Integrand size = 35, antiderivative size = 521 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 b B+63 b^3 B+15 a^3 C+5 a b^2 (49 A+29 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac {2 \sqrt {a+b} \left (15 a^3 (7 B-C)+b^3 (35 A-63 B+25 C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 \left (35 A b^2+56 a b B+15 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 b B+5 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]
-2/105*(a-b)*(161*B*a^2*b+63*B*b^3+15*a^3*C+5*a*b^2*(49*A+29*C))*cot(d*x+c )*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(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^2/d +2/105*(15*a^3*(7*B-C)+b^3*(35*A-63*B+25*C)+a^2*b*(315*A-161*B+135*C)-a*b^ 2*(245*A-119*B+145*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^( 1/2),((a+b)/(a-b))^(1/2))*(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/d-2*a^2*A*cot(d*x+c)*EllipticPi((a+b*sec(d*x+ c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d *x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+2/35*(7*B*b+5*C*a)*( a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c) /d+2/105*(35*A*b^2+56*B*a*b+15*C*a^2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan( d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(1401\) vs. \(2(521)=1042\).
Time = 25.74 (sec) , antiderivative size = 1401, normalized size of antiderivative = 2.69 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \]
(-4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqr t[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(245*a^2*A*b^2*Tan[(c + d*x)/2] + 245*a*A *b^3*Tan[(c + d*x)/2] + 161*a^3*b*B*Tan[(c + d*x)/2] + 161*a^2*b^2*B*Tan[( c + d*x)/2] + 63*a*b^3*B*Tan[(c + d*x)/2] + 63*b^4*B*Tan[(c + d*x)/2] + 15 *a^4*C*Tan[(c + d*x)/2] + 15*a^3*b*C*Tan[(c + d*x)/2] + 145*a^2*b^2*C*Tan[ (c + d*x)/2] + 145*a*b^3*C*Tan[(c + d*x)/2] - 490*a^2*A*b^2*Tan[(c + d*x)/ 2]^3 - 322*a^3*b*B*Tan[(c + d*x)/2]^3 - 126*a*b^3*B*Tan[(c + d*x)/2]^3 - 3 0*a^4*C*Tan[(c + d*x)/2]^3 - 290*a^2*b^2*C*Tan[(c + d*x)/2]^3 + 245*a^2*A* b^2*Tan[(c + d*x)/2]^5 - 245*a*A*b^3*Tan[(c + d*x)/2]^5 + 161*a^3*b*B*Tan[ (c + d*x)/2]^5 - 161*a^2*b^2*B*Tan[(c + d*x)/2]^5 + 63*a*b^3*B*Tan[(c + d* x)/2]^5 - 63*b^4*B*Tan[(c + d*x)/2]^5 + 15*a^4*C*Tan[(c + d*x)/2]^5 - 15*a ^3*b*C*Tan[(c + d*x)/2]^5 + 145*a^2*b^2*C*Tan[(c + d*x)/2]^5 - 145*a*b^3*C *Tan[(c + d*x)/2]^5 - 210*a^3*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[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)] - 210*a^3*A*b*EllipticPi[-1, ArcS in[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[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)] + (a + b)*(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*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 ...
Time = 2.25 (sec) , antiderivative size = 532, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4544, 27, 3042, 4544, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.
| \(\displaystyle \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {2}{7} \int \frac {1}{2} (a+b \sec (c+d x))^{3/2} \left ((7 b B+5 a C) \sec ^2(c+d x)+(7 A b+5 C b+7 a B) \sec (c+d x)+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int (a+b \sec (c+d x))^{3/2} \left ((7 b B+5 a C) \sec ^2(c+d x)+(7 A b+5 C b+7 a B) \sec (c+d x)+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left ((7 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(7 A b+5 C b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left (35 A a^2+\left (15 C a^2+56 b B a+35 A b^2+25 b^2 C\right ) \sec ^2(c+d x)+\left (35 B a^2+70 A b a+40 b C a+21 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left (35 A a^2+\left (15 C a^2+56 b B a+35 A b^2+25 b^2 C\right ) \sec ^2(c+d x)+\left (35 B a^2+70 A b a+40 b C a+21 b^2 B\right ) \sec (c+d x)\right )dx+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (35 A a^2+\left (15 C a^2+56 b B a+35 A b^2+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (35 B a^2+70 A b a+40 b C a+21 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4544 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {105 A a^3+\left (15 C a^3+161 b B a^2+5 b^2 (49 A+29 C) a+63 b^3 B\right ) \sec ^2(c+d x)+\left (105 B a^3+45 b (7 A+3 C) a^2+119 b^2 B a+5 b^3 (7 A+5 C)\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {105 A a^3+\left (15 C a^3+161 b B a^2+5 b^2 (49 A+29 C) a+63 b^3 B\right ) \sec ^2(c+d x)+\left (105 B a^3+45 b (7 A+3 C) a^2+119 b^2 B a+5 b^3 (7 A+5 C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {105 A a^3+\left (15 C a^3+161 b B a^2+5 b^2 (49 A+29 C) a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (105 B a^3+45 b (7 A+3 C) a^2+119 b^2 B a+5 b^3 (7 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {105 A a^3+\left (105 B a^3-15 C a^3-161 b B a^2+45 b (7 A+3 C) a^2+119 b^2 B a-5 b^2 (49 A+29 C) a-63 b^3 B+5 b^3 (7 A+5 C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {105 A a^3+\left (105 B a^3-15 C a^3-161 b B a^2+45 b (7 A+3 C) a^2+119 b^2 B a-5 b^2 (49 A+29 C) a-63 b^3 B+5 b^3 (7 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 A \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 a^3 A \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {210 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {210 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \cot (c+d x) \left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \tan (c+d x) \left (15 a^2 C+56 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (-\frac {210 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \cot (c+d x) \left (15 a^3 (7 B-C)+a^2 b (315 A-161 B+135 C)-a b^2 (245 A-119 B+145 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (15 a^3 C+161 a^2 b B+5 a b^2 (49 A+29 C)+63 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )+\frac {2 (5 a C+7 b B) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
(2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*(7*b*B + 5*a*C)* (a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b] *(161*a^2*b*B + 63*b^3*B + 15*a^3*C + 5*a*b^2*(49*A + 29*C))*Cot[c + d*x]* EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) ])/(b^2*d) + (2*Sqrt[a + b]*(15*a^3*(7*B - C) + b^3*(35*A - 63*B + 25*C) + a^2*b*(315*A - 161*B + 135*C) - a*b^2*(245*A - 119*B + 145*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) - (210*a^2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, Ar cSin[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))])/d)/3 + (2 *(35*A*b^2 + 56*a*b*B + 15*a^2*C + 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[ c + d*x])/(3*d))/5)/7
3.10.52.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[( a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m )*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(6503\) vs. \(2(478)=956\).
Time = 14.65 (sec) , antiderivative size = 6504, normalized size of antiderivative = 12.48
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(6504\) |
| default | \(\text {Expression too large to display}\) | \(6563\) |
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
integral((C*b^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*sec(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c ))*sqrt(b*sec(d*x + c) + a), x)
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]