Integrand size = 41, antiderivative size = 285 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b^3 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^5 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{8 a^5 d}-\frac {\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 d}+\frac {\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac {(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac {A \sec ^3(c+d x) \tan (c+d x)}{4 a d} \]
1/8*(8*A*b^4-4*B*a^3*b-8*B*a*b^3+4*a^2*b^2*(A+2*C)+a^4*(3*A+4*C))*arctanh( sin(d*x+c))/a^5/d-2*b^3*(A*b^2-a*(B*b-C*a))*arctan((a-b)^(1/2)*tan(1/2*d*x +1/2*c)/(a+b)^(1/2))/a^5/d/(a-b)^(1/2)/(a+b)^(1/2)-1/3*(3*A*b^3-2*B*a^3-3* B*a*b^2+a^2*b*(2*A+3*C))*tan(d*x+c)/a^4/d+1/8*(4*A*b^2-4*B*a*b+a^2*(3*A+4* C))*sec(d*x+c)*tan(d*x+c)/a^3/d-1/3*(A*b-B*a)*sec(d*x+c)^2*tan(d*x+c)/a^2/ d+1/4*A*sec(d*x+c)^3*tan(d*x+c)/a/d
Time = 2.62 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.42 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {96 b^3 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-6 \left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+a \left (21 a^3 A+12 a A b^2-12 a^2 b B+12 a^3 C+4 \left (-9 A b^3+10 a^3 B+9 a b^2 B-a^2 b (10 A+9 C)\right ) \cos (c+d x)+3 a \left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \cos (2 (c+d x))-8 a^2 A b \cos (3 (c+d x))-12 A b^3 \cos (3 (c+d x))+8 a^3 B \cos (3 (c+d x))+12 a b^2 B \cos (3 (c+d x))-12 a^2 b C \cos (3 (c+d x))\right ) \sec ^3(c+d x) \tan (c+d x)}{48 a^5 d} \]
((96*b^3*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqr t[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 6*(8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4* a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/ 2]] + 6*(8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + a*(21*a^3*A + 12*a*A*b^ 2 - 12*a^2*b*B + 12*a^3*C + 4*(-9*A*b^3 + 10*a^3*B + 9*a*b^2*B - a^2*b*(10 *A + 9*C))*Cos[c + d*x] + 3*a*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Cos[2* (c + d*x)] - 8*a^2*A*b*Cos[3*(c + d*x)] - 12*A*b^3*Cos[3*(c + d*x)] + 8*a^ 3*B*Cos[3*(c + d*x)] + 12*a*b^2*B*Cos[3*(c + d*x)] - 12*a^2*b*C*Cos[3*(c + d*x)])*Sec[c + d*x]^3*Tan[c + d*x])/(48*a^5*d)
Time = 2.21 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.09, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.439, Rules used = {3042, 3534, 25, 3042, 3534, 25, 3042, 3534, 25, 3042, 3534, 27, 3042, 3480, 3042, 3138, 218, 4257}
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.
| \(\displaystyle \int \frac {\sec ^5(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\int -\frac {\left (-3 A b \cos ^2(c+d x)-a (3 A+4 C) \cos (c+d x)+4 (A b-a B)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\int \frac {\left (-3 A b \cos ^2(c+d x)-a (3 A+4 C) \cos (c+d x)+4 (A b-a B)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\int \frac {-3 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{4 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {\int -\frac {\left (-8 b (A b-a B) \cos ^2(c+d x)+a (A b+8 a B) \cos (c+d x)+3 \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-8 b (A b-a B) \cos ^2(c+d x)+a (A b+8 a B) \cos (c+d x)+3 \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right )\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-8 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (A b+8 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {\left (-3 b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos ^2(c+d x)+a \left (-3 (3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos (c+d x)+8 \left (-2 B a^3+b (2 A+3 C) a^2-3 b^2 B a+3 A b^3\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\int \frac {\left (-3 b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos ^2(c+d x)+a \left (-3 (3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos (c+d x)+8 \left (-2 B a^3+b (2 A+3 C) a^2-3 b^2 B a+3 A b^3\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\int \frac {-3 b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a \left (-3 (3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+8 \left (-2 B a^3+b (2 A+3 C) a^2-3 b^2 B a+3 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {\int -\frac {3 \left ((3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos (c+d x) a+8 A b^4\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \int \frac {\left ((3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \cos (c+d x) a+8 A b^4\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \int \frac {(3 A+4 C) a^4-4 b B a^3+4 b^2 (A+2 C) a^2-8 b^3 B a+b \left ((3 A+4 C) a^2-4 b B a+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^4}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {\left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right ) \int \sec (c+d x)dx}{a}-\frac {8 b^3 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {\left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {8 b^3 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {\left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {16 b^3 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {\left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {16 b^3 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^3(c+d x)}{4 a d}-\frac {\frac {4 (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 \tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{2 a d}-\frac {\frac {8 \tan (c+d x) \left (-2 a^3 B+a^2 b (2 A+3 C)-3 a b^2 B+3 A b^3\right )}{a d}-\frac {3 \left (\frac {\left (a^4 (3 A+4 C)-4 a^3 b B+4 a^2 b^2 (A+2 C)-8 a b^3 B+8 A b^4\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {16 b^3 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}}{4 a}\) |
(A*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d) - ((4*(A*b - a*B)*Sec[c + d*x]^2*T an[c + d*x])/(3*a*d) - ((3*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Sec[c + d *x]*Tan[c + d*x])/(2*a*d) - ((-3*((-16*b^3*(A*b^2 - a*(b*B - a*C))*ArcTan[ (Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4* C))*ArcTanh[Sin[c + d*x]])/(a*d)))/a + (8*(3*A*b^3 - 2*a^3*B - 3*a*b^2*B + a^2*b*(2*A + 3*C))*Tan[c + d*x])/(a*d))/(2*a))/(3*a))/(4*a)
3.10.85.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, 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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(565\) vs. \(2(266)=532\).
Time = 0.78 (sec) , antiderivative size = 566, normalized size of antiderivative = 1.99
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {-3 a A -2 A b +2 B a}{6 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7 A \,a^{2}+4 a A b +4 A \,b^{2}-4 B \,a^{2}-4 B a b +4 a^{2} C}{8 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (3 A \,a^{4}+4 A \,a^{2} b^{2}+8 A \,b^{4}-4 B \,a^{3} b -8 B a \,b^{3}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a^{5}}-\frac {-5 A \,a^{3}-8 A \,a^{2} b -4 a A \,b^{2}-8 A \,b^{3}+8 B \,a^{3}+4 B \,a^{2} b +8 B a \,b^{2}-4 a^{3} C -8 a^{2} b C}{8 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {-3 a A -2 A b +2 B a}{6 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-7 A \,a^{2}-4 a A b -4 A \,b^{2}+4 B \,a^{2}+4 B a b -4 a^{2} C}{8 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-3 A \,a^{4}-4 A \,a^{2} b^{2}-8 A \,b^{4}+4 B \,a^{3} b +8 B a \,b^{3}-4 a^{4} C -8 C \,a^{2} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a^{5}}-\frac {-5 A \,a^{3}-8 A \,a^{2} b -4 a A \,b^{2}-8 A \,b^{3}+8 B \,a^{3}+4 B \,a^{2} b +8 B a \,b^{2}-4 a^{3} C -8 a^{2} b C}{8 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b^{3} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(566\) |
| default | \(\frac {-\frac {A}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {-3 a A -2 A b +2 B a}{6 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7 A \,a^{2}+4 a A b +4 A \,b^{2}-4 B \,a^{2}-4 B a b +4 a^{2} C}{8 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (3 A \,a^{4}+4 A \,a^{2} b^{2}+8 A \,b^{4}-4 B \,a^{3} b -8 B a \,b^{3}+4 a^{4} C +8 C \,a^{2} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a^{5}}-\frac {-5 A \,a^{3}-8 A \,a^{2} b -4 a A \,b^{2}-8 A \,b^{3}+8 B \,a^{3}+4 B \,a^{2} b +8 B a \,b^{2}-4 a^{3} C -8 a^{2} b C}{8 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{4 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {-3 a A -2 A b +2 B a}{6 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {-7 A \,a^{2}-4 a A b -4 A \,b^{2}+4 B \,a^{2}+4 B a b -4 a^{2} C}{8 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-3 A \,a^{4}-4 A \,a^{2} b^{2}-8 A \,b^{4}+4 B \,a^{3} b +8 B a \,b^{3}-4 a^{4} C -8 C \,a^{2} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a^{5}}-\frac {-5 A \,a^{3}-8 A \,a^{2} b -4 a A \,b^{2}-8 A \,b^{3}+8 B \,a^{3}+4 B \,a^{2} b +8 B a \,b^{2}-4 a^{3} C -8 a^{2} b C}{8 a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b^{3} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(566\) |
| risch | \(\text {Expression too large to display}\) | \(1293\) |
1/d*(-1/4*A/a/(tan(1/2*d*x+1/2*c)+1)^4-1/6*(-3*A*a-2*A*b+2*B*a)/a^2/(tan(1 /2*d*x+1/2*c)+1)^3-1/8*(7*A*a^2+4*A*a*b+4*A*b^2-4*B*a^2-4*B*a*b+4*C*a^2)/a ^3/(tan(1/2*d*x+1/2*c)+1)^2+1/8*(3*A*a^4+4*A*a^2*b^2+8*A*b^4-4*B*a^3*b-8*B *a*b^3+4*C*a^4+8*C*a^2*b^2)/a^5*ln(tan(1/2*d*x+1/2*c)+1)-1/8*(-5*A*a^3-8*A *a^2*b-4*A*a*b^2-8*A*b^3+8*B*a^3+4*B*a^2*b+8*B*a*b^2-4*C*a^3-8*C*a^2*b)/a^ 4/(tan(1/2*d*x+1/2*c)+1)+1/4*A/a/(tan(1/2*d*x+1/2*c)-1)^4-1/6*(-3*A*a-2*A* b+2*B*a)/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/8*(-7*A*a^2-4*A*a*b-4*A*b^2+4*B*a^ 2+4*B*a*b-4*C*a^2)/a^3/(tan(1/2*d*x+1/2*c)-1)^2+1/8/a^5*(-3*A*a^4-4*A*a^2* b^2-8*A*b^4+4*B*a^3*b+8*B*a*b^3-4*C*a^4-8*C*a^2*b^2)*ln(tan(1/2*d*x+1/2*c) -1)-1/8*(-5*A*a^3-8*A*a^2*b-4*A*a*b^2-8*A*b^3+8*B*a^3+4*B*a^2*b+8*B*a*b^2- 4*C*a^3-8*C*a^2*b)/a^4/(tan(1/2*d*x+1/2*c)-1)-2*b^3*(A*b^2-B*a*b+C*a^2)/a^ 5/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)) )
Time = 28.22 (sec) , antiderivative size = 993, normalized size of antiderivative = 3.48 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="fricas")
[-1/48*(24*(C*a^2*b^3 - B*a*b^4 + A*b^5)*sqrt(-a^2 + b^2)*cos(d*x + c)^4*l og((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2) *(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2* a*b*cos(d*x + c) + a^2)) - 3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A + 4*C)*a^4* b^2 - 4*B*a^3*b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*cos(d*x + c )^4*log(sin(d*x + c) + 1) + 3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A + 4*C)*a^4 *b^2 - 4*B*a^3*b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 2*(6*A*a^6 - 6*A*a^4*b^2 + 8*(2*B*a^6 - (2*A + 3*C)*a^5*b + B*a^4*b^2 - (A - 3*C)*a^3*b^3 - 3*B*a^2*b^4 + 3*A*a*b^5)*c os(d*x + c)^3 + 3*((3*A + 4*C)*a^6 - 4*B*a^5*b + (A - 4*C)*a^4*b^2 + 4*B*a ^3*b^3 - 4*A*a^2*b^4)*cos(d*x + c)^2 + 8*(B*a^6 - A*a^5*b - B*a^4*b^2 + A* a^3*b^3)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d*cos(d*x + c)^4), - 1/48*(48*(C*a^2*b^3 - B*a*b^4 + A*b^5)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^4 - 3*((3*A + 4*C)* a^6 - 4*B*a^5*b + (A + 4*C)*a^4*b^2 - 4*B*a^3*b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + 3*((3*A + 4*C) *a^6 - 4*B*a^5*b + (A + 4*C)*a^4*b^2 - 4*B*a^3*b^3 + 4*(A - 2*C)*a^2*b^4 + 8*B*a*b^5 - 8*A*b^6)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 2*(6*A*a^6 - 6*A*a^4*b^2 + 8*(2*B*a^6 - (2*A + 3*C)*a^5*b + B*a^4*b^2 - (A - 3*C)*a^3* b^3 - 3*B*a^2*b^4 + 3*A*a*b^5)*cos(d*x + c)^3 + 3*((3*A + 4*C)*a^6 - 4*...
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="maxima")
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 878 vs. \(2 (266) = 532\).
Time = 0.36 (sec) , antiderivative size = 878, normalized size of antiderivative = 3.08 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
1/24*(3*(3*A*a^4 + 4*C*a^4 - 4*B*a^3*b + 4*A*a^2*b^2 + 8*C*a^2*b^2 - 8*B*a *b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 3*(3*A*a^4 + 4*C* a^4 - 4*B*a^3*b + 4*A*a^2*b^2 + 8*C*a^2*b^2 - 8*B*a*b^3 + 8*A*b^4)*log(abs (tan(1/2*d*x + 1/2*c) - 1))/a^5 + 48*(C*a^2*b^3 - B*a*b^4 + A*b^5)*(pi*flo or(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2* c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^5) + 2*( 15*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a ^3*tan(1/2*d*x + 1/2*c)^7 + 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*B*a^2*b *tan(1/2*d*x + 1/2*c)^7 + 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*t an(1/2*d*x + 1/2*c)^7 - 24*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^3*tan(1 /2*d*x + 1/2*c)^7 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^2*b*tan(1/2*d*x + 1/ 2*c)^5 + 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b*tan(1/2*d*x + 1/2* c)^5 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*a*b^2*tan(1/2*d*x + 1/2*c) ^5 - 72*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 40 *B*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*A*a^2 *b*tan(1/2*d*x + 1/2*c)^3 + 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b *tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*a*b^2*t an(1/2*d*x + 1/2*c)^3 + 72*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^3*tan(1/2 *d*x + 1/2*c) + 24*B*a^3*tan(1/2*d*x + 1/2*c) + 12*C*a^3*tan(1/2*d*x + ...
Time = 12.55 (sec) , antiderivative size = 9543, normalized size of antiderivative = 33.48 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^7*(5*A*a^3 + 8*A*b^3 - 8*B*a^3 + 4*C*a^3 + 4*A*a*b^2 + 8*A*a^2*b - 8*B*a*b^2 - 4*B*a^2*b + 8*C*a^2*b))/(4*a^4) + (tan(c/2 + (d* x)/2)^3*(9*A*a^3 + 72*A*b^3 - 40*B*a^3 - 12*C*a^3 - 12*A*a*b^2 + 40*A*a^2* b - 72*B*a*b^2 + 12*B*a^2*b + 72*C*a^2*b))/(12*a^4) - (tan(c/2 + (d*x)/2)^ 5*(72*A*b^3 - 9*A*a^3 - 40*B*a^3 + 12*C*a^3 + 12*A*a*b^2 + 40*A*a^2*b - 72 *B*a*b^2 - 12*B*a^2*b + 72*C*a^2*b))/(12*a^4) + (tan(c/2 + (d*x)/2)*(5*A*a ^3 - 8*A*b^3 + 8*B*a^3 + 4*C*a^3 + 4*A*a*b^2 - 8*A*a^2*b + 8*B*a*b^2 - 4*B *a^2*b - 8*C*a^2*b))/(4*a^4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d* x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (atan((((( tan(c/2 + (d*x)/2)*(9*A^2*a^11 - 128*A^2*b^11 + 16*C^2*a^11 + 256*A^2*a*b^ 10 - 27*A^2*a^10*b - 48*C^2*a^10*b - 256*A^2*a^2*b^9 + 256*A^2*a^3*b^8 - 2 56*A^2*a^4*b^7 + 256*A^2*a^5*b^6 - 216*A^2*a^6*b^5 + 136*A^2*a^7*b^4 - 81* A^2*a^8*b^3 + 51*A^2*a^9*b^2 - 128*B^2*a^2*b^9 + 256*B^2*a^3*b^8 - 256*B^2 *a^4*b^7 + 256*B^2*a^5*b^6 - 208*B^2*a^6*b^5 + 112*B^2*a^7*b^4 - 48*B^2*a^ 8*b^3 + 16*B^2*a^9*b^2 - 128*C^2*a^4*b^7 + 256*C^2*a^5*b^6 - 256*C^2*a^6*b ^5 + 256*C^2*a^7*b^4 - 208*C^2*a^8*b^3 + 112*C^2*a^9*b^2 + 24*A*C*a^11 + 2 56*A*B*a*b^10 - 24*A*B*a^10*b - 72*A*C*a^10*b - 32*B*C*a^10*b - 512*A*B*a^ 2*b^9 + 512*A*B*a^3*b^8 - 512*A*B*a^4*b^7 + 464*A*B*a^5*b^6 - 368*A*B*a^6* b^5 + 264*A*B*a^7*b^4 - 152*A*B*a^8*b^3 + 72*A*B*a^9*b^2 - 256*A*C*a^2*b^9 + 512*A*C*a^3*b^8 - 512*A*C*a^4*b^7 + 512*A*C*a^5*b^6 - 464*A*C*a^6*b^...