Integrand size = 42, antiderivative size = 565 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-255 a^3 b^2 C-3705 a b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (6 a b^3 (209 B-505 C)-3 b^4 (539 B-225 C)-a^3 b (110 B-30 C)-15 a^2 b^2 (121 B-19 C)+40 a^4 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac {2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-285 a^2 b^2 C-675 b^4 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-335 a b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}-\frac {2 \left (22 a b B-8 a^2 C-81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac {2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d} \]
2/3465*(a-b)*(110*B*a^4*b-3069*B*a^2*b^3-1617*B*b^5-40*C*a^5-255*C*a^3*b^2 -3705*C*a*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-2/3465*(a-b)*(6*a*b^3*(209*B-505*C)-3*b^4*(539*B- 225*C)-a^3*b*(110*B-30*C)-15*a^2*b^2*(121*B-19*C)+40*a^4*C)*cot(d*x+c)*Ell ipticF((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-2/34 65*(110*B*a^2*b-539*B*b^3-40*C*a^3-335*C*a*b^2)*(a+b*sec(d*x+c))^(3/2)*tan (d*x+c)/b^2/d-2/693*(22*B*a*b-8*C*a^2-81*C*b^2)*(a+b*sec(d*x+c))^(5/2)*tan (d*x+c)/b^2/d+2/99*(11*B*b-4*C*a)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b^2/d+ 2/11*C*sec(d*x+c)*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b/d-2/3465*(110*B*a^3* b-1254*B*a*b^3-40*C*a^4-285*C*a^2*b^2-675*C*b^4)*(a+b*sec(d*x+c))^(1/2)*ta n(d*x+c)/b^2/d
Leaf count is larger than twice the leaf count of optimal. \(4227\) vs. \(2(565)=1130\).
Time = 26.99 (sec) , antiderivative size = 4227, normalized size of antiderivative = 7.48 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(-110*a^4*b*B + 3069*a^2*b^ 3*B + 1617*b^5*B + 40*a^5*C + 255*a^3*b^2*C + 3705*a*b^4*C)*Sin[c + d*x])/ (3465*b^3) + (2*Sec[c + d*x]^4*(11*b^2*B*Sin[c + d*x] + 23*a*b*C*Sin[c + d *x]))/99 + (2*Sec[c + d*x]^3*(209*a*b*B*Sin[c + d*x] + 113*a^2*C*Sin[c + d *x] + 81*b^2*C*Sin[c + d*x]))/693 + (2*Sec[c + d*x]^2*(825*a^2*b*B*Sin[c + d*x] + 539*b^3*B*Sin[c + d*x] + 15*a^3*C*Sin[c + d*x] + 1145*a*b^2*C*Sin[ c + d*x]))/(3465*b) + (2*Sec[c + d*x]*(55*a^3*b*B*Sin[c + d*x] + 1793*a*b^ 3*B*Sin[c + d*x] - 20*a^4*C*Sin[c + d*x] + 1025*a^2*b^2*C*Sin[c + d*x] + 6 75*b^4*C*Sin[c + d*x]))/(3465*b^2) + (2*b^2*C*Sec[c + d*x]^4*Tan[c + d*x]) /11))/(d*(b + a*Cos[c + d*x])^2) - (2*((2*a^4*B)/(63*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (31*a^2*b*B)/(35*Sqrt[b + a*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) - (7*b^3*B)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] ]) - (17*a^3*C)/(231*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^5 *C)/(693*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (247*a*b^2*C)/ (231*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (124*a^3*B*Sqrt[Sec[c + d*x]])/(315*Sqrt[b + a*Cos[c + d*x]]) + (2*a^5*B*Sqrt[Sec[c + d*x]])/(63 *b^2*Sqrt[b + a*Cos[c + d*x]]) + (38*a*b^2*B*Sqrt[Sec[c + d*x]])/(105*Sqrt [b + a*Cos[c + d*x]]) - (8*a^6*C*Sqrt[Sec[c + d*x]])/(693*b^3*Sqrt[b + a*C os[c + d*x]]) - (7*a^4*C*Sqrt[Sec[c + d*x]])/(99*b*Sqrt[b + a*Cos[c + d*x] ]) - (26*a^2*b*C*Sqrt[Sec[c + d*x]])/(231*Sqrt[b + a*Cos[c + d*x]]) + (...
Time = 2.69 (sec) , antiderivative size = 583, normalized size of antiderivative = 1.03, number of steps used = 22, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4560, 3042, 4521, 27, 3042, 4570, 27, 3042, 4490, 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.
\(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4560 |
\(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} (B+C \sec (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4521 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 b B-4 a C) \sec ^2(c+d x)+9 b C \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left ((11 b B-4 a C) \sec ^2(c+d x)+9 b C \sec (c+d x)+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left ((11 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+9 b C \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 b B-10 a C)-\left (-8 C a^2+22 b B a-81 b^2 C\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (b (77 b B-10 a C)-\left (-8 C a^2+22 b B a-81 b^2 C\right ) \sec (c+d x)\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (b (77 b B-10 a C)+\left (8 C a^2-22 b B a+81 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {\frac {2}{7} \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 C a^2+143 b B a+135 b^2 C\right )-\left (-40 C a^3+110 b B a^2-335 b^2 C a-539 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {1}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b \left (-10 C a^2+143 b B a+135 b^2 C\right )-\left (-40 C a^3+110 b B a^2-335 b^2 C a-539 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {1}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b \left (-10 C a^2+143 b B a+135 b^2 C\right )+\left (40 C a^3-110 b B a^2+335 b^2 C a+539 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 C a^3+605 b B a^2+1010 b^2 C a+539 b^3 B\right )-\left (-40 C a^4+110 b B a^3-285 b^2 C a^2-1254 b^3 B a-675 b^4 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-10 C a^3+605 b B a^2+1010 b^2 C a+539 b^3 B\right )-\left (-40 C a^4+110 b B a^3-285 b^2 C a^2-1254 b^3 B a-675 b^4 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-10 C a^3+605 b B a^2+1010 b^2 C a+539 b^3 B\right )+\left (40 C a^4-110 b B a^3+285 b^2 C a^2+1254 b^3 B a+675 b^4 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (10 C a^4+1705 b B a^3+3315 b^2 C a^2+2871 b^3 B a+675 b^4 C\right )-\left (-40 C a^5+110 b B a^4-255 b^2 C a^3-3069 b^3 B a^2-3705 b^4 C a-1617 b^5 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (10 C a^4+1705 b B a^3+3315 b^2 C a^2+2871 b^3 B a+675 b^4 C\right )-\left (-40 C a^5+110 b B a^4-255 b^2 C a^3-3069 b^3 B a^2-3705 b^4 C a-1617 b^5 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (10 C a^4+1705 b B a^3+3315 b^2 C a^2+2871 b^3 B a+675 b^4 C\right )+\left (40 C a^5-110 b B a^4+255 b^2 C a^3+3069 b^3 B a^2+3705 b^4 C a+1617 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (40 a^4 C-a^3 b (110 B-30 C)-15 a^2 b^2 (121 B-19 C)+6 a b^3 (209 B-505 C)-3 b^4 (539 B-225 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-40 a^5 C+110 a^4 b B-255 a^3 b^2 C-3069 a^2 b^3 B-3705 a b^4 C-1617 b^5 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (40 a^4 C-a^3 b (110 B-30 C)-15 a^2 b^2 (121 B-19 C)+6 a b^3 (209 B-505 C)-3 b^4 (539 B-225 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-40 a^5 C+110 a^4 b B-255 a^3 b^2 C-3069 a^2 b^3 B-3705 a b^4 C-1617 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (-\left (-40 a^5 C+110 a^4 b B-255 a^3 b^2 C-3069 a^2 b^3 B-3705 a b^4 C-1617 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (40 a^4 C-a^3 b (110 B-30 C)-15 a^2 b^2 (121 B-19 C)+6 a b^3 (209 B-505 C)-3 b^4 (539 B-225 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (-40 a^5 C+110 a^4 b B-255 a^3 b^2 C-3069 a^2 b^3 B-3705 a b^4 C-1617 b^5 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (40 a^4 C-a^3 b (110 B-30 C)-15 a^2 b^2 (121 B-19 C)+6 a b^3 (209 B-505 C)-3 b^4 (539 B-225 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-40 a^4 C+110 a^3 b B-285 a^2 b^2 C-1254 a b^3 B-675 b^4 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-40 a^3 C+110 a^2 b B-335 a b^2 C-539 b^3 B\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\right )-\frac {2 \left (-8 a^2 C+22 a b B-81 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}}{9 b}+\frac {2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}}{11 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d}\) |
(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d) + ((2* (11*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d) + ((-2*( 22*a*b*B - 8*a^2*C - 81*b^2*C)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7 *d) + ((-2*(110*a^2*b*B - 539*b^3*B - 40*a^3*C - 335*a*b^2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*(((2*(a - b)*Sqrt[a + b]*(110*a^4* b*B - 3069*a^2*b^3*B - 1617*b^5*B - 40*a^5*C - 255*a^3*b^2*C - 3705*a*b^4* 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 - b)*Sqrt[a + b]*(6*a*b^3*(209*B - 505* C) - 3*b^4*(539*B - 225*C) - a^3*b*(110*B - 30*C) - 15*a^2*b^2*(121*B - 19 *C) + 40*a^4*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqr t[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*(110*a^3*b*B - 1254*a*b^3*B - 40*a^4*C - 285*a^2*b^2*C - 675*b^4*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d *x])/(3*d)))/5)/7)/(9*b))/(11*b)
3.9.29.3.1 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[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(6883\) vs. \(2(523)=1046\).
Time = 110.84 (sec) , antiderivative size = 6884, normalized size of antiderivative = 12.18
method | result | size |
parts | \(\text {Expression too large to display}\) | \(6884\) |
default | \(\text {Expression too large to display}\) | \(6974\) |
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
integral((C*b^2*sec(d*x + c)^6 + B*a^2*sec(d*x + c)^3 + (2*C*a*b + B*b^2)* sec(d*x + c)^5 + (C*a^2 + 2*B*a*b)*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a ), x)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]