Integrand size = 35, antiderivative size = 408 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (3 b^3 (A-C)+8 a^3 C+6 a^2 b C-a b^2 (A+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
2/3*(3*b^4*(A-C)-8*a^4*C+a^2*b^2*(A+15*C))*cot(d*x+c)*EllipticE((a+b*sec(d *x+c))^(1/2)/(a+b)^(1/2),((a+b)/(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/(a^2-b^2)/d/(a+b)^(1/2)-2/3*(3*b^3* (A-C)+8*a^3*C+6*a^2*b*C-a*b^2*(A+9*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c ))^(1/2)/(a+b)^(1/2),((a+b)/(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/(a^2-b^2)/d/(a+b)^(1/2)+2/3*a*(A*b^2+C* a^2)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(3*A*b^4-5*a^4* C+a^2*b^2*(A+9*C))*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3819\) vs. \(2(408)=816\).
Time = 30.02 (sec) , antiderivative size = 3819, normalized size of antiderivative = 9.36 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-4*(a^2*A*b^ 2 + 3*A*b^4 - 8*a^4*C + 15*a^2*b^2*C - 3*b^4*C)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) + (4*(A*b^2*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(3*b*(-a^2 + b^ 2)*(b + a*Cos[c + d*x])^2) + (8*(a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d* x] - 2*a^4*C*Sin[c + d*x] + 4*a^2*b^2*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2) ^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) - (4*(b + a*Cos[c + d*x])^2*((2*a^2*A)/(3*(-a^2 + b^2)^2*Sq rt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*b^2)/((-a^2 + b^2)^2*Sqr t[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (10*a^2*C)/((-a^2 + b^2)^2*Sqr t[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^4*C)/(3*b^2*(-a^2 + b^2) ^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*b^2*C)/((-a^2 + b^2)^ 2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^3*A*Sqrt[Sec[c + d*x ]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (2*a*A*b*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^5*C*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (34*a^3*C*Sqrt[ Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (6*a*b*C*Sq rt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3*A*Cos [2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d *x]]) + (2*a*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt [b + a*Cos[c + d*x]]) - (16*a^5*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/...
Time = 1.59 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4579, 27, 3042, 4568, 27, 25, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4579 |
| \(\displaystyle \frac {2 \int -\frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)-a \left (-2 C a^2+A b^2+3 b^2 C\right ) \sec (c+d x)+3 b \left (C a^2+A b^2\right )\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}+\frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)-a \left (-2 C a^2+A b^2+3 b^2 C\right ) \sec (c+d x)+3 b \left (C a^2+A b^2\right )\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 b \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2-a \left (-2 C a^2+A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 b \left (C a^2+A b^2\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {2 \int -\frac {\sec (c+d x) \left (2 a \left (2 A b^2-\left (a^2-3 b^2\right ) C\right ) b^2+\left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\int -\frac {\sec (c+d x) \left (2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {\int \frac {\sec (c+d x) \left (2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-b \left (-8 C a^4+b^2 (A+15 C) a^2+3 b^4 (A-C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {-b \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\left (b (a-b) \left (8 a^3 C+6 a^2 b C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {-b \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (b (a-b) \left (8 a^3 C+6 a^2 b C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {-b \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C+6 a^2 b C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b \left (a^2-b^2\right )}-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C+6 a^2 b C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}\) |
(2*a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x ])^(3/2)) - (-(((2*(a - b)*Sqrt[a + b]*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2* (A + 15*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*(a - b)*Sqrt[a + b]*(3*b^3*(A - C) + 8*a^3*C + 6*a^2*b*C - a*b^2*(A + 9*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqr t[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)/(b*(a^2 - b^2) )) - (2*(3*A*b^4 - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[c + d*x])/((a^2 - b^2) *d*Sqrt[a + b*Sec[c + d*x]]))/(3*b^2*(a^2 - b^2))
3.8.49.3.1 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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] )^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(cs c[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*(A*b^2 + a^2* C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*Simp[b*(m + 1)*(a^2*C + A*b^2) - a*(A*b^2*(m + 2) + C*(a^2 + b^ 2*(m + 1)))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(7451\) vs. \(2(378)=756\).
Time = 16.01 (sec) , antiderivative size = 7452, normalized size of antiderivative = 18.26
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(7452\) |
| default | \(\text {Expression too large to display}\) | \(7521\) |
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a)/(b ^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]