Integrand size = 42, antiderivative size = 475 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (18 a^3 b B-246 a b^3 B-8 a^4 C-33 a^2 b^2 C-147 b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 B-49 C)-3 a b^2 (57 B-13 C)-6 a^2 b (3 B-C)+8 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 \left (18 a^2 b B-75 b^3 B-8 a^3 C-39 a b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}-\frac {2 \left (18 a b B-8 a^2 C-49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B-4 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d} \]
2/315*(a-b)*(18*B*a^3*b-246*B*a*b^3-8*C*a^4-33*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^ 4/d-2/315*(a-b)*(3*b^3*(25*B-49*C)-3*a*b^2*(57*B-13*C)-6*a^2*b*(3*B-C)+8*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^3/d-2/315*(18*B*a*b-8*C*a^2-49*C*b^2)*(a+b*sec(d*x+c))^(3/2)*t an(d*x+c)/b^2/d+2/63*(9*B*b-4*C*a)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d +2/9*C*sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d-2/315*(18*B*a^2*b- 75*B*b^3-8*C*a^3-39*C*a*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d
Time = 22.24 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.38 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 (a+b) \left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (8 a^3 C-6 a^2 b (3 B+C)+3 a b^2 (57 B+13 C)+3 b^3 (25 B+49 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^3 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {2 \left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \sin (c+d x)}{315 b^3}+\frac {2}{63} \sec ^3(c+d x) (9 b B \sin (c+d x)+10 a C \sin (c+d x))+\frac {2 \sec ^2(c+d x) \left (72 a b B \sin (c+d x)+3 a^2 C \sin (c+d x)+49 b^2 C \sin (c+d x)\right )}{315 b}+\frac {2 \sec (c+d x) \left (9 a^2 b B \sin (c+d x)+75 b^3 B \sin (c+d x)-4 a^3 C \sin (c+d x)+88 a b^2 C \sin (c+d x)\right )}{315 b^2}+\frac {2}{9} b C \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \]
(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*(a + b)*(-18*a^3*b*B + 246*a*b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4*C)*Sqr t[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2* b*(a + b)*(8*a^3*C - 6*a^2*b*(3*B + C) + 3*a*b^2*(57*B + 13*C) + 3*b^3*(25 *B + 49*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x] )/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b )/(a + b)] + (-18*a^3*b*B + 246*a*b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4 *C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) )/(315*b^3*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^ (3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*((2*(-18*a^3*b*B + 246*a *b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^3) + (2* Sec[c + d*x]^3*(9*b*B*Sin[c + d*x] + 10*a*C*Sin[c + d*x]))/63 + (2*Sec[c + d*x]^2*(72*a*b*B*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d *x]))/(315*b) + (2*Sec[c + d*x]*(9*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + d*x] - 4*a^3*C*Sin[c + d*x] + 88*a*b^2*C*Sin[c + d*x]))/(315*b^2) + (2*b* C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x]))
Time = 2.25 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.03, 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, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4521 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 b B-4 a C) \sec ^2(c+d x)+7 b C \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 b B-4 a C) \sec ^2(c+d x)+7 b C \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left ((9 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+7 b C \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 b B-2 a C)-\left (-8 C a^2+18 b B a-49 b^2 C\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 b B-2 a C)-\left (-8 C a^2+18 b B a-49 b^2 C\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b (15 b B-2 a C)+\left (8 C a^2-18 b B a+49 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )-\left (-8 C a^3+18 b B a^2-39 b^2 C a-75 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )-\left (-8 C a^3+18 b B a^2-39 b^2 C a-75 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )+\left (8 C a^3-18 b B a^2+39 b^2 C a+75 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )-\left (-8 C a^4+18 b B a^3-33 b^2 C a^2-246 b^3 B a-147 b^4 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )-\left (-8 C a^4+18 b B a^3-33 b^2 C a^2-246 b^3 B a-147 b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )+\left (8 C a^4-18 b B a^3+33 b^2 C a^2+246 b^3 B a+147 b^4 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\) |
(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(9*b*d) + ((2*( 9*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-2*(18 *a*b*B - 8*a^2*C - 49*b^2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d ) + (3*(((2*(a - b)*Sqrt[a + b]*(18*a^3*b*B - 246*a*b^3*B - 8*a^4*C - 33*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)]*Sqr t[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(3* b^3*(25*B - 49*C) - 3*a*b^2*(57*B - 13*C) - 6*a^2*b*(3*B - C) + 8*a^3*C)*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*b*B - 75*b^3*B - 8*a^3*C - 39*a*b^2*C )*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/(7*b))/(9*b)
3.9.22.3.1 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. \(5583\) vs. \(2(437)=874\).
Time = 39.27 (sec) , antiderivative size = 5584, normalized size of antiderivative = 11.76
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5584\) |
| default | \(\text {Expression too large to display}\) | \(5647\) |
int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
integral((C*b*sec(d*x + c)^5 + B*a*sec(d*x + c)^3 + (C*a + B*b)*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]