Integrand size = 35, antiderivative size = 393 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {4 a (a-b) \sqrt {a+b} \left (35 A b^2+24 a^2 C+22 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^5 d}+\frac {2 \sqrt {a+b} \left (48 a^3 C-12 a^2 b C+5 b^3 (7 A+5 C)+2 a b^2 (35 A+22 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}+\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^3 d}-\frac {12 a C \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{35 b^2 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{7 b d} \]
4/105*a*(a-b)*(35*A*b^2+24*C*a^2+22*C*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^5/d+2/105*(48*a^3*C-12*a^ 2*b*C+5*b^3*(7*A+5*C)+2*a*b^2*(35*A+22*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+2/105*(24*C*a^2+5*b^2 *(7*A+5*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/d-12/35*a*C*sec(d*x+c)*( a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/7*C*sec(d*x+c)^2*(a+b*sec(d*x+c)) ^(1/2)*tan(d*x+c)/b/d
Leaf count is larger than twice the leaf count of optimal. \(3255\) vs. \(2(393)=786\).
Time = 25.58 (sec) , antiderivative size = 3255, normalized size of antiderivative = 8.28 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]
(Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*((-8*a*(35*A*b^2 + 24*a^2*C + 22*b^2*C)*Sin[c + d*x])/(105*b^4) + (4*Sec[c + d*x]*(35*A*b^ 2*Sin[c + d*x] + 24*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^3) - (24*a*C*Sec[c + d*x]*Tan[c + d*x])/(35*b^2) + (4*C*Sec[c + d*x]^2*Tan[c + d*x])/(7*b)))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[a + b*Sec[c + d*x] ]) + (8*((4*a*A)/(3*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (32*a ^3*C)/(35*b^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (88*a*C)/(105 *b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*Sqrt[Sec[c + d*x]]) /(3*Sqrt[b + a*Cos[c + d*x]]) + (4*a^2*A*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]]) + (10*C*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x ]]) + (32*a^4*C*Sqrt[Sec[c + d*x]])/(35*b^4*Sqrt[b + a*Cos[c + d*x]]) + (6 4*a^2*C*Sqrt[Sec[c + d*x]])/(105*b^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^2*A* Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]]) + (3 2*a^4*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(35*b^4*Sqrt[b + a*Cos[c + d* x]]) + (88*a^2*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*b^2*Sqrt[b + a* Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + C*Sec[c + d*x]^ 2)*(2*a*(a + b)*(35*A*b^2 + 24*a^2*C + 22*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Co s[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Ellip ticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-48*a^3*C - 12*a^2*b* C + 5*b^3*(7*A + 5*C) - 2*a*b^2*(35*A + 22*C))*Sqrt[Cos[c + d*x]/(1 + C...
Time = 1.68 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4591, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4591 |
| \(\displaystyle \frac {2 \int \frac {\sec ^2(c+d x) \left (-6 a C \sec ^2(c+d x)+b (7 A+5 C) \sec (c+d x)+4 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-6 a C \sec ^2(c+d x)+b (7 A+5 C) \sec (c+d x)+4 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (-6 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (7 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sec (c+d x) \left (12 C a^2-2 b C \sec (c+d x) a-\left (24 C a^2+5 b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (12 C a^2-2 b C \sec (c+d x) a-\left (24 C a^2+5 b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (12 C a^2-2 b C \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (-24 C a^2-5 b^2 (7 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (b \left (-12 C a^2+35 A b^2+25 b^2 C\right )-2 a \left (24 C a^2+35 A b^2+22 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^2+35 A b^2+25 b^2 C\right )-2 a \left (24 C a^2+35 A b^2+22 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^2+35 A b^2+25 b^2 C\right )-2 a \left (24 C a^2+35 A b^2+22 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {-\frac {-\frac {\left (48 a^3 C-12 a^2 b C+2 a b^2 (35 A+22 C)+5 b^3 (7 A+5 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-2 a \left (24 a^2 C+35 A b^2+22 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {-\frac {\left (48 a^3 C-12 a^2 b C+2 a b^2 (35 A+22 C)+5 b^3 (7 A+5 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a \left (24 a^2 C+35 A b^2+22 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {2 \sqrt {a+b} \left (48 a^3 C-12 a^2 b C+2 a b^2 (35 A+22 C)+5 b^3 (7 A+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-2 a \left (24 a^2 C+35 A b^2+22 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {-\frac {-\frac {2 \left (24 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\frac {4 a (a-b) \sqrt {a+b} \left (24 a^2 C+35 A b^2+22 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 \sqrt {a+b} \left (48 a^3 C-12 a^2 b C+2 a b^2 (35 A+22 C)+5 b^3 (7 A+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}}{5 b}-\frac {12 a C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\) |
(2*C*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*b*d) + ((-12 *a*C*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) - (-1/3*( (4*a*(a - b)*Sqrt[a + b]*(35*A*b^2 + 24*a^2*C + 22*b^2*C)*Cot[c + d*x]*Ell ipticE[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*Sqrt[a + b]*(48*a^3*C - 12*a^2*b*C + 5*b^3*(7*A + 5*C) + 2*a* b^2*(35*A + 22*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))/b - (2*(24*a^2*C + 5*b^2*(7*A + 5*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b))/(7*b)
3.8.35.3.1 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_)]*((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_)]^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))*C sc[e + f*x] - a*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, 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. \(3472\) vs. \(2(359)=718\).
Time = 31.93 (sec) , antiderivative size = 3473, normalized size of antiderivative = 8.84
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3473\) |
| default | \(\text {Expression too large to display}\) | \(3513\) |
2/3*A/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(2*Elli pticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b*cos(d*x+c)^2- EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^2*cos(d*x+c )^2-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(co s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*cos (d*x+c)^2-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+ c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a *b*cos(d*x+c)^2+4*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 1/2)*a*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 ))^(1/2)*b^2*cos(d*x+c)-4*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1 /2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+ c)+1))^(1/2)*a^2*cos(d*x+c)-4*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b) )^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos( d*x+c)+1))^(1/2)*a*b*cos(d*x+c)+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+ b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c), ((a-b)/(a+b))^(1/2))*a*b-(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+...
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]