2*d 
*x+1/2*c),-2*b/(a-b),2^(1/2)))-2*a*(B*b-C*a)/b^2*(-1/2/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)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*cos(1/2...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ 
3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+b*cos(d*x+c) 
)**3,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ 
3,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^ 
3,x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^3*sqrt 
(cos(d*x + c))), x)
 

Mupad [F(-1)]

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

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b*cos(c + 
 d*x))^3),x)
 

Output:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b*cos(c + 
 d*x))^3), x)
 

Reduce [F]

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

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x)
 

Output:

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