Integrand size = 43, antiderivative size = 229 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+9 B-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \]
1/10*(A+9*B-49*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic E(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(A+3*B-13*C)*(cos(1/2*d*x+1/2*c)^2 )^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d-1/1 0*(A+9*B-49*C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)-1/5*(A-B+C)*sin(d*x+c)/d/ (a+a*cos(d*x+c))^3/cos(d*x+c)^(1/2)+1/15*(2*A+3*B-8*C)*sin(d*x+c)/a/d/(a+a *cos(d*x+c))^2/cos(d*x+c)^(1/2)+1/6*(A+3*B-13*C)*sin(d*x+c)/d/(a^3+a^3*cos (d*x+c))/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 14.14 (sec) , antiderivative size = 1851, normalized size of antiderivative = 8.08 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]
(-4*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C* Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c] ]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) - (4*B*Cos[c/2 + (d*x) /2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c ]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d *x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + C ot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[ c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c ]^2]*(a + a*Sec[c + d*x])^3) + (52*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hyperge ometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*S qrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d* x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*( (4*(20*C - A*Cos[c] - 9*B*Cos[c] + 29*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/ (5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/...
Time = 1.43 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{5/2} (a \sec (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {a (A-B+11 C)+5 a (A+B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (A-B+11 C)+5 a (A+B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (A-B+11 C)+5 a (A+B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {(A-6 B+41 C) a^2+3 (2 A+3 B-8 C) \cos (c+d x) a^2}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {(A-6 B+41 C) a^2+3 (2 A+3 B-8 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int -\frac {3 a^3 (A+9 B-49 C)-5 a^3 (A+3 B-13 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {\int \frac {3 a^3 (A+9 B-49 C)-5 a^3 (A+3 B-13 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {\int \frac {3 a^3 (A+9 B-49 C)-5 a^3 (A+3 B-13 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx-5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )-5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-5 a^3 (A+3 B-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 a^3 (A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]) ^3) + ((2*a*(2*A + 3*B - 8*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(a + a *Cos[c + d*x])^2) + ((5*a^2*(A + 3*B - 13*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])) - ((-10*a^3*(A + 3*B - 13*C)*EllipticF[(c + d *x)/2, 2])/d + 3*a^3*(A + 9*B - 49*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/(2*a^2))/(3*a^2))/(10*a^2)
3.13.37.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(788\) vs. \(2(261)=522\).
Time = 3.39 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.45
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, method=_RETURNVERBOSE)
1/60*(-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6 5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*EllipticE(cos(1/2*d*x+1/2* c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))+147*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2 *c)^2*cos(1/2*d*x+1/2*c)-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(cos(1/2*d*x +1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*Elliptic E(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+9*B-49*C)*sin(1/2*d*x+1/2*c)^8-2*(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+147*B-817*C)*sin(1/2* d*x+1/2*c)^6+6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (A + 9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, A + 33 \, B - 188 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (A - 9 \, B + 59 \, C\right )} \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^3,x, algorithm="fricas")
-1/60*(2*(3*(A + 9*B - 49*C)*cos(d*x + c)^3 + 2*(2*A + 33*B - 188*C)*cos(d *x + c)^2 - 5*(A - 9*B + 59*C)*cos(d*x + c) - 60*C)*sqrt(cos(d*x + c))*sin (d*x + c) + 5*(sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*( I*A + 3*I*B - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*co s(d*x + c)^2 + sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c))*weierstrassPIn verse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 3*I*B + 1 3*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) + 3*(sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(- I*A - 9*I*B + 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*c os(d*x + c)^2 + sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c))*weierstrassZ eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3* (sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c) ^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^3,x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^3,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3* cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]