Integrand size = 33, antiderivative size = 475 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (18 a^3 A b-246 a A b^3-8 a^4 B-33 a^2 b^2 B-147 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 A-49 B)-3 a b^2 (57 A-13 B)-6 a^2 b (3 A-B)+8 a^3 B\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 \left (18 a^2 A b-75 A b^3-8 a^3 B-39 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}-\frac {2 \left (18 a A b-8 a^2 B-49 b^2 B\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 A b-4 a B) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 B \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d} \]
2/315*(a-b)*(18*A*a^3*b-246*A*a*b^3-8*B*a^4-33*B*a^2*b^2-147*B*b^4)*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^ 4/d-2/315*(a-b)*(3*b^3*(25*A-49*B)-3*a*b^2*(57*A-13*B)-6*a^2*b*(3*A-B)+8*B *a^3)*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^3/d-2/315*(18*A*a*b-8*B*a^2-49*B*b^2)*(a+b*sec(d*x+c))^(3/2)*t an(d*x+c)/b^2/d+2/63*(9*A*b-4*B*a)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d +2/9*B*sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d-2/315*(18*A*a^2*b- 75*A*b^3-8*B*a^3-39*B*a*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d
Leaf count is larger than twice the leaf count of optimal. \(3766\) vs. \(2(475)=950\).
Time = 31.44 (sec) , antiderivative size = 3766, normalized size of antiderivative = 7.93 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*((2*(-18*a^3*A*b + 246*a*A*b^3 + 8*a^4*B + 33*a^2*b^2*B + 147*b^4*B)*Sin[c + d*x])/(315*b^3) + (2*Sec[c + d *x]^3*(9*A*b*Sin[c + d*x] + 10*a*B*Sin[c + d*x]))/63 + (2*Sec[c + d*x]^2*( 72*a*A*b*Sin[c + d*x] + 3*a^2*B*Sin[c + d*x] + 49*b^2*B*Sin[c + d*x]))/(31 5*b) + (2*Sec[c + d*x]*(9*a^2*A*b*Sin[c + d*x] + 75*A*b^3*Sin[c + d*x] - 4 *a^3*B*Sin[c + d*x] + 88*a*b^2*B*Sin[c + d*x]))/(315*b^2) + (2*b*B*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])) - (2*((2*a^3*A)/(35*b*S qrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (82*a*A*b)/(105*Sqrt[b + a*C os[c + d*x]]*Sqrt[Sec[c + d*x]]) - (11*a^2*B)/(105*Sqrt[b + a*Cos[c + d*x] ]*Sqrt[Sec[c + d*x]]) - (8*a^4*B)/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[S ec[c + d*x]]) - (7*b^2*B)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (31*a^2*A*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) + (2*a^4*A *Sqrt[Sec[c + d*x]])/(35*b^2*Sqrt[b + a*Cos[c + d*x]]) + (5*A*b^2*Sqrt[Sec [c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (8*a^5*B*Sqrt[Sec[c + d*x]])/( 315*b^3*Sqrt[b + a*Cos[c + d*x]]) - (31*a^3*B*Sqrt[Sec[c + d*x]])/(315*b*S qrt[b + a*Cos[c + d*x]]) + (13*a*b*B*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*C os[c + d*x]]) - (82*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) + (2*a^4*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(35*b^ 2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^5*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]] )/(315*b^3*Sqrt[b + a*Cos[c + d*x]]) - (11*a^3*B*Cos[2*(c + d*x)]*Sqrt[...
Time = 2.10 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4521, 27, 3042, 4570, 27, 3042, 4490, 27, 3042, 4490, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4521 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 A b-4 a B) \sec ^2(c+d x)+7 b B \sec (c+d x)+2 a B\right )dx}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 A b-4 a B) \sec ^2(c+d x)+7 b B \sec (c+d x)+2 a B\right )dx}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left ((9 A b-4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+7 b B \csc \left (c+d x+\frac {\pi }{2}\right )+2 a B\right )dx}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 A b-2 a B)-\left (-8 B a^2+18 A b a-49 b^2 B\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 A b-2 a B)-\left (-8 B a^2+18 A b a-49 b^2 B\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b (15 A b-2 a B)+\left (8 B a^2-18 A b a+49 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 B a^2+57 A b a+49 b^2 B\right )-\left (-8 B a^3+18 A b a^2-39 b^2 B a-75 A b^3\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 B a^2+57 A b a+49 b^2 B\right )-\left (-8 B a^3+18 A b a^2-39 b^2 B a-75 A b^3\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-2 B a^2+57 A b a+49 b^2 B\right )+\left (8 B a^3-18 A b a^2+39 b^2 B a+75 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (2 B a^3+153 A b a^2+186 b^2 B a+75 A b^3\right )-\left (-8 B a^4+18 A b a^3-33 b^2 B a^2-246 A b^3 a-147 b^4 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (2 B a^3+153 A b a^2+186 b^2 B a+75 A b^3\right )-\left (-8 B a^4+18 A b a^3-33 b^2 B a^2-246 A b^3 a-147 b^4 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 B a^3+153 A b a^2+186 b^2 B a+75 A b^3\right )+\left (8 B a^4-18 A b a^3+33 b^2 B a^2+246 A b^3 a+147 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 B-6 a^2 b (3 A-B)-3 a b^2 (57 A-13 B)+3 b^3 (25 A-49 B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-8 a^4 B+18 a^3 A b-33 a^2 b^2 B-246 a A b^3-147 b^4 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 B-6 a^2 b (3 A-B)-3 a b^2 (57 A-13 B)+3 b^3 (25 A-49 B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-8 a^4 B+18 a^3 A b-33 a^2 b^2 B-246 a A b^3-147 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left (-8 a^4 B+18 a^3 A b-33 a^2 b^2 B-246 a A b^3-147 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 B-6 a^2 b (3 A-B)-3 a b^2 (57 A-13 B)+3 b^3 (25 A-49 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 B+18 a^3 A b-33 a^2 b^2 B-246 a A b^3-147 b^4 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 B-6 a^2 b (3 A-B)-3 a b^2 (57 A-13 B)+3 b^3 (25 A-49 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 B+18 a^2 A b-39 a b^2 B-75 A b^3\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 B+18 a A b-49 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 A b-4 a B) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 B \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
(2*B*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(9*b*d) + ((2*( 9*A*b - 4*a*B)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-2*(18 *a*A*b - 8*a^2*B - 49*b^2*B)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d ) + (3*(((2*(a - b)*Sqrt[a + b]*(18*a^3*A*b - 246*a*A*b^3 - 8*a^4*B - 33*a ^2*b^2*B - 147*b^4*B)*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)]*Sqr t[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(3* b^3*(25*A - 49*B) - 3*a*b^2*(57*A - 13*B) - 6*a^2*b*(3*A - B) + 8*a^3*B)*C ot[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))/3 - (2*(18*a^2*A*b - 75*A*b^3 - 8*a^3*B - 39*a*b^2*B )*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/(7*b))/(9*b)
3.4.56.3.1 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_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* (m + n))), x] + Simp[d^2/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B , m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] && !IGtQ[m, 1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(5581\) vs. \(2(437)=874\).
Time = 39.86 (sec) , antiderivative size = 5582, normalized size of antiderivative = 11.75
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5582\) |
| default | \(\text {Expression too large to display}\) | \(5647\) |
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integral((B*b*sec(d*x + c)^5 + A*a*sec(d*x + c)^3 + (B*a + A*b)*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]