Integrand size = 35, antiderivative size = 209 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(A-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {2 (A-4 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/10*(A-49*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(A-13*C)*Inv erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-1/10*(A-49*C)*sin(d*x+c)/a^3/d/c os(d*x+c)^(1/2)-1/5*(A+C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^3 +2/15*(A-4*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^2+1/6*(A-13 *C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a^3+a^3*cos(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.37 (sec) , antiderivative size = 1237, normalized size of antiderivative = 5.92 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^ 3),x]
Output:
(-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 + C*Sec[c + d*x]^2)*S ec[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 + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*S ec[c + d*x])^3) + (52*C*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 + 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 + 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 + C*Sec[c + d *x]^2)*((-4*(-20*C + A*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] - 29*C*Sin[(d*x)/2]))/ (5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]) )/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + 11*C*Sin[(d*x )/2]))/(15*d) + (32*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (8*(A + 11*C)*Sec[ c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/ 2])/(5*d)))/(Sqrt[Cos[c + d*x]]*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[ c + d*x])^3) - (2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x +...
Time = 1.37 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4602, 3042, 3521, 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.
| \(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sec (c+d x)^2}{\cos (c+d x)^{5/2} (a \sec (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 4602 |
| \(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {a (A+11 C)+5 a (A-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (A+11 C)+5 a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (A+11 C)+5 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {(A+41 C) a^2+6 (A-4 C) \cos (c+d x) a^2}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(A+41 C) a^2+6 (A-4 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {3 a^3 (A-49 C)-5 a^3 (A-13 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {\int \frac {3 a^3 (A-49 C)-5 a^3 (A-13 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {\int \frac {3 a^3 (A-49 C)-5 a^3 (A-13 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx-5 a^3 (A-13 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )-5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)}-\frac {3 a^3 (A-49 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 a^3 (A-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}\) |
Input:
Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]
Output:
-1/5*((A + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3) + ((4*a*(A - 4*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x ])^2) + ((5*a^2*(A - 13*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[ c + d*x])) - ((-10*a^3*(A - 13*C)*EllipticF[(c + d*x)/2, 2])/d + 3*a^3*(A - 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)
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) I nt[(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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_.) + (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 + A*Cos [e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(678\) vs. \(2(192)=384\).
Time = 4.43 (sec) , antiderivative size = 679, normalized size of antiderivative = 3.25
Input:
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x,method=_RETUR NVERBOSE)
Output:
1/60*(-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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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))-65*C*Elliptic F(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*(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)*(2*sin(1/2*d*x+1/2*c)^2 -1)^(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))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*Elli pticE(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*(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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))-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)))*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-4 9*C)*sin(1/2*d*x+1/2*c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) ^(1/2)*(13*A-817*C)*sin(1/2*d*x+1/2*c)^6+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1 /2*d*x+1/2*c)^2)^(1/2)*(A-124*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2* c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-439*C)*sin(1/2*d*x+1/2*c)^2)/a^3/cos(1 /2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.46 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algori thm="fricas")
Output:
-1/60*(2*(3*(A - 49*C)*cos(d*x + c)^3 + 4*(A - 94*C)*cos(d*x + c)^2 - 5*(A + 59*C)*cos(d*x + c) - 60*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2) *(I*A - 13*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(I*A - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 13*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 13*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt( 2)*(-I*A + 13*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + c) ^3 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 13*I*C)*co s(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3* (sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A + 49*I*C)*cos(d* x + c)^3 + 3*sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 49*I *C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A - 49*I*C)*cos(d*x + c)^4 + 3*s qrt(2)*(I*A - 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 49*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 49*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierst rassPInverse(-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. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**3,x)
Output:
Timed out
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algori thm="maxima")
Output:
Timed out
\[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algori thm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(5/2 )), x)
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \] Input:
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^3),x)
Output:
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^3), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{3}}d x \right ) c}{a^{3}} \] Input:
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*sec(c + d*x)**3 + 3*cos(c + d*x)* *3*sec(c + d*x)**2 + 3*cos(c + d*x)**3*sec(c + d*x) + cos(c + d*x)**3),x)* a + int((sqrt(cos(c + d*x))*sec(c + d*x)**2)/(cos(c + d*x)**3*sec(c + d*x) **3 + 3*cos(c + d*x)**3*sec(c + d*x)**2 + 3*cos(c + d*x)**3*sec(c + d*x) + cos(c + d*x)**3),x)*c)/a**3