Integrand size = 43, antiderivative size = 180 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (6 a b B+b^2 (3 A+C)+a^2 (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b (3 b B+4 a C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]
2*(B*a^2-B*b^2+2*a*b*(A-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*(6*B*a*b+b^2*(3*A+C)+a^2*(A+ 3*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))/d+2/3*C*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(3/2) +2/3*b*(3*B*b+4*C*a)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*a^2*(A-C)*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 = 12.27 (sec) , antiderivative size = 2307, normalized size of antiderivative = 12.82 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
Integrate[Cos[c + d*x]^(3/2)*(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*(2*a*A*b + a^2*B - 2*b^2*B - 4*a*b*C + 2*a*A*b*Cos[2*c] + a^ 2*B*Cos[2*c])*Csc[c]*Sec[c])/d + (4*a^2*A*Cos[d*x]*Sin[c])/(3*d) + (4*a^2* A*Cos[c]*Sin[d*x])/(3*d) + (4*b^2*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (4*Sec[c]*Sec[c + d*x]*(b^2*C*Sin[c] + 3*b^2*B*Sin[d*x] + 6*a*b*C*Sin[d* x]))/(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])) - (4*a^2*A*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 - A rcTan[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*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*A*b^2 *Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcT an[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[-(S qrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Ar cTan[Cot[c]]]])/(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]) - (8*a*b*B*Cos[c + d*x]^4*Csc[c]*Hyp ergeometricPFQ[{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[Co...
Time = 1.19 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.99, 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, 3510, 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.
| \(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^{3/2} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {2}{3} \int \frac {(b+a \cos (c+d x)) \left (3 a (A-C) \cos ^2(c+d x)+(3 A b+C b+3 a B) \cos (c+d x)+3 b B+4 a C\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {(b+a \cos (c+d x)) \left (3 a (A-C) \cos ^2(c+d x)+(3 A b+C b+3 a B) \cos (c+d x)+3 b B+4 a C\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 A b+C b+3 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 b B+4 a C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-2 \int -\frac {3 (A-C) \cos ^2(c+d x) a^2+4 C a^2+6 b B a+3 A b^2+b^2 C+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 (A-C) \cos ^2(c+d x) a^2+4 C a^2+6 b B a+3 A b^2+b^2 C+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {3 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+4 C a^2+6 b B a+3 A b^2+b^2 C+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {3 \left ((A+3 C) a^2+6 b B a+b^2 (3 A+C)+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+b^2 (3 A+C)+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(A+3 C) a^2+6 b B a+b^2 (3 A+C)+3 \left (B a^2+2 b (A-C) a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \int \sqrt {\cos (c+d x)}dx+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{3} \left (\left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 (A+3 C)+6 a b B+b^2 (3 A+C)\right )}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A-C)-b^2 B\right )}{d}+\frac {2 a^2 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d}+\frac {2 b (4 a C+3 b B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
(2*C*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((6*( a^2*B - b^2*B + 2*a*b*(A - C))*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*b*B + b^2*(3*A + C) + a^2*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/d + (2*b*(3*b* B + 4*a*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*a^2*(A - C)*Sqrt[Cos[ c + d*x]]*Sin[c + d*x])/d)/3
3.14.2.3.1 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[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, 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. \(1302\) vs. \(2(218)=436\).
Time = 3.52 (sec) , antiderivative size = 1303, normalized size of antiderivative = 7.24
int(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d *x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(6*A*EllipticE( cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/ 2*c)^2-1)^(1/2)*a*b-6*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x +1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b-6*C*EllipticE(cos(1/ 2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*a*b+8*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^2-8*A*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2+2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2 *c)^2*a^2+2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2-A*EllipticF(cos( 1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*a^2-3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2 *c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^2+3*B*EllipticE(cos(1/2*d* x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^ (1/2)*a^2-3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^2-3*C*EllipticF(cos(1/2*d*x+1/2* c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)* a^2-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( 2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^2-24*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/ 2*c)^4*a*b+12*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^2*a*b-12*B*cos(1/2*d *x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2+6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.62 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} {\left (-i \, {\left (A + 3 \, C\right )} a^{2} - 6 i \, B a b - i \, {\left (3 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, {\left (A + 3 \, C\right )} a^{2} + 6 i \, B a b + i \, {\left (3 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, {\left (A - C\right )} a b + i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, {\left (A - C\right )} a b - i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (A a^{2} \cos \left (d x + c\right )^{2} + C b^{2} + 3 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \]
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
1/3*(sqrt(2)*(-I*(A + 3*C)*a^2 - 6*I*B*a*b - I*(3*A + C)*b^2)*cos(d*x + c) ^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I* (A + 3*C)*a^2 + 6*I*B*a*b + I*(3*A + C)*b^2)*cos(d*x + c)^2*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-I*B*a^2 - 2*I*(A - C)*a*b + I*B*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(I*B*a^2 + 2*I*(A - C)*a*b - I*B*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInve rse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(A*a^2*cos(d*x + c)^2 + C*b ^2 + 3*(2*C*a*b + B*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d *cos(d*x + c)^2)
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(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)^(3/2), x)
\[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(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)^(3/2), x)
Time = 20.35 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.49 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,C\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(A*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*B*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*b^2*ellipticF (c/2 + (d*x)/2, 2))/d + (2*C*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (4*A*a*b *ellipticE(c/2 + (d*x)/2, 2))/d + (4*B*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*b^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*co s(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*b^2*sin(c + d*x)*hypergeom ([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^ 2)^(1/2)) + (4*C*a*b*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))