Integrand size = 40, antiderivative size = 387 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (21 a^2 b B+63 b^3 B-6 a^3 C+82 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (a b (21 B-57 C)-b^2 (63 B-25 C)-6 a^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac {2 \left (21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}+\frac {2 (7 b B-2 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]
-2/105*(a-b)*(21*B*a^2*b+63*B*b^3-6*C*a^3+82*C*a*b^2)*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^3/d-2/105*(a-b )*(a*b*(21*B-57*C)-b^2*(63*B-25*C)-6*C*a^2)*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^2/d+2/35*(7*B*b-2*C*a)*( a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+ c)/b/d+2/105*(21*B*a*b-6*C*a^2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c) /b/d
Time = 17.49 (sec) , antiderivative size = 547, normalized size of antiderivative = 1.41 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 (a+b) \left (-21 a^2 b B-63 b^3 B+6 a^3 C-82 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) \left (-6 a^2 C+3 a b (7 B+19 C)+b^2 (63 B+25 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-21 a^2 b B-63 b^3 B+6 a^3 C-82 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^2 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {2 \left (-21 a^2 b B-63 b^3 B+6 a^3 C-82 a b^2 C\right ) \sin (c+d x)}{105 b^2}+\frac {2}{35} \sec ^2(c+d x) (7 b B \sin (c+d x)+8 a C \sin (c+d x))+\frac {2 \sec (c+d x) \left (42 a b B \sin (c+d x)+3 a^2 C \sin (c+d x)+25 b^2 C \sin (c+d x)\right )}{105 b}+\frac {2}{7} b C \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*(a + b)*(-21*a^2*b*B - 63*b^3*B + 6*a^3*C - 82*a*b^2*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]))]*E llipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-6*a^2* C + 3*a*b*(7*B + 19*C) + b^2*(63*B + 25*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)] + (-21*a^2*b*B - 63*b^3*B + 6*a ^3*C - 82*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Ta n[(c + d*x)/2]))/(105*b^2*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*(-21 *a^2*b*B - 63*b^3*B + 6*a^3*C - 82*a*b^2*C)*Sin[c + d*x])/(105*b^2) + (2*S ec[c + d*x]^2*(7*b*B*Sin[c + d*x] + 8*a*C*Sin[c + d*x]))/35 + (2*Sec[c + d *x]*(42*a*b*B*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]) )/(105*b) + (2*b*C*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x] ))
Time = 1.67 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 3042, 4498, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} (5 b C+(7 b B-2 a C) \sec (c+d x))dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (5 b C+(7 b B-2 a C) \sec (c+d x))dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (5 b C+(7 b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (21 b B+19 a C)+\left (-6 C a^2+21 b B a+25 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (21 b B+19 a C)+\left (-6 C a^2+21 b B a+25 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (21 b B+19 a C)+\left (-6 C a^2+21 b B a+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (51 C a^2+84 b B a+25 b^2 C\right )+\left (-6 C a^3+21 b B a^2+82 b^2 C a+63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (51 C a^2+84 b B a+25 b^2 C\right )+\left (-6 C a^3+21 b B a^2+82 b^2 C a+63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (51 C a^2+84 b B a+25 b^2 C\right )+\left (-6 C a^3+21 b B a^2+82 b^2 C a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\left (-6 a^3 C+21 a^2 b B+82 a b^2 C+63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-6 a^2 C+a b (21 B-57 C)-b^2 (63 B-25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\left (-6 a^3 C+21 a^2 b B+82 a b^2 C+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-6 a^2 C+a b (21 B-57 C)-b^2 (63 B-25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\left (-6 a^3 C+21 a^2 b B+82 a b^2 C+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-6 a^2 C+a b (21 B-57 C)-b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2 \left (-6 a^2 C+21 a b B+25 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \left (-6 a^2 C+a b (21 B-57 C)-b^2 (63 B-25 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (-6 a^3 C+21 a^2 b B+82 a b^2 C+63 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
(2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((2*(7*b*B - 2*a*C )*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + b]*(21*a^2*b*B + 63*b^3*B - 6*a^3*C + 82*a*b^2*C)*Cot[c + d*x]*EllipticE[A rcSin[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]*(a*b*(21*B - 57*C) - b^2*(63*B - 25*C) - 6*a^2*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))/3 + (2*(21*a*b*B - 6*a^2*C + 25*b^2*C)*Sqrt[a + b* Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5)/(7*b)
3.9.23.3.1 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_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*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*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[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]
Leaf count of result is larger than twice the leaf count of optimal. \(4325\) vs. \(2(353)=706\).
Time = 24.45 (sec) , antiderivative size = 4326, normalized size of antiderivative = 11.18
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4326\) |
| default | \(\text {Expression too large to display}\) | \(4370\) |
-2/5*B/d/b*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-3*a*b^ 2*tan(d*x+c)-3*sin(d*x+c)*cos(d*x+c)*a*b^2-2*EllipticE(cot(d*x+c)-csc(d*x+ c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)-3*sin(d*x+c)*b^3-3*a*b^2* sin(d*x+c)-b^3*tan(d*x+c)-2*a^2*b*cos(d*x+c)*sin(d*x+c)+3*(1/(a+b)*(b+a*co s(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic F(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^3-(1/(a+b)*(b+a*cos(d*x+c)) /(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x +c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3-3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d *x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc (d*x+c),((a-b)/(a+b))^(1/2))*b^3-6*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/ (a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*a*b^2*cos(d*x+c)+2*EllipticF(cot(d*x+c)-csc(d*x+c),( (a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)+8*EllipticF(cot(d*x+c)-csc(d* x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)* (cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*cos(d*x+c)-EllipticE(cot(d*x+c)-cs c(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 /2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)^2-3*EllipticE(cot(d *x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*...
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(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)^4 + B*a*sec(d*x + c)^2 + (C*a + B*b)*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \]