Integrand size = 41, antiderivative size = 137 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(3 A+5 B+19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec (c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(7 A+B-9 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]
1/32*(3*A+5*B+19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c)) ^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x +c))^(5/2)+1/16*(7*A+B-9*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.91 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.37 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (2 \sqrt {2} (5 B+19 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\cos (c+d x) (5 B-13 C+(B-9 C) \cos (c+d x)) \sqrt {1-\sec (c+d x)}+16 A \cos ^4\left (\frac {1}{2} (c+d x)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt {1-\sec (c+d x)}\right ) \sec ^2(c+d x) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se c[c + d*x])^(5/2),x]
((2*Sqrt[2]*(5*B + 19*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4 + Cos[c + d*x]*(5*B - 13*C + (B - 9*C)*Cos[c + d*x])*Sqrt[1 - Se c[c + d*x]] + 16*A*Cos[(c + d*x)/2]^4*Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c + d*x])/2]*Sqrt[1 - Sec[c + d*x]])*Sec[c + d*x]^2*Tan[c + d*x])/(16* d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 0.66 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3042, 4566, 27, 3042, 4488, 3042, 4282, 216}
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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4566 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) (2 a (3 A+B-C)-a (A-B-7 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) (2 a (3 A+B-C)-a (A-B-7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a (3 A+B-C)-a (A-B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4488 |
| \(\displaystyle \frac {\frac {1}{4} (3 A+5 B+19 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx+\frac {a (7 A+B-9 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} (3 A+5 B+19 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {a (7 A+B-9 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {a (7 A+B-9 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(3 A+5 B+19 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{2 d}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {(3 A+5 B+19 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}+\frac {a (7 A+B-9 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*((A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2) ) + (((3*A + 5*B + 19*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a *Sec[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) + (a*(7*A + B - 9*C)*Tan[c + d*x]) /(2*d*(a + a*Sec[c + d*x])^(3/2)))/(8*a^2)
3.6.30.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[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[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(a*b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
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*A - b*B + a*C))*Cot[e + f*x]*Csc[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[Csc[e + f*x] *(a + b*Csc[e + f*x])^(m + 1)*Simp[a*B - b*C - 2*A*b*(m + 1) - (b*B*(m + 2) - a*(A*(m + 2) - C*(m - 1)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f , A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(469\) vs. \(2(118)=236\).
Time = 1.16 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.43
| method | result | size |
| default | \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 A \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}-2 B \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}+2 C \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}-5 A \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 B \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+11 C \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+5 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+19 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 a^{3} d}\) | \(470\) |
| parts | \(\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}+\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}+\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+11 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+19 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\) | \(583\) |
int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,me thod=_RETURNVERBOSE)
1/32/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^ 2*csc(d*x+c)^2-1)^(1/2)*(2*A*(1-cos(d*x+c))^3*((1-cos(d*x+c))^2*csc(d*x+c) ^2-1)^(1/2)*csc(d*x+c)^3-2*B*(1-cos(d*x+c))^3*((1-cos(d*x+c))^2*csc(d*x+c) ^2-1)^(1/2)*csc(d*x+c)^3+2*C*(1-cos(d*x+c))^3*((1-cos(d*x+c))^2*csc(d*x+c) ^2-1)^(1/2)*csc(d*x+c)^3-5*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot (d*x+c)+csc(d*x+c))-3*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+ c)+csc(d*x+c))+11*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+c sc(d*x+c))+3*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^ (1/2))+5*B*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2 ))+19*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 511, normalized size of antiderivative = 3.73 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 5 \, B + 19 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (7 \, A + B - 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A + 5 \, B - 13 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 5 \, B + 19 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 5 \, B + 19 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (7 \, A + B - 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A + 5 \, B - 13 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="fricas")
[-1/64*(sqrt(2)*((3*A + 5*B + 19*C)*cos(d*x + c)^3 + 3*(3*A + 5*B + 19*C)* cos(d*x + c)^2 + 3*(3*A + 5*B + 19*C)*cos(d*x + c) + 3*A + 5*B + 19*C)*sqr t(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos( d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d* x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((7*A + B - 9*C)*cos(d*x + c)^2 + (3*A + 5*B - 13*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d *x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((3*A + 5*B + 19*C)*cos(d*x + c)^3 + 3*(3*A + 5*B + 19*C)*cos(d*x + c)^2 + 3*(3*A + 5*B + 19*C)*cos(d*x + c) + 3*A + 5 *B + 19*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((7*A + B - 9*C)*cos(d*x + c)^2 + (3*A + 5*B - 13*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos (d*x + c) + a^3*d)]
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2), x)
Timed out. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="maxima")
Time = 1.65 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.34 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (5 \, A a^{5} + 3 \, B a^{5} - 11 \, C a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (3 \, A + 5 \, B + 19 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="giac")
-1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C*a ^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(5*A*a^5 + 3* B*a^5 - 11*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)* (3*A + 5*B + 19*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/ 2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]