Integrand size = 43, antiderivative size = 442 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (5 A b+4 a B-8 b C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 b d}+\frac {\sqrt {a+b} (b (5 A+8 B-8 C)+2 a (A+2 B+8 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 d}-\frac {\sqrt {a+b} \left (3 A b^2+12 a b B+4 a^2 (A+2 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac {(3 A b+4 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \]
1/2*A*cos(d*x+c)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/4*(a-b)*(5*A*b+4*B* a-8*C*b)*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/d+1/4*(b*(5*A+8*B-8*C)+2*a*(A+2*B+8*C))*cot(d*x+c)*Elliptic F((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)/d-1/4*(3*A*b^2+ 12*B*a*b+4*a^2*(A+2*C))*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b) ^(1/2),(a+b)/a,((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)/a/d+1/4*(3*A*b+4*B*a)*sin(d*x+c)*(a+b *sec(d*x+c))^(1/2)/d
Leaf count is larger than twice the leaf count of optimal. \(4506\) vs. \(2(442)=884\).
Time = 25.96 (sec) , antiderivative size = 4506, normalized size of antiderivative = 10.19 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
((Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(4*b*C*Sin[c + d*x] + (a*A*Sin[2 *(c + d*x)])/2))/(d*(b + a*Cos[c + d*x])) - (Sqrt[Cos[c + d*x]*Sec[(c + d* x)/2]^2]*((a^2*A)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*b^2 )/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a*b*B)/(Sqrt[b + a*Co s[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^2*C)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) - (2*b^2*C)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (7*a*A*b*Sqrt[Sec[c + d*x]])/(4*Sqrt[b + a*Cos[c + d*x]]) + (a^2*B*Sqrt[ Sec[c + d*x]])/Sqrt[b + a*Cos[c + d*x]] + (2*b^2*B*Sqrt[Sec[c + d*x]])/Sqr t[b + a*Cos[c + d*x]] + (2*a*b*C*Sqrt[Sec[c + d*x]])/Sqrt[b + a*Cos[c + d* x]] + (5*a*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(4*Sqrt[b + a*Cos[c + d*x]]) + (a^2*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/Sqrt[b + a*Cos[c + d* x]] - (2*a*b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/Sqrt[b + a*Cos[c + d*x ]])*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(-(a* (a + b)*(5*A*b + 4*a*B - 8*b*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b )/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^ 2)/(a + b)]) + b*(a + b)*(3*A*b + 2*a*(A + 2*B - 4*C))*EllipticF[ArcSin[Ta n[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + (3*A*b^2 + 12*a*b*B + 4*a^2*(A + 2*C) )*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*a*Elli pticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[(c + d*x)/2]...
Time = 1.88 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4582, 27, 3042, 4582, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (-b (A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+3 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (-b (A-4 C) \sec ^2(c+d x)+2 (2 b B+a (A+2 C)) \sec (c+d x)+3 A b+4 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (-b (A-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 (2 b B+a (A+2 C)) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b+4 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \sec ^2(c+d x)+2 b (4 b B+a (A+8 C)) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \sec ^2(c+d x)+2 b (4 b B+a (A+8 C)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2-b (5 A b-8 C b+4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 b (4 b B+a (A+8 C)) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2+(b (5 A b-8 C b+4 a B)+2 b (4 b B+a (A+8 C))) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b (4 a B+5 A b-8 b C) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \frac {4 (A+2 C) a^2+12 b B a+3 A b^2+(b (5 A b-8 C b+4 a B)+2 b (4 b B+a (A+8 C))) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (b (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-b (4 a B+5 A b-8 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-\frac {2 \sqrt {a+b} \cot (c+d x) \left (4 a^2 (A+2 C)+12 a b B+3 A b^2\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 \sqrt {a+b} \cot (c+d x) (2 a (A+2 B+8 C)+b (5 A+8 B-8 C)) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (4 a B+5 A b-8 b C) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}\right )+\frac {(4 a B+3 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d}\) |
(A*Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d) + (((2*(a - b)*Sqrt[a + b]*(5*A*b + 4*a*B - 8*b*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*d) + (2*Sqrt[a + b]*(b*(5*A + 8*B - 8*C) + 2*a*(A + 2*B + 8*C))*Cot[c + d*x]*EllipticF[Ar cSin[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))])/d - (2*Sq rt[a + b]*(3*A*b^2 + 12*a*b*B + 4*a^2*(A + 2*C))*Cot[c + d*x]*EllipticPi[( a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) ])/(a*d))/2 + ((3*A*b + 4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d)/4
3.10.47.3.1 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[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C 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, C}, x] && NeQ[a^2 - b^2, 0]
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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4871\) vs. \(2(401)=802\).
Time = 2.74 (sec) , antiderivative size = 4872, normalized size of antiderivative = 11.02
int(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
-1/4/d*(24*B*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(cos (d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1 /2)*a*b*cos(d*x+c)^2+48*B*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b) )^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+co s(d*x+c)))^(1/2)*a*b*cos(d*x+c)-4*A*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b) /(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)) /(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)^2+10*A*EllipticE(cot(d*x+c)-csc(d*x+ c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*co s(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2*cos(d*x+c)-8*A*EllipticF(cot(d*x+c)-cs c(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*( b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)+8*B*EllipticE(cot(d*x +c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/( a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)+32*C*EllipticPi (cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c))) ^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*cos(d*x+c)-16*C *EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2*cos(d*x+ c)+4*B*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1 +cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*co s(d*x+c)^2+16*C*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2)...
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
integral((C*b*cos(d*x + c)^2*sec(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2*s ec(d*x + c)^2 + A*a*cos(d*x + c)^2 + (B*a + A*b)*cos(d*x + c)^2*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/ 2)*cos(d*x + c)^2, x)
\[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/ 2)*cos(d*x + c)^2, x)
Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]