\(\int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\) [380]

Optimal result

Integrand size = 33, antiderivative size = 345 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=-\frac {\left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (15 a^4 A b-6 a^2 A b^3+3 A b^5-3 a^5 B-10 a^3 b^2 B+a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^2 (a-b)^2 b (a+b)^3 d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

-1/4*(9*A*a^2*b-3*A*b^3-5*B*a^3-B*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1 
/2))/a^2/(a^2-b^2)^2/d-1/4*(7*A*a^2*b-A*b^3-3*B*a^3-3*B*a*b^2)*InverseJaco 
biAM(1/2*d*x+1/2*c,2^(1/2))/a/b/(a^2-b^2)^2/d+1/4*(15*A*a^4*b-6*A*a^2*b^3+ 
3*A*b^5-3*B*a^5-10*B*a^3*b^2+B*a*b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a 
+b),2^(1/2))/a^2/(a-b)^2/b/(a+b)^3/d+1/2*b*(A*b-B*a)*cos(d*x+c)^(1/2)*sin( 
d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1/4*b*(9*A*a^2*b-3*A*b^3-5*B*a^3-B 
*a*b^2)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.39 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 b \sqrt {\cos (c+d x)} \left (a \left (-11 a^2 A b+5 A b^3+7 a^3 B-a b^2 B\right )+b \left (-9 a^2 A b+3 A b^3+5 a^3 B+a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {\left (16 a^4 A-19 a^2 A b^2+9 A b^4-9 a^3 b B+3 a b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 a \left (-4 a^2 A b+A b^3+2 a^3 B+a b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}+\frac {\left (-9 a^2 A b+3 A b^3+5 a^3 B+a b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a^2 d} \] Input:

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

Output:

((-2*b*Sqrt[Cos[c + d*x]]*(a*(-11*a^2*A*b + 5*A*b^3 + 7*a^3*B - a*b^2*B) + 
 b*(-9*a^2*A*b + 3*A*b^3 + 5*a^3*B + a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/ 
((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + (((16*a^4*A - 19*a^2*A*b^2 + 9*A* 
b^4 - 9*a^3*b*B + 3*a*b^3*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a 
 + b) + (8*a*(-4*a^2*A*b + A*b^3 + 2*a^3*B + a*b^2*B)*((a + b)*EllipticF[( 
c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) 
 + ((-9*a^2*A*b + 3*A*b^3 + 5*a^3*B + a*b^2*B)*(-2*a*b*EllipticE[ArcSin[Sq 
rt[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*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*a^2*d)
 

Rubi [A] (verified)

Time = 2.26 (sec) , antiderivative size = 336, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3479, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\int \frac {8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a-4 \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \cos (c+d x) a+3 A b^4-b \left (-5 B a^3+9 A b a^2-b^2 B a-3 A b^3\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a-4 \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \cos (c+d x) a+3 A b^4-b \left (-5 B a^3+9 A b a^2-b^2 B a-3 A b^3\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a-4 \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^4-b \left (-5 B a^3+9 A b a^2-b^2 B a-3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {-\left (\left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \int \sqrt {\cos (c+d x)}dx\right )-\frac {\int -\frac {b \left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right )-a b \left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right )-a b \left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \int \sqrt {\cos (c+d x)}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right )-a b \left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right )-a b \left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx-a \left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {2 \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-a \left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 a \left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}-\frac {2 \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {b \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\frac {2 \left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 a \left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}-\frac {2 \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

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

Output:

(b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b* 
Cos[c + d*x])^2) + (((-2*(9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*Ellipti 
cE[(c + d*x)/2, 2])/d + ((-2*a*(7*a^2*A*b - A*b^3 - 3*a^3*B - 3*a*b^2*B)*E 
llipticF[(c + d*x)/2, 2])/d + (2*(15*a^4*A*b - 6*a^2*A*b^3 + 3*A*b^5 - 3*a 
^5*B - 10*a^3*b^2*B + a*b^4*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/ 
((a + b)*d))/b)/(2*a*(a^2 - b^2)) + (b*(9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a* 
b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d* 
x])))/(4*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 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. \(1743\) vs. \(2(336)=672\).

Time = 7.03 (sec) , antiderivative size = 1744, normalized size of antiderivative = 5.06

method	result	size
default	\(\text {Expression too large to display}\)	\(1744\)

Input:

int((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b*(-b^2/a/ 
(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/(a+b)*(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))-1/2*b/(a^2-b^2)/a 
*(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))+1/2*b/(a^2-b^2)/a*(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)*Ellipti 
cE(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)*(-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))+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)*(-2*co 
s(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)))+2*(A*b-B*a)/b*(- 
1/2*b^2/a/(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*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)-7/8/(a+b)/(a^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*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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