Integrand size = 43, antiderivative size = 552 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 b \sqrt {a+b} d}-\frac {\left (15 A b^2+a b (5 A-12 B)-2 a^2 (A+2 B-4 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^3 \sqrt {a+b} d}-\frac {\sqrt {a+b} \left (15 A b^2-12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^4 d}-\frac {(5 A b-4 a B) \sin (c+d x)}{4 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (15 A b^3+4 a^3 B-12 a b^2 B-a^2 (7 A b-8 b C)\right ) \tan (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:
1/4*(15*A*b^3+4*B*a^3-12*B*a*b^2-a^2*(7*A*b-8*C*b))*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-4*C))*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)*(15*A*b^2- 12*B*a*b+4*a^2*(A+2*C))*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*(15*A*b^3+4*B*a^3-12*B*a*b^2-a^2*(7*A*b-8*C*b))*tan(d*x+c)/a^3/(a^2-b^2) /d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2489\) vs. \(2(552)=1104\).
Time = 17.21 (sec) , antiderivative size = 2489, normalized size of antiderivative = 4.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x])^(3/2),x]
Output:
(((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((4*b*(A*b^2 - a*b*B + a^2*C)*Sin[ c + d*x])/(a^3*(-a^2 + b^2)) + (4*(A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d* x] + a^2*b^2*C*Sin[c + d*x]))/(a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])) + (A* Sin[2*(c + d*x)])/(2*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)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*T an[(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*Ta n[(c + d*x)/2] + 4*a^4*B*Tan[(c + d*x)/2] + 4*a^3*b*B*Tan[(c + d*x)/2] - 1 2*a^2*b^2*B*Tan[(c + d*x)/2] - 12*a*b^3*B*Tan[(c + d*x)/2] + 8*a^3*b*C*Tan [(c + d*x)/2] + 8*a^2*b^2*C*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 - 16*a^3*b*C*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^3*b*C*Tan[(c + d*x)/2]^5 - 8*a^2*b^2*C*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 +...
Time = 2.73 (sec) , antiderivative size = 587, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4592, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (A+2 C) \sec (c+d x)+5 A b-4 a B\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{2 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-3 A b \sec ^2(c+d x)-2 a (A+2 C) \sec (c+d x)+5 A b-4 a B\right )}{(a+b \sec (c+d x))^{3/2}}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 A b-4 a B}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{4 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2-12 b B a+6 A b \sec (c+d x) a+15 A b^2-b (5 A b-4 a B) \sec ^2(c+d x)}{2 (a+b \sec (c+d x))^{3/2}}dx}{a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2-12 b B a+6 A b \sec (c+d x) a+15 A b^2-b (5 A b-4 a B) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {4 (A+2 C) a^2-12 b B a+6 A b \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^2-b (5 A b-4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {-b \left (4 B a^3-(7 A b-8 b C) a^2-12 b^2 B a+15 A b^3\right ) \sec ^2(c+d x)-2 a b \left (-\left ((A-4 C) a^2\right )-4 b B a+5 A b^2\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2-12 b B a+15 A b^2\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {-b \left (4 B a^3-(7 A b-8 b C) a^2-12 b^2 B a+15 A b^3\right ) \sec ^2(c+d x)-2 a b \left (-\left ((A-4 C) a^2\right )-4 b B a+5 A b^2\right ) \sec (c+d x)+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2-12 b B a+15 A b^2\right )}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {-b \left (4 B a^3-(7 A b-8 b C) a^2-12 b^2 B a+15 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b \left (-\left ((A-4 C) a^2\right )-4 b B a+5 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+\left (a^2-b^2\right ) \left (4 (A+2 C) a^2-12 b B a+15 A b^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\left (a^2-b^2\right ) \left (4 (A+2 C) a^2-12 b B a+15 A b^2\right )+\left (b \left (4 B a^3-(7 A b-8 b C) a^2-12 b^2 B a+15 A b^3\right )-2 a b \left (-\left ((A-4 C) a^2\right )-4 b B a+5 A b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\left (a^2-b^2\right ) \left (4 (A+2 C) a^2-12 b B a+15 A b^2\right )+\left (b \left (4 B a^3-(7 A b-8 b C) a^2-12 b^2 B a+15 A b^3\right )-2 a b \left (-\left ((A-4 C) a^2\right )-4 b B a+5 A b^2\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) \left (-2 a^2 (A+2 B-4 C)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (-2 a^2 (A+2 B-4 C)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b (a-b) \left (-2 a^2 (A+2 B-4 C)+a (5 A b-12 b B)+15 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a (5 A b-12 b B)+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {(5 A b-4 a B) \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-2 a^2 (A+2 B-4 C)+a (5 A b-12 b B)+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \left (4 a^2 (A+2 C)-12 a b B+15 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}+\frac {2 b \tan (c+d x) \left (4 a^3 B-a^2 (7 A b-8 b C)-12 a b^2 B+15 A b^3\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}}{4 a}\) |
Input:
Int[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(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]*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Cot[c + d*x ]*EllipticE[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 + a*(5*A*b - 12*b*B) - 2*a^2 *(A + 2*B - 4*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/S qrt[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)*(15*A*b^2 - 12*a*b*B + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[S qrt[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*(15*A*b^3 + 4*a^3*B - 12*a*b^2*B - a^2*(7*A*b - 8*b*C))*Tan [c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(2*a))/(4*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(2898\) vs. \(2(507)=1014\).
Time = 15.68 (sec) , antiderivative size = 2899, normalized size of antiderivative = 5.25
Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, method=_RETURNVERBOSE)
Output:
-1/4/d/a^3/(a+b)/(a-b)*((16*cos(d*x+c)^2+32*cos(d*x+c)+16)*C*(1/(a+b)*(b+a *cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4*E llipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(8*cos(d*x+c)^2+1 6*cos(d*x+c)+8)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x +c)/(cos(d*x+c)+1))^(1/2)*a^4*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/ (a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)*(5*cos(d*x+c)-2)*A*a^3*b+sin(d*x+c)*co s(d*x+c)*(2*cos(d*x+c)^2+2*cos(d*x+c)+7)*A*a^2*b^2+sin(d*x+c)*cos(d*x+c)*( -5*cos(d*x+c)+10)*a*A*b^3+sin(d*x+c)*cos(d*x+c)*(4*cos(d*x+c)-8)*B*a^2*b^2 +(15*cos(d*x+c)^2+30*cos(d*x+c)+15)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c )+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*EllipticE(-csc(d*x+c)+co t(d*x+c),((a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*B*(1/(a+b)*( b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^ 4*EllipticE(-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*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+ c)/(cos(d*x+c)+1))^(1/2)*a^4*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)*C*(1/(a+b)*(b+a*cos(d*x+c))/(co s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4*EllipticF(-csc(d* x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-30*cos(d*x+c)^2-60*cos(d*x+c)-30)*A *(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*b^4*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))-...
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + 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*s ec(d*x + c) + a^2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cos(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))* *(3/2),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a + b*s ec(c + d*x))**(3/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^2/(b*sec(d* x + c) + a)^(3/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^2/(b*sec(d* x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((cos(c + d*x)^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(3/2),x)
Output:
int((cos(c + d*x)^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(3/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) a \] Input:
int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(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)**2)/(sec(c + d* x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*c + int((sqrt(sec(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x))/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a* b + a**2),x)*b + int((sqrt(sec(c + d*x)*b + a)*cos(c + d*x)**2)/(sec(c + d *x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*a