Integrand size = 35, antiderivative size = 460 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]
-2/5*(2*a^2*b^2*(5*A-4*C)+16*a^4*C-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)-2/5*(16*a^3*C+ 12*a^2*b*C+2*a*b^2*(5*A+2*C)+b^3*(5*A+3*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+C*a^2)*sec( d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2/5*a*(5*A*b^2+8* C*a^2-3*C*b^2)*(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+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
Leaf count is larger than twice the leaf count of optimal. \(3853\) vs. \(2(460)=920\).
Time = 29.57 (sec) , antiderivative size = 3853, normalized size of antiderivative = 8.38 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((4*(-10*a^2*A*b^2 + 5*A*b^ 4 - 16*a^4*C + 8*a^2*b^2*C + 3*b^4*C)*Sin[c + d*x])/(5*b^4*(-a^2 + b^2)) + (4*(a^2*A*b^2*Sin[c + d*x] + a^4*C*Sin[c + d*x]))/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d*x])) - (12*a*C*Tan[c + d*x])/(5*b^3) + (4*C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^ (3/2)) + (4*(b + a*Cos[c + d*x])*((4*a^2*A)/(b*(-a^2 + b^2)*Sqrt[b + a*Cos [c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*A*b)/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (32*a^4*C)/(5*b^3*(-a^2 + b^2)*Sqrt[b + a*Cos[ c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^2*C)/(5*b*(-a^2 + b^2)*Sqrt[b + a*Co s[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*b*C)/(5*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (4*a*A*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sq rt[b + a*Cos[c + d*x]]) + (4*a^3*A*Sqrt[Sec[c + d*x]])/(b^2*(-a^2 + b^2)*S qrt[b + a*Cos[c + d*x]]) - (8*a*C*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)*Sqrt [b + a*Cos[c + d*x]]) + (32*a^5*C*Sqrt[Sec[c + d*x]])/(5*b^4*(-a^2 + b^2)* Sqrt[b + a*Cos[c + d*x]]) - (24*a^3*C*Sqrt[Sec[c + d*x]])/(5*b^2*(-a^2 + b ^2)*Sqrt[b + a*Cos[c + d*x]]) - (2*a*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]] )/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*A*Cos[2*(c + d*x)]*Sqrt [Sec[c + d*x]])/(b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (6*a*C*Cos[2 *(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (32*a^5*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*b^4*(-a^2 + b^2)*Sq...
Time = 2.04 (sec) , antiderivative size = 472, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4587, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4587 |
| \(\displaystyle -\frac {2 \int \frac {\sec ^2(c+d x) \left (-\left (\left (5 A b^2+\left (6 a^2-b^2\right ) C\right ) \sec ^2(c+d x)\right )-a b (A+C) \sec (c+d x)+4 \left (C a^2+A b^2\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (6 C a^2+5 A b^2-b^2 C\right ) \sec ^2(c+d x)\right )-a b (A+C) \sec (c+d x)+4 \left (C a^2+A b^2\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-6 C a^2-5 A b^2+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-a b (A+C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 \left (C a^2+A b^2\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \sec ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \sec ^2(c+d x)-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a \left (8 C a^2+5 A b^2-3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 C a^2+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 C a^2+b^2 (5 A-C)\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )+\left (16 C a^4+2 b^2 (5 A-4 C) a^2-b^4 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle -\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (16 a^3 C+12 a^2 b C+2 a b^2 (5 A+2 C)+b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}}{b \left (a^2-b^2\right )}\) |
(-2*(A*b^2 + a^2*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 + 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) - (((-2*(a - b)*Sqrt[a + b]*(2*a^2* b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSi n[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 - b)*Sqrt[a + b]*(16*a^3*C + 12*a^2*b*C + 2*a*b^2*(5*A + 2*C) + b^3*(5 *A + 3*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*a*(5*A*b^2 + 8*a^2*C - 3*b^2*C)*S qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(b*d))/(5*b))/(b*(a^2 - b^2))
3.8.42.3.1 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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d) *(A*b^2 + 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^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(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, 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. \(5108\) vs. \(2(428)=856\).
Time = 29.12 (sec) , antiderivative size = 5109, normalized size of antiderivative = 11.11
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5109\) |
| default | \(\text {Expression too large to display}\) | \(5163\) |
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((C*sec(d*x + c)^5 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b ^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]