Integrand size = 43, antiderivative size = 346 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^4 \left (a^2-b^2\right ) d}+\frac {b \left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^4 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \] Output:
(5*A*b^3+2*B*a^3-3*B*a*b^2-a^2*b*(4*A-C))*EllipticE(sin(1/2*d*x+1/2*c),2^( 1/2))/a^3/(a^2-b^2)/d-1/3*(15*A*b^4+12*B*a^3*b-9*B*a*b^3-a^2*b^2*(16*A-3*C )-2*a^4*(A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^4/(a^2-b^2)/d+b* (5*A*b^4+5*B*a^3*b-3*B*a*b^3-a^2*b^2*(7*A-C)-3*a^4*C)*EllipticPi(sin(1/2*d *x+1/2*c),2*a/(a+b),2^(1/2))/a^4/(a-b)/(a+b)^2/d-1/3*(5*A*b^2-3*B*a*b-a^2* (2*A-3*C))*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/(a^2-b^2)/d+(A*b^2-a*(B*b-C*a)) *cos(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))
Time = 2.88 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {4 \sqrt {\cos (c+d x)} \left (2 A+\frac {3 b \left (A b^2+a (-b B+a C)\right )}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (5 A b^3+6 a^3 B-3 a b^2 B-a^2 b (8 A+3 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (2 A b^2-3 a b B+a^2 (A+3 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (5 A b^3+2 a^3 B-3 a b^2 B+a^2 b (-4 A+C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{12 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
Output:
(4*Sqrt[Cos[c + d*x]]*(2*A + (3*b*(A*b^2 + a*(-(b*B) + a*C)))/((-a^2 + b^2 )*(b + a*Cos[c + d*x])))*Sin[c + d*x] - ((2*(5*A*b^3 + 6*a^3*B - 3*a*b^2*B - a^2*b*(8*A + 3*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(2*A*b^2 - 3*a*b*B + a^2*(A + 3*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + (6*(5*A*b^3 + 2 *a^3*B - 3*a*b^2*B + a^2*b*(-4*A + C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + ( a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d *x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(12*a^2*d)
Time = 2.36 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3528, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{(a+b \sec (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{(a \cos (c+d x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right ) \cos ^2(c+d x)\right )-2 a (A b+C b-a B) \cos (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{2 (b+a \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\left (\left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right ) \cos ^2(c+d x)\right )-2 a (A b+C b-a B) \cos (c+d x)+3 \left (A b^2-a (b B-a C)\right )\right )}{b+a \cos (c+d x)}dx}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\left ((2 A-3 C) a^2+3 b B a-5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (A b^2-a (b B-a C)\right )\right )}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {\frac {2 \int -\frac {-3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right ) \cos ^2(c+d x)-2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \cos (c+d x)+b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right ) \cos ^2(c+d x)-2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \cos (c+d x)+b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 \left (2 B a^3-b (4 A-C) a^2-3 b^2 B a+5 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a \left ((A+3 C) a^2-3 b B a+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {-\frac {-\frac {3 \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx}{a}-\frac {\int -\frac {a b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )+\left (-2 (A+3 C) a^4+12 b B a^3-b^2 (16 A-3 C) a^2-9 b^3 B a+15 A b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )+\left (-2 (A+3 C) a^4+12 b B a^3-b^2 (16 A-3 C) a^2-9 b^3 B a+15 A b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {3 \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx}{a}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )+\left (-2 (A+3 C) a^4+12 b B a^3-b^2 (16 A-3 C) a^2-9 b^3 B a+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {3 \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {a b \left (-\left ((2 A-3 C) a^2\right )-3 b B a+5 A b^2\right )+\left (-2 (A+3 C) a^4+12 b B a^3-b^2 (16 A-3 C) a^2-9 b^3 B a+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a d}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a d}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a d}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{a d}-\frac {3 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a d}}{3 a}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a d}-\frac {\frac {\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{a d}-\frac {6 b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a d}}{3 a}}{2 a \left (a^2-b^2\right )}\) |
Input:
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Se c[c + d*x])^2,x]
Output:
((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(a*(a^2 - b^2)*d *(b + a*Cos[c + d*x])) + (-1/3*((-6*(5*A*b^3 + 2*a^3*B - 3*a*b^2*B - a^2*b *(4*A - C))*EllipticE[(c + d*x)/2, 2])/(a*d) + ((2*(15*A*b^4 + 12*a^3*b*B - 9*a*b^3*B - a^2*b^2*(16*A - 3*C) - 2*a^4*(A + 3*C))*EllipticF[(c + d*x)/ 2, 2])/(a*d) - (6*b*(5*A*b^4 + 5*a^3*b*B - 3*a*b^3*B - a^2*b^2*(7*A - C) - 3*a^4*C)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d))/a)/a - (2*(5*A*b^2 - 3*a*b*B - a^2*(2*A - 3*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x]) /(3*a*d))/(2*a*(a^2 - b^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{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] && GtQ[m, 0] && LtQ[n, -1]
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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(1122\) vs. \(2(343)=686\).
Time = 8.84 (sec) , antiderivative size = 1123, normalized size of antiderivative = 3.25
Input:
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/3/a^4*(4*A*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2-2*A*cos(1/2*d*x+1/2*c)*sin(1/2* d*x+1/2*c)^2*a^2+A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/ 2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2+9*A*EllipticF(cos(1/2*d*x +1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*b^2+6*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b-6*B*EllipticF(cos(1/2*d*x+1/2*c ),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a *b-3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2+3*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2)/(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*b^2*(A*b^2-B*a*b+C*a^ 2)/a^4*(a^2/b/(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*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/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 *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*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)*(-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)...
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)*sec(d*x + c)^2 + B*cos(d*x + c)*sec(d*x + c) + A* cos(d*x + c))*sqrt(cos(d*x + c))/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+ c))**2,x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*se c(d*x + c) + a)^2, x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \] Input:
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co s(c + d*x))^2,x)
Output:
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co s(c + d*x))^2, x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) a \] Input:
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)
Output:
int((sqrt(cos(c + d*x))*cos(c + d*x)*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(cos(c + d*x))*cos(c + d*x) *sec(c + d*x))/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*b + i nt((sqrt(cos(c + d*x))*cos(c + d*x))/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x )*a*b + a**2),x)*a