\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [383]

Optimal result

Integrand size = 33, antiderivative size = 531 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {\left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 b \sqrt {a+b} d}-\frac {\left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt {a+b} d}-\frac {\sqrt {a+b} \left (4 a^2 A+15 A b^2-12 a b B\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^4 d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {b \left (7 a^2 A b-15 A b^3-4 a^3 B+12 a b^2 B\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:

-1/4*(7*A*a^2*b-15*A*b^3-4*B*a^3+12*B*a*b^2)*cot(d*x+c)*EllipticE((a+b*sec 
(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^( 
1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/b/(a+b)^(1/2)/d-1/4*(15*A*b^2+a*b 
*(5*A-12*B)-2*a^2*(A+2*B))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+ 
b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d* 
x+c))/(a-b))^(1/2)/a^3/(a+b)^(1/2)/d-1/4*(a+b)^(1/2)*(4*A*a^2+15*A*b^2-12* 
B*a*b)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,(( 
a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b) 
)^(1/2)/a^4/d-1/4*(5*A*b-4*B*a)*sin(d*x+c)/a^2/d/(a+b*sec(d*x+c))^(1/2)+1/ 
2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(1/2)-1/4*b*(7*A*a^2*b-15*A 
*b^3-4*B*a^3+12*B*a*b^2)*tan(d*x+c)/a^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1962\) vs. \(2(531)=1062\).

Time = 14.80 (sec) , antiderivative size = 1962, normalized size of antiderivative = 3.69 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2) 
,x]
 

Output:

((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*b^2*(A*b - a*B)*Sin[c + d*x])/( 
a^3*(-a^2 + b^2)) + (2*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x]))/(a^3*( 
a^2 - b^2)*(b + a*Cos[c + d*x])) + (A*Sin[2*(c + d*x)])/(4*a^2)))/(d*(a + 
b*Sec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*Sq 
rt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x 
)/2]^2)]*(-7*a^3*A*b*Tan[(c + d*x)/2] - 7*a^2*A*b^2*Tan[(c + d*x)/2] + 15* 
a*A*b^3*Tan[(c + d*x)/2] + 15*A*b^4*Tan[(c + d*x)/2] + 4*a^4*B*Tan[(c + d* 
x)/2] + 4*a^3*b*B*Tan[(c + d*x)/2] - 12*a^2*b^2*B*Tan[(c + d*x)/2] - 12*a* 
b^3*B*Tan[(c + d*x)/2] + 14*a^3*A*b*Tan[(c + d*x)/2]^3 - 30*a*A*b^3*Tan[(c 
 + d*x)/2]^3 - 8*a^4*B*Tan[(c + d*x)/2]^3 + 24*a^2*b^2*B*Tan[(c + d*x)/2]^ 
3 - 7*a^3*A*b*Tan[(c + d*x)/2]^5 + 7*a^2*A*b^2*Tan[(c + d*x)/2]^5 + 15*a*A 
*b^3*Tan[(c + d*x)/2]^5 - 15*A*b^4*Tan[(c + d*x)/2]^5 + 4*a^4*B*Tan[(c + d 
*x)/2]^5 - 4*a^3*b*B*Tan[(c + d*x)/2]^5 - 12*a^2*b^2*B*Tan[(c + d*x)/2]^5 
+ 12*a*b^3*B*Tan[(c + d*x)/2]^5 + 8*a^4*A*EllipticPi[-1, ArcSin[Tan[(c + d 
*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan 
[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 22*a^2*A*b^2*EllipticPi 
[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^ 
2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 3 
0*A*b^4*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - 
 Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*...
 

Rubi [A] (verified)

Time = 2.58 (sec) , antiderivative size = 566, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4522, 27, 3042, 4592, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 A a^2-12 b B a+6 A b \sec (c+d x) a+15 A b^2-b (5 A b-4 a B) \sec ^2(c+d x)}{2 (a+b \sec (c+d x))^{3/2}}dx}{a}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 A a^2-12 b B a+6 A b \sec (c+d x) a+15 A b^2-b (5 A b-4 a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \int -\frac {b \left (-4 B a^3+7 A b a^2+12 b^2 B a-15 A b^3\right ) \sec ^2(c+d x)+2 a b \left (A a^2+4 b B a-5 A b^2\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 A a^2-12 b B a+15 A b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (-4 B a^3+7 A b a^2+12 b^2 B a-15 A b^3\right ) \sec ^2(c+d x)+2 a b \left (A a^2+4 b B a-5 A b^2\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 A a^2-12 b B a+15 A b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (-4 B a^3+7 A b a^2+12 b^2 B a-15 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a b \left (A a^2+4 b B a-5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+\left (a^2-b^2\right ) \left (4 A a^2-12 b B a+15 A b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\left (a^2-b^2\right ) \left (4 A a^2-12 b B a+15 A b^2\right )+\left (2 a b \left (A a^2+4 b B a-5 A b^2\right )-b \left (-4 B a^3+7 A b a^2+12 b^2 B a-15 A b^3\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {\left (a^2-b^2\right ) \left (4 A a^2-12 b B a+15 A b^2\right )+\left (2 a b \left (A a^2+4 b B a-5 A b^2\right )-b \left (-4 B a^3+7 A b a^2+12 b^2 B a-15 A b^3\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) \left (4 a^2 A-12 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) \left (-2 a^2 (A+2 B)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) \left (4 a^2 A-12 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (-2 a^2 (A+2 B)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b (a-b) \left (-2 a^2 (A+2 B)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 A-12 a b B+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 (A+2 B)+a (5 A b-12 b B)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 A-12 a b B+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 (A+2 B)+a (5 A b-12 b B)+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \left (4 a^2 A-12 a b B+15 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (-4 a^3 B+7 a^2 A b+12 a b^2 B-15 A b^3\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\)

Input:

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

Output:

(A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + b*Sec[c + d*x]]) - (((5*A*b 
- 4*a*B)*Sin[c + d*x])/(a*d*Sqrt[a + b*Sec[c + d*x]]) - (((-2*(a - b)*Sqrt 
[a + b]*(7*a^2*A*b - 15*A*b^3 - 4*a^3*B + 12*a*b^2*B)*Cot[c + d*x]*Ellipti 
cE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b* 
(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d 
) - (2*(a - b)*Sqrt[a + b]*(15*A*b^2 - 2*a^2*(A + 2*B) + a*(5*A*b - 12*b*B 
))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a 
 + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + 
 d*x]))/(a - b))])/d - (2*Sqrt[a + b]*(a^2 - b^2)*(4*a^2*A + 15*A*b^2 - 12 
*a*b*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]] 
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[ 
-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) - (2*b*(7*a^2*A 
*b - 15*A*b^3 - 4*a^3*B + 12*a*b^2*B)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[ 
a + b*Sec[c + d*x]]))/(2*a))/(4*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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

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

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4522

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim 
p[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* 
n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* 
x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] 
 && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A 
, B, C}, x] && NeQ[a^2 - b^2, 0]
 

rule 4548

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - 
a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 
 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^( 
m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x 
] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, 
 b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
 

rule 4592

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d 
*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m 
*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* 
Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d 
, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2242\) vs. \(2(486)=972\).

Time = 12.45 (sec) , antiderivative size = 2243, normalized size of antiderivative = 4.22

method	result	size
default	\(\text {Expression too large to display}\)	\(2243\)

Input:

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

Output:

1/4/d/a^3/(a+b)/(a-b)*(4*B*a^3*b*cos(d*x+c)*sin(d*x+c)+sin(d*x+c)*cos(d*x+ 
c)^2*(2+2*cos(d*x+c))*a^4*A+sin(d*x+c)*cos(d*x+c)*(-5*cos(d*x+c)+2)*A*a^3* 
b+15*A*b^4*cos(d*x+c)*sin(d*x+c)+(12*cos(d*x+c)^2+24*cos(d*x+c)+12)*B*(cos 
(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1 
/2)*a*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d* 
x+c)^2-4*cos(d*x+c)-2)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*c 
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(( 
a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*A*(cos(d*x+c)/(1+cos(d* 
x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^2*Ellip 
ticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(10*cos(d*x+c)^2+20*cos(d 
*x+c)+10)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1 
+cos(d*x+c)))^(1/2)*a*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^( 
1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*b*EllipticF(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*B*(c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^ 
(1/2)*a^2*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-22*c 
os(d*x+c)^2-44*cos(d*x+c)-22)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b) 
*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b^2*EllipticPi(-csc(d*x+c)+cot 
(d*x+c),-1,((a-b)/(a+b))^(1/2))+(24*cos(d*x+c)^2+48*cos(d*x+c)+24)*B*(c...
 

Fricas [F]

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

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

Output:

integral((B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)*sqrt(b*sec(d*x 
 + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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