Integrand size = 42, antiderivative size = 485 \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (24 a^3 b B+57 a b^3 B-16 a^4 C-24 a^2 b^2 C+147 b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^5 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 B-49 C)+18 a b^2 (B-2 C)+12 a^2 b (2 B-C)-16 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^4 d}-\frac {2 \left (12 a^2 b B-75 b^3 B-8 a^3 C-13 a b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^3 d}+\frac {2 \left (9 a b B-6 a^2 C+49 b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B+a C) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{63 b d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{9 d} \]
-2/315*(a-b)*(24*B*a^3*b+57*B*a*b^3-16*C*a^4-24*C*a^2*b^2+147*C*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 ^5/d-2/315*(a-b)*(3*b^3*(25*B-49*C)+18*a*b^2*(B-2*C)+12*a^2*b*(2*B-C)-16*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))*(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*(12*B*a^2*b-75*B*b^3-8*C*a^3-13*C*a*b^2)*(a+b*sec(d* x+c))^(1/2)*tan(d*x+c)/b^3/d+2/315*(9*B*a*b-6*C*a^2+49*C*b^2)*sec(d*x+c)*( a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/63*(9*B*b+C*a)*sec(d*x+c)^2*(a+b* sec(d*x+c))^(1/2)*tan(d*x+c)/b/d+2/9*C*sec(d*x+c)^3*(a+b*sec(d*x+c))^(1/2) *tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(3734\) vs. \(2(485)=970\).
Time = 25.14 (sec) , antiderivative size = 3734, normalized size of antiderivative = 7.70 \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Sqrt[a + b*Sec[c + d*x]]*((2*(24*a^3*b*B + 57*a*b^3*B - 16*a^4*C - 24*a^2 *b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^4) + (2*Sec[c + d*x]^3*(9*b*B*Sin [c + d*x] + a*C*Sin[c + d*x]))/(63*b) + (2*Sec[c + d*x]^2*(9*a*b*B*Sin[c + d*x] - 6*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/(315*b^2) + (2*Sec[ c + d*x]*(-12*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + d*x] + 8*a^3*C*Sin[c + d*x] + 13*a*b^2*C*Sin[c + d*x]))/(315*b^3) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/d + (2*((-19*a*B)/(105*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d *x]]) - (8*a^3*B)/(105*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a^4*C)/(315*b^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a^2* C)/(105*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (7*b*C)/(15*Sqrt[ b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^4*B*Sqrt[Sec[c + d*x]])/(10 5*b^3*Sqrt[b + a*Cos[c + d*x]]) - (17*a^2*B*Sqrt[Sec[c + d*x]])/(105*b*Sqr t[b + a*Cos[c + d*x]]) + (5*b*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (4*a*C*Sqrt[Sec[c + d*x]])/(35*Sqrt[b + a*Cos[c + d*x]]) + (16*a ^5*C*Sqrt[Sec[c + d*x]])/(315*b^4*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*C*Sqr t[Sec[c + d*x]])/(63*b^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^4*B*Cos[2*(c + d *x)]*Sqrt[Sec[c + d*x]])/(105*b^3*Sqrt[b + a*Cos[c + d*x]]) - (19*a^2*B*Co s[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*b*Sqrt[b + a*Cos[c + d*x]]) - (7*a *C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (1 6*a^5*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(315*b^4*Sqrt[b + a*Cos[c ...
Time = 2.49 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4519, 27, 3042, 4590, 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.
| \(\displaystyle \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^4(c+d x) \sqrt {a+b \sec (c+d x)} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4519 |
| \(\displaystyle \frac {2}{9} \int \frac {\sec ^3(c+d x) \left ((9 b B+a C) \sec ^2(c+d x)+(9 a B+7 b C) \sec (c+d x)+6 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \int \frac {\sec ^3(c+d x) \left ((9 b B+a C) \sec ^2(c+d x)+(9 a B+7 b C) \sec (c+d x)+6 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left ((9 b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(9 a B+7 b C) \csc \left (c+d x+\frac {\pi }{2}\right )+6 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4590 |
| \(\displaystyle \frac {1}{9} \left (\frac {2 \int \frac {\sec ^2(c+d x) \left (\left (-6 C a^2+9 b B a+49 b^2 C\right ) \sec ^2(c+d x)+b (45 b B+47 a C) \sec (c+d x)+4 a (9 b B+a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (\frac {\int \frac {\sec ^2(c+d x) \left (\left (-6 C a^2+9 b B a+49 b^2 C\right ) \sec ^2(c+d x)+b (45 b B+47 a C) \sec (c+d x)+4 a (9 b B+a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (-6 C a^2+9 b B a+49 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (45 b B+47 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a (9 b B+a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {2 \int \frac {\sec (c+d x) \left (-3 \left (-8 C a^3+12 b B a^2-13 b^2 C a-75 b^3 B\right ) \sec ^2(c+d x)+b \left (2 C a^2+207 b B a+147 b^2 C\right ) \sec (c+d x)+2 a \left (-6 C a^2+9 b B a+49 b^2 C\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\int \frac {\sec (c+d x) \left (-3 \left (-8 C a^3+12 b B a^2-13 b^2 C a-75 b^3 B\right ) \sec ^2(c+d x)+b \left (2 C a^2+207 b B a+147 b^2 C\right ) \sec (c+d x)+2 a \left (-6 C a^2+9 b B a+49 b^2 C\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 \left (-8 C a^3+12 b B a^2-13 b^2 C a-75 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left (2 C a^2+207 b B a+147 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (-6 C a^2+9 b B a+49 b^2 C\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (b \left (-4 C a^3+6 b B a^2+111 b^2 C a+75 b^3 B\right )+\left (-16 C a^4+24 b B a^3-24 b^2 C a^2+57 b^3 B a+147 b^4 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (-4 C a^3+6 b B a^2+111 b^2 C a+75 b^3 B\right )+\left (-16 C a^4+24 b B a^3-24 b^2 C a^2+57 b^3 B a+147 b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-4 C a^3+6 b B a^2+111 b^2 C a+75 b^3 B\right )+\left (-16 C a^4+24 b B a^3-24 b^2 C a^2+57 b^3 B a+147 b^4 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {\left (-16 a^4 C+24 a^3 b B-24 a^2 b^2 C+57 a b^3 B+147 b^4 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-16 a^3 C+12 a^2 b (2 B-C)+18 a b^2 (B-2 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {\left (-16 a^4 C+24 a^3 b B-24 a^2 b^2 C+57 a b^3 B+147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-16 a^3 C+12 a^2 b (2 B-C)+18 a b^2 (B-2 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {\frac {\left (-16 a^4 C+24 a^3 b B-24 a^2 b^2 C+57 a b^3 B+147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^3 C+12 a^2 b (2 B-C)+18 a b^2 (B-2 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}+\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{9} \left (\frac {\frac {2 \left (-6 a^2 C+9 a b B+49 b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}+\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^3 C+12 a^2 b (2 B-C)+18 a b^2 (B-2 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^4 C+24 a^3 b B-24 a^2 b^2 C+57 a b^3 B+147 b^4 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{b}-\frac {2 \left (-8 a^3 C+12 a^2 b B-13 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}}{7 b}+\frac {2 (a C+9 b B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\right )+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)}}{9 d}\) |
(2*C*Sec[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + ((2*(9* b*B + a*C)*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*b*d) + ((2*(9*a*b*B - 6*a^2*C + 49*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]*(24*a^3*b*B + 57*a*b^3*B - 16*a^4*C - 24*a^2*b^2*C + 147*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]*(3*b^3*(25*B - 49*C) + 18*a*b^2*(B - 2*C) + 12*a^2*b*(2*B - C) - 16*a^3*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))/b - (2*(12*a^2*b*B - 75*b^3*B - 8*a ^3*C - 13*a*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(b*d))/(5*b))/(7 *b))/9
3.9.14.3.1 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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(m + n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n - 1)*Simp[a*B*(n - 1) + (b*B*(m + n - 1) + a*A*(m + n))*Csc[e + f*x] + (a* B*m + A*b*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B }, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[0, m, 1] && GtQ[n, 0 ]
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]
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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 )*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc [e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(5581\) vs. \(2(447)=894\).
Time = 37.07 (sec) , antiderivative size = 5582, normalized size of antiderivative = 11.51
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5582\) |
| default | \(\text {Expression too large to display}\) | \(5647\) |
int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="fricas")
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="maxima")
\[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^3(c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^3} \,d x \]