Integrand size = 43, antiderivative size = 186 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (2 A b+a B-6 b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^2 (A-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]
2/5*(10*B*a*b+5*b^2*(A-C)+a^2*(3*A+5*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))/d+2/3*(B*a^2+3*B*b^2+ 2*a*b*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(s in(1/2*d*x+1/2*c),2^(1/2))/d+2/5*a^2*(A-5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d +2*C*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*a*(2*A*b+B*a-6*C *b)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 19.11 (sec) , antiderivative size = 2421, normalized size of antiderivative = 13.02 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(Cos[c + d*x]^(9/2)*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-2*(3*a^2*A + 5*A*b^2 + 10*a*b*B + 5*a^2*C - 10*b^2*C + 3*a^2*A *Cos[2*c] + 5*A*b^2*Cos[2*c] + 10*a*b*B*Cos[2*c] + 5*a^2*C*Cos[2*c])*Csc[c ]*Sec[c])/(5*d) + (4*a*(2*A*b + a*B)*Cos[d*x]*Sin[c])/(3*d) + (2*a^2*A*Cos [2*d*x]*Sin[2*c])/(5*d) + (4*a*(2*A*b + a*B)*Cos[c]*Sin[d*x])/(3*d) + (4*b ^2*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (2*a^2*A*Cos[2*c]*Sin[2*d*x])/(5*d) ))/((b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x ])) - (8*a*A*b*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^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[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*C os[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*a^2*B*Cos[c + d *x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]] ]^2]*(a + b*Sec[c + d*x])^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[-(Sqrt[1 + Co t[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c ]]]])/(3*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*b^2*B*Cos[c + d*x]^4*Csc[c]*Hypergeomet ricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d...
Time = 1.21 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3512, 27, 3042, 3502, 27, 3042, 3227, 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.
\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{5/2} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle 2 \int \frac {(b+a \cos (c+d x)) \left (a (A-5 C) \cos ^2(c+d x)+(A b-C b+a B) \cos (c+d x)+b B+4 a C\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(b+a \cos (c+d x)) \left (a (A-5 C) \cos ^2(c+d x)+(A b-C b+a B) \cos (c+d x)+b B+4 a C\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(A b-C b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+b B+4 a C\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {2}{5} \int \frac {5 a (2 A b-6 C b+a B) \cos ^2(c+d x)+\left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)+5 b (b B+4 a C)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5 a (2 A b-6 C b+a B) \cos ^2(c+d x)+\left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)+5 b (b B+4 a C)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {5 a (2 A b-6 C b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b (b B+4 a C)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )+3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )+3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (B a^2+2 b (A+3 C) a+3 b^2 B\right )+3 \left ((3 A+5 C) a^2+10 b B a+5 b^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right ) \int \sqrt {\cos (c+d x)}dx\right )+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{d}\right )+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{d}\right )+\frac {10 a \sin (c+d x) \sqrt {\cos (c+d x)} (a B+2 A b-6 b C)}{3 d}\right )+\frac {2 a^2 (A-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{d \sqrt {\cos (c+d x)}}\) |
(2*a^2*(A - 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*C*(b + a*Cos[ c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (((6*(10*a*b*B + 5*b^2* (A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/d + (10*(a^2*B + 3*b ^2*B + 2*a*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d)/3 + (10*a*(2*A*b + a *B - 6*b*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/5
3.14.1.3.1 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[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_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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. \(678\) vs. \(2(224)=448\).
Time = 432.11 (sec) , antiderivative size = 679, normalized size of antiderivative = 3.65
method | result | size |
default | \(\frac {\frac {16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{2}}{5}-\frac {16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2}}{5}-\frac {16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a b}{3}-\frac {8 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2}}{3}+\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}}{5}+\frac {8 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b}{3}-\frac {4 a A b \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}+\frac {6 A \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}}{5}+2 A \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+\frac {4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}}{3}-\frac {2 B \,a^{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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3}-2 B \,b^{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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+4 B \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}-4 C a b \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-2 C \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{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}\) | \(679\) |
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
2/15*(24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^2-24*A*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4*a^2-40*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4 *a*b-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2+6*A*cos(1/2*d*x+1/2* c)*sin(1/2*d*x+1/2*c)^2*a^2+20*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a *b-10*a*A*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)* EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+15 *A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE (cos(1/2*d*x+1/2*c),2^(1/2))*b^2+10*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c )^2*a^2-5*B*a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2 ))+30*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+30*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 1/2*c)^2*b^2-30*C*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*(sin(1/2*d*x+1/2*c)^2 )^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2))*a^2-15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2 )*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^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 = 292, normalized size of antiderivative = 1.57 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, {\left (A + 3 \, C\right )} a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, {\left (A + 3 \, C\right )} a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{2} - 10 i \, B a b - 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{2} + 10 i \, B a b + 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 15 \, C b^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )} \]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(I*B*a^2 + 2*I*(A + 3*C)*a*b + 3*I*B*b^2)*cos(d*x + c)*we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B* a^2 - 2*I*(A + 3*C)*a*b - 3*I*B*b^2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-I*(3*A + 5*C)*a^2 - 10*I*B *a*b - 5*I*(A - C)*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPIn verse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(I*(3*A + 5*C)*a^ 2 + 10*I*B*a*b + 5*I*(A - C)*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^2*cos(d* x + c)^2 + 15*C*b^2 + 5*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sqrt(cos(d*x + c)) *sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*c os(d*x + c)^(5/2), x)
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*c os(d*x + c)^(5/2), x)
Time = 19.18 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.40 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(B*a^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2 , 2))/3))/d + (2*A*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*C*a^2*ellipticE (c/2 + (d*x)/2, 2))/d + (2*B*b^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*a*b *((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/ 3))/d + (4*B*a*b*ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a*b*ellipticF(c/2 + (d*x)/2, 2))/d - (2*A*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) + (2*C*b^2*sin( c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2 )*(sin(c + d*x)^2)^(1/2))