Integrand size = 43, antiderivative size = 236 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}} \]
-2/5*(5*A*b^2-5*B*a*b+5*C*a^2+3*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^3/d+2/3*(B*b-C*a)*(co s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c), 2^(1/2))/b^2/d-2*a*(A*b^2-a*(B*b-C*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/ 2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/b^3/(a+b)/d+ 2/5*C*sin(d*x+c)/b/d/cos(d*x+c)^(5/2)+2/3*(B*b-C*a)*sin(d*x+c)/b^2/d/cos(d *x+c)^(3/2)+2/5*(5*A*b^2-5*B*a*b+5*C*a^2+3*C*b^2)*sin(d*x+c)/b^3/d/cos(d*x +c)^(1/2)
Time = 5.96 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {\frac {2 \left (-45 a^2 b B-10 b^3 B+45 a^3 C+a b^2 (45 A+19 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {2 b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {6 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}-\frac {2 \left (10 b (b B-a C) \sin (c+d x)+3 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (2 (c+d x))+6 b^2 C \tan (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}}{30 b^3 d} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])),x]
-1/30*((2*(-45*a^2*b*B - 10*b^3*B + 45*a^3*C + a*b^2*(45*A + 19*C))*Ellipt icPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (2*b*(15*A*b^2 - 20*a*b*B + 20*a^2*C + 9*b^2*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/ (a + b), (c + d*x)/2, 2])/(a + b)))/a + (6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*E llipticF[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*b*Sqrt[Sin[c + d*x]^2] ) - (2*(10*b*(b*B - a*C)*Sin[c + d*x] + 3*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3 *b^2*C)*Sin[2*(c + d*x)] + 6*b^2*C*Tan[c + d*x]))/Cos[c + d*x]^(3/2))/(b^3 *d)
Time = 2.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4600, 3042, 3534, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{5/2} (a+b \sec (c+d x))}dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {2 \int \frac {3 a C \cos ^2(c+d x)+b (5 A+3 C) \cos (c+d x)+5 (b B-a C)}{2 \cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))}dx}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 a C \cos ^2(c+d x)+b (5 A+3 C) \cos (c+d x)+5 (b B-a C)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))}dx}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 (b B-a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \int \frac {5 a (b B-a C) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {5 a (b B-a C) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {5 a (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {\frac {2 \int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \cos (c+d x)+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \cos (c+d x)+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {\frac {-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (b (b B-a C) \cos (c+d x) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \cos (c+d x) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \left (a b (b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \left (a b (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {\frac {5 \left (\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}+\frac {\frac {5 \left (\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*C*Sin[c + d*x])/(5*b*d*Cos[c + d*x]^(5/2)) + ((10*(b*B - a*C)*Sin[c + d *x])/(3*b*d*Cos[c + d*x]^(3/2)) + (((-6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b ^2*C)*EllipticE[(c + d*x)/2, 2])/d + (5*((2*a*b*(b*B - a*C)*EllipticF[(c + d*x)/2, 2])/d - (6*a^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/((a + b)*d)))/a)/b + (6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3* b^2*C)*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(3*b))/(5*b)
3.14.23.3.1 Defintions of rubi rules used
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[(-(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])))
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. \(772\) vs. \(2(298)=596\).
Time = 4.29 (sec) , antiderivative size = 773, normalized size of antiderivative = 3.28
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x,me thod=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*C/b/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(co s(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)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c )^4+12*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)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c) ^2*cos(1/2*d*x+1/2*c)-3*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))*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(B*b-C*a)/b^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/ 2)^2+1/3*(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)))+2*(A*b^2-B*a*b+C*a^2)/b^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2* d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-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))+2*(A*b^2- B*a*b+C*a^2)*a^2/b^3/(a^2-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)*E llipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*...
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c) ),x, algorithm="fricas")
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c) ),x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c) ),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*co s(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )} \,d x \]