Integrand size = 43, antiderivative size = 201 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \left (10 a b B-5 a^2 (A-C)+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^2 B+b^2 B+2 a b (3 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b (5 b B+4 a C) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2+10 a b B+4 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (b+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:
-2/5*(10*B*a*b-5*a^2*(A-C)+b^2*(5*A+3*C))*EllipticE(sin(1/2*d*x+1/2*c),2^( 1/2))/d+2/3*(3*B*a^2+B*b^2+2*a*b*(3*A+C))*InverseJacobiAM(1/2*d*x+1/2*c,2^ (1/2))/d+2/15*b*(5*B*b+4*C*a)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*A*b^2+1 0*B*a*b+4*C*a^2+3*C*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/5*C*(b+a*cos(d*x+ c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.53 (sec) , antiderivative size = 2427, normalized size of antiderivative = 12.07 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:
Integrate[Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(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*(5*a^2*A - 10*A*b^2 - 20*a*b*B - 10*a^2*C - 6*b^2*C + 5*a^2* A*Cos[2*c])*Csc[c]*Sec[c])/(5*d) + (4*b^2*C*Sec[c]*Sec[c + d*x]^3*Sin[d*x] )/(5*d) + (4*Sec[c]*Sec[c + d*x]^2*(3*b^2*C*Sin[c] + 5*b^2*B*Sin[d*x] + 10 *a*b*C*Sin[d*x]))/(15*d) + (4*Sec[c]*Sec[c + d*x]*(5*b^2*B*Sin[c] + 10*a*b *C*Sin[c] + 15*A*b^2*Sin[d*x] + 30*a*b*B*Sin[d*x] + 15*a^2*C*Sin[d*x] + 9* b^2*C*Sin[d*x]))/(15*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]*Hypergeometric PFQ[{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 - Ar cTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[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]) - (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*S ec[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 + S in[d*x - ArcTan[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]) - (4*b^2*B*Cos[c + d*x]^4 *Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^...
Time = 1.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, 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, 3500, 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.
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4600 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {2}{5} \int \frac {(b+a \cos (c+d x)) \left (a (5 A-C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+4 a C\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(b+a \cos (c+d x)) \left (a (5 A-C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 b B+4 a C\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(5 A b+3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 b B+4 a C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int -\frac {3 a^2 (5 A-C) \cos ^2(c+d x)+5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right ) \cos (c+d x)+3 \left (4 C a^2+10 b B a+5 A b^2+3 b^2 C\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 a^2 (5 A-C) \cos ^2(c+d x)+5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right ) \cos (c+d x)+3 \left (4 C a^2+10 b B a+5 A b^2+3 b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 a^2 (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (4 C a^2+10 b B a+5 A b^2+3 b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )-3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )-3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {5 \left (3 B a^2+2 b (3 A+C) a+b^2 B\right )-3 \left (-5 (A-C) a^2+10 b B a+b^2 (5 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)}dx+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (3 a^2 B+2 a b (3 A+C)+b^2 B\right )}{d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^2 (A-C)+10 a b B+b^2 (5 A+3 C)\right )}{d}+\frac {6 \sin (c+d x) \left (4 a^2 C+10 a b B+5 A b^2+3 b^2 C\right )}{d \sqrt {\cos (c+d x)}}\right )+\frac {2 b (4 a C+5 b B) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[ c + d*x]^2),x]
Output:
(2*C*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((2*b *(5*b*B + 4*a*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((-6*(10*a*b*B - 5*a^2*(A - C) + b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/d + (10*(3*a^ 2*B + b^2*B + 2*a*b*(3*A + C))*EllipticF[(c + d*x)/2, 2])/d + (6*(5*A*b^2 + 10*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/3)/5
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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. \(972\) vs. \(2(188)=376\).
Time = 6.89 (sec) , antiderivative size = 973, normalized size of antiderivative = 4.84
Input:
int(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*a^2*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 2*(A*b^2+2*B*a*b+C*a^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(- 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 *cos(1/2*d*x+1/2*c)-(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2))+2/5*C*b^2/(8*sin(1/2*d*x+1/ 2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c )^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/ 2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2* d*x+1/2*c)-3*(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))*(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*b*(B*b+2*C*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d* x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(s in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d *x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 2)))-2*a^2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.59 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, B a^{2} + 2 i \, {\left (3 \, A + C\right )} a b + i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{2} - 2 i \, {\left (3 \, A + C\right )} a b - i \, B b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, {\left (A - C\right )} a^{2} + 10 i \, B a b + i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, {\left (A - C\right )} a^{2} - 10 i \, B a b - i \, {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, C b^{2} + 3 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
Output:
-1/15*(5*sqrt(2)*(3*I*B*a^2 + 2*I*(3*A + C)*a*b + I*B*b^2)*cos(d*x + c)^3* weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3* I*B*a^2 - 2*I*(3*A + C)*a*b - I*B*b^2)*cos(d*x + c)^3*weierstrassPInverse( -4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(-5*I*(A - C)*a^2 + 10*I *B*a*b + I*(5*A + 3*C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(5*I*(A - C )*a^2 - 10*I*B*a*b - I*(5*A + 3*C)*b^2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*C*b^ 2 + 3*(5*C*a^2 + 10*B*a*b + (5*A + 3*C)*b^2)*cos(d*x + c)^2 + 5*(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)^3)
\[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**(1/2)*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+ c)**2),x)
Output:
Integral((a + b*sec(c + d*x))**2*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)* sqrt(cos(c + d*x)), x)
\[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*s qrt(cos(d*x + c)), x)
\[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*s qrt(cos(d*x + c)), x)
Time = 15.57 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.54 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6\,C\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,C\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,C\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,B\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(cos(c + d*x)^(1/2)*(a + b/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos( c + d*x)^2),x)
Output:
(6*C*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*C* a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 ) + 20*C*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*A*a^2* ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (4*A*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*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)) + (2*B*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*B*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))
\[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a^{3}+\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) b^{2} c +2 \left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a b c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b^{3}+\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a^{2} c +3 \left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a^{2} b \] Input:
int(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
int(sqrt(cos(c + d*x)),x)*a**3 + int(sqrt(cos(c + d*x))*sec(c + d*x)**4,x) *b**2*c + 2*int(sqrt(cos(c + d*x))*sec(c + d*x)**3,x)*a*b*c + int(sqrt(cos (c + d*x))*sec(c + d*x)**3,x)*b**3 + int(sqrt(cos(c + d*x))*sec(c + d*x)** 2,x)*a**2*c + 3*int(sqrt(cos(c + d*x))*sec(c + d*x)**2,x)*a*b**2 + 3*int(s qrt(cos(c + d*x))*sec(c + d*x),x)*a**2*b