Integrand size = 43, antiderivative size = 510 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^5 \sqrt {a+b} d}+\frac {2 \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]
2/15*(40*B*a^3*b-25*B*a*b^3-6*a^2*b^2*(5*A-4*C)-48*a^4*C+3*b^4*(5*A+3*C))* 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)/b^5/d/(a +b)^(1/2)+2/15*(a^2*b*(40*B-36*C)-48*a^3*C-6*a*b^2*(5*A-5*B+2*C)-b^3*(15*A -5*B+9*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)/b^4/d/(a+b)^(1/2)-2*(A*b^2-a*(B*b-C*a))*sec(d*x+c)^2*tan(d*x+c)/b/(a^2- b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/15*(20*B*a^2*b-5*B*b^3-3*a*b^2*(5*A-3*C)-2 4*a^3*C)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A*b^2-5* B*a*b+6*C*a^2-C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2 -b^2)/d
Time = 24.40 (sec) , antiderivative size = 887, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^3(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 {4 (b+a \cos (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left ((a+b) \left (-40 a^3 b B+25 a b^3 B+6 a^2 b^2 (5 A-4 C)+48 a^4 C-3 b^4 (5 A+3 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) \left (-48 a^3 C-6 a b^2 (5 A+5 B+2 C)+b^3 (15 A+5 B+9 C)+4 a^2 b (10 B+9 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (-40 a^3 b B+25 a b^3 B+6 a^2 b^2 (5 A-4 C)+48 a^4 C-3 b^4 (5 A+3 C)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{15 b^4 \left (-a^2+b^2\right ) d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {(b+a \cos (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (-30 a^2 A b^2+15 A b^4+40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \sin (c+d x)}{15 b^4 \left (-a^2+b^2\right )}+\frac {4 \sec (c+d x) (5 b B \sin (c+d x)-9 a C \sin (c+d x))}{15 b^3}+\frac {4 \left (a^2 A b^2 \sin (c+d x)-a^3 b B \sin (c+d x)+a^4 C \sin (c+d x)\right )}{b^3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {4 C \sec (c+d x) \tan (c+d x)}{5 b^2}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{3/2}} \]
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x])^(3/2),x]
(4*(b + a*Cos[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt [(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a + b)*(-40*a^3*b*B + 25*a*b^3*B + 6*a^2 *b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(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) ] + b*(a + b)*(-48*a^3*C - 6*a*b^2*(5*A + 5*B + 2*C) + b^3*(15*A + 5*B + 9 *C) + 4*a^2*b*(10*B + 9*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(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)] + (-40*a^3*b*B + 25 *a*b^3*B + 6*a^2*b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Ta n[(c + d*x)/2]^2))))/(15*b^4*(-a^2 + b^2)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(1 + Tan [(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x )/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + ((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(-30*a^2*A*b^2 + 15*A*b^4 + 40*a^3*b*B - 25 *a*b^3*B - 48*a^4*C + 24*a^2*b^2*C + 9*b^4*C)*Sin[c + d*x])/(15*b^4*(-a^2 + b^2)) + (4*Sec[c + d*x]*(5*b*B*Sin[c + d*x] - 9*a*C*Sin[c + d*x]))/(15*b ^3) + (4*(a^2*A*b^2*Sin[c + d*x] - a^3*b*B*Sin[c + d*x] + a^4*C*Sin[c + d* x]))/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (4*C*Sec[c + d*x]*Tan[c ...
Time = 2.32 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4586, 27, 3042, 4580, 27, 3042, 4570, 27, 3042, 4493, 3042, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4586 |
| \(\displaystyle -\frac {2 \int \frac {\sec ^2(c+d x) \left (-\left (\left (6 C a^2-5 b B a+5 A b^2-b^2 C\right ) \sec ^2(c+d x)\right )+b (b B-a (A+C)) \sec (c+d x)+4 \left (A b^2-a (b B-a C)\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (6 C a^2-5 b B a+5 A b^2-b^2 C\right ) \sec ^2(c+d x)\right )+b (b B-a (A+C)) \sec (c+d x)+4 \left (A b^2-a (b B-a C)\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-6 C a^2+5 b B a-5 A b^2+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (b B-a (A+C)) \csc \left (c+d x+\frac {\pi }{2}\right )+4 \left (A b^2-a (b B-a C)\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \sec ^2(c+d x)-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \sec ^2(c+d x)-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle -\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}}{5 b}}{b \left (a^2-b^2\right )}\) |
(-2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d* Sqrt[a + b*Sec[c + d*x]]) - ((-2*(5*A*b^2 - 5*a*b*B + 6*a^2*C - b^2*C)*Sec [c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) - (-1/3*((-2*(a - b)*Sqrt[a + b]*(40*a^3*b*B - 25*a*b^3*B - 6*a^2*b^2*(5*A - 4*C) - 48*a^4* C + 3*b^4*(5*A + 3*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)]*Sq rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(a ^2*b*(40*B - 36*C) - 48*a^3*C - 6*a*b^2*(5*A - 5*B + 2*C) - b^3*(15*A - 5* B + 9*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/b + (2*(20*a^2*b*B - 5*b^3*B - 3*a*b^2*(5 *A - 3*C) - 24*a^3*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b ))/(b*(a^2 - b^2))
3.10.64.3.1 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[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_)]*(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[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B 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} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, 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[(-d)*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1)) ), x] + Simp[d/(b*(a^2 - b^2)*(m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*( d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) + b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C }, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(7392\) vs. \(2(478)=956\).
Time = 39.37 (sec) , antiderivative size = 7393, normalized size of antiderivative = 14.50
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(7393\) |
| default | \(\text {Expression too large to display}\) | \(7452\) |
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, method=_RETURNVERBOSE)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="fricas")
integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*s ec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*s ec(c + d*x))**(3/2), x)
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="maxima")
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d* x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]