Integrand size = 43, antiderivative size = 180 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {(7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(10 A-5 B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}-\frac {(7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
-(7*A-4*B+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin (1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(10*A-5*B+2*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^2/d-1/3*(7 *A-4*B+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos (d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+1/3*(10*A-5*B+2*C)*sin(d*x+c )*cos(d*x+c)^(1/2)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.79 (sec) , antiderivative size = 1710, normalized size of antiderivative = 9.50 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
(-40*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(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[-(S qrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Ar cTan[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqr t[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) + (20*B*Cos[c/2 + (d*x)/2]^4*Csc[c /2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[ c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqr t[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])^2) - (8*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(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]]]]*S qrt[-(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])^2) + (Cos[c/2 + (d*x)/2]^4*Sqr t[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*(3*A - 2*B + C + 4*A*Cos[c] - 2*B*Cos[c])*Csc[c])/d + (16*A*Cos[d*x]*Sin[c])/(3*d) + (8* Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2] - 2*B*Sin[(d*x)/2] + C*Si...
Time = 1.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3456, 27, 3042, 3227, 3042, 3115, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{(a \sec (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{(a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int -\frac {\cos ^{\frac {3}{2}}(c+d x) (a (5 A-5 B-C)-3 a (3 A-B+C) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a (5 A-5 B-C)-3 a (3 A-B+C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (5 A-5 B-C)-3 a (3 A-B+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int 3 \sqrt {\cos (c+d x)} \left (a^2 (7 A-4 B+C)-a^2 (10 A-5 B+2 C) \cos (c+d x)\right )dx}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \sqrt {\cos (c+d x)} \left (a^2 (7 A-4 B+C)-a^2 (10 A-5 B+2 C) \cos (c+d x)\right )dx}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a^2 (7 A-4 B+C)-a^2 (10 A-5 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (7 A-4 B+C) \int \sqrt {\cos (c+d x)}dx-a^2 (10 A-5 B+2 C) \int \cos ^{\frac {3}{2}}(c+d x)dx\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (7 A-4 B+C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (10 A-5 B+2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (7 A-4 B+C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (7 A-4 B+C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {2 a^2 (7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {2 a^2 (7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a^2 (10 A-5 B+2 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )}{a^2}+\frac {2 (7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^2) - ((2*(7*A - 4*B + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*((2*a^2*(7*A - 4*B + C)*EllipticE[(c + d*x)/2, 2])/d - a^2*(1 0*A - 5*B + 2*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d* x]]*Sin[c + d*x])/(3*d))))/a^2)/(6*a^2)
3.13.26.3.1 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[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, 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] && (In tegerQ[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. \(471\) vs. \(2(218)=436\).
Time = 4.11 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.62
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (10 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (10 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+16 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-76 A +24 B -12 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (84 A -34 B +16 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-25 A +11 B -5 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(472\) |
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(sin(1/2* d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(10*A*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*Elli pticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 ))+2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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)^2+2*(sin(1/2*d*x+1/2*c )^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(10*A*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*EllipticF(cos (1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*El lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 2)))*cos(1/2*d*x+1/2*c)+16*A*sin(1/2*d*x+1/2*c)^8+(-76*A+24*B-12*C)*sin(1/ 2*d*x+1/2*c)^6+(84*A-34*B+16*C)*sin(1/2*d*x+1/2*c)^4+(-25*A+11*B-5*C)*sin( 1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-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 higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.26 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (13 \, A - 6 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 10 \, A - 5 \, B + 2 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, 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 (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, 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 (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, 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 )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="fricas")
1/6*(2*(2*A*cos(d*x + c)^2 + (13*A - 6*B + 3*C)*cos(d*x + c) + 10*A - 5*B + 2*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-10*I*A + 5*I*B - 2*I*C )*cos(d*x + c)^2 - 2*sqrt(2)*(10*I*A - 5*I*B + 2*I*C)*cos(d*x + c) + sqrt( 2)*(-10*I*A + 5*I*B - 2*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I* sin(d*x + c)) + (sqrt(2)*(10*I*A - 5*I*B + 2*I*C)*cos(d*x + c)^2 - 2*sqrt( 2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c) + sqrt(2)*(10*I*A - 5*I*B + 2*I* C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2) *(7*I*A - 4*I*B + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(7*I*A - 4*I*B + I*C)*co s(d*x + c) + sqrt(2)*(7*I*A - 4*I*B + I*C))*weierstrassZeta(-4, 0, weierst rassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-7*I*A + 4*I*B - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-7*I*A + 4*I*B - I*C)*cos(d*x + c) + sqrt(2)*(-7*I*A + 4*I*B - I*C))*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^ 2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*se c(d*x + c) + a)^2, x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]