Integrand size = 34, antiderivative size = 446 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b \sqrt {a+b} \left (a^2-b^2\right )^2 d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \]
-2/15*(23*B*a^2*b+9*B*b^3-3*C*a^3-29*C*a*b^2)*cot(d*x+c)*EllipticE((a+b*se c(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-b)^2/b^2/(a+b)^(5/2)/d+2/15*(3*a^ 2*(5*B+C)-8*a*b*(B+3*C)+b^2*(9*B+5*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/(a^2-b^2)^2/d/(a+b)^(1/2)-2/5*(B*b-C*a)*t an(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(5/2)-2/15*(8*B*a*b-3*C*a^2-5*C*b^2 )*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(3/2)-2/15*(23*B*a^2*b+9*B*b^3 -3*C*a^3-29*C*a*b^2)*tan(d*x+c)/(a^2-b^2)^3/d/(a+b*sec(d*x+c))^(1/2)
Time = 22.17 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.62 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {(b+a \cos (c+d x))^4 \sec ^4(c+d x) \left (-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \sin (c+d x)}{15 b \left (-a^2+b^2\right )^3}-\frac {2 \left (b^3 B \sin (c+d x)-a b^2 C \sin (c+d x)\right )}{5 a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))^3}-\frac {2 \left (-14 a^2 b^2 B \sin (c+d x)+6 b^4 B \sin (c+d x)+9 a^3 b C \sin (c+d x)-a b^3 C \sin (c+d x)\right )}{15 a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {2 \left (-34 a^4 b B \sin (c+d x)+5 a^2 b^3 B \sin (c+d x)-3 b^5 B \sin (c+d x)+9 a^5 C \sin (c+d x)+25 a^3 b^2 C \sin (c+d x)-2 a b^4 C \sin (c+d x)\right )}{15 a^2 \left (a^2-b^2\right )^3 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{7/2}}-\frac {2 (b+a \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-23 a^2 b B-9 b^3 B+3 a^3 C+29 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (b^2 (9 B-5 C)+8 a b (B-3 C)+3 a^2 (5 B-C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-23 a^2 b B-9 b^3 B+3 a^3 C+29 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b \left (-a^2+b^2\right )^3 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{7/2}} \]
((b + a*Cos[c + d*x])^4*Sec[c + d*x]^4*((-2*(23*a^2*b*B + 9*b^3*B - 3*a^3* C - 29*a*b^2*C)*Sin[c + d*x])/(15*b*(-a^2 + b^2)^3) - (2*(b^3*B*Sin[c + d* x] - a*b^2*C*Sin[c + d*x]))/(5*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^3) - ( 2*(-14*a^2*b^2*B*Sin[c + d*x] + 6*b^4*B*Sin[c + d*x] + 9*a^3*b*C*Sin[c + d *x] - a*b^3*C*Sin[c + d*x]))/(15*a^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (2*(-34*a^4*b*B*Sin[c + d*x] + 5*a^2*b^3*B*Sin[c + d*x] - 3*b^5*B*Sin[c + d*x] + 9*a^5*C*Sin[c + d*x] + 25*a^3*b^2*C*Sin[c + d*x] - 2*a*b^4*C*Sin [c + d*x]))/(15*a^2*(a^2 - b^2)^3*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(7/2)) - (2*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(7/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29 *a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/ ((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/ (a + b)] + 2*b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C ))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b )*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b) ] + (-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos [c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b*(-a^2 + b^2)^3*d*Sq rt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(7/2))
Time = 1.76 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {3042, 4548, 27, 3042, 4548, 27, 3042, 4548, 27, 3042, 4546, 27, 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.
| \(\displaystyle \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle -\frac {2 \int -\frac {5 a (a B-b C) \sec (c+d x)-3 a (b B-a C) \sec ^2(c+d x)}{2 (a+b \sec (c+d x))^{5/2}}dx}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 a (a B-b C) \sec (c+d x)-3 a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}}dx}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a (a B-b C) \csc \left (c+d x+\frac {\pi }{2}\right )-3 a (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {3 a^2 \left (5 B a^2-8 b C a+3 b^2 B\right ) \sec (c+d x)-a^2 \left (-3 C a^2+8 b B a-5 b^2 C\right ) \sec ^2(c+d x)}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 \left (5 B a^2-8 b C a+3 b^2 B\right ) \sec (c+d x)-a^2 \left (-3 C a^2+8 b B a-5 b^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 a^2 \left (5 B a^2-8 b C a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )-a^2 \left (-3 C a^2+8 b B a-5 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {\frac {-\frac {2 \int -\frac {\left (-3 C a^3+23 b B a^2-29 b^2 C a+9 b^3 B\right ) \sec ^2(c+d x) a^3+\left (15 B a^3-27 b C a^2+17 b^2 B a-5 b^3 C\right ) \sec (c+d x) a^3}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (-3 C a^3+23 b B a^2-29 b^2 C a+9 b^3 B\right ) \sec ^2(c+d x) a^3+\left (15 B a^3-27 b C a^2+17 b^2 B a-5 b^3 C\right ) \sec (c+d x) a^3}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (-3 C a^3+23 b B a^2-29 b^2 C a+9 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 a^3+\left (15 B a^3-27 b C a^2+17 b^2 B a-5 b^3 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {\frac {\frac {a^3 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\left (a^3 \left (15 B a^3-27 b C a^2+17 b^2 B a-5 b^3 C\right )-a^3 \left (-3 C a^3+23 b B a^2-29 b^2 C a+9 b^3 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {a^3 (a-b) \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+a^3 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^3 (a-b) \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a^3 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {a^3 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^3 (a-b) \sqrt {a+b} \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 a^3 (a-b) \sqrt {a+b} \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 a^3 (a-b) \sqrt {a+b} \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a \left (a^2-b^2\right )}-\frac {2 a^2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 a \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\) |
(-2*(b*B - a*C)*Tan[c + d*x])/(5*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(5/2)) + ((-2*a*(8*a*b*B - 3*a^2*C - 5*b^2*C)*Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (((-2*a^3*(a - b)*Sqrt[a + b]*(23*a^2*b*B + 9*b ^3*B - 3*a^3*C - 29*a*b^2*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^2*d) + (2*a^3*(a - b)*Sqrt [a + b]*(3*a^2*(5*B + C) - 8*a*b*(B + 3*C) + b^2*(9*B + 5*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))/(a*(a^2 - b^2)) - (2*a^2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a *b^2*C)*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*a*(a^2 - b^2)))/(5*a*(a^2 - b^2))
3.9.54.3.1 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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(9223\) vs. \(2(412)=824\).
Time = 17.39 (sec) , antiderivative size = 9224, normalized size of antiderivative = 20.68
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(9224\) |
| default | \(\text {Expression too large to display}\) | \(9320\) |
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^4 *sec(d*x + c)^4 + 4*a*b^3*sec(d*x + c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^ 3*b*sec(d*x + c) + a^4), x)
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]