Integrand size = 43, antiderivative size = 214 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {4 a^2 (5 B+4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (3 A+2 B+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]
2/15*a^2*(15*A+25*B+17*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*C*(a+a*sec(d*x +c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/15*(5*B+4*C)*(a^2+a^2*sec(d*x+c))*s in(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2*(5*B+4*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))*cos(d*x+c)^(1/2 )*sec(d*x+c)^(1/2)/d+4/3*a^2*(3*A+2*B+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))*cos(d*x+c)^(1/2)*sec( d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.47 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.24 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-90 i B \cos (c+d x)-72 i C \cos (c+d x)-30 i B \cos (3 (c+d x))-24 i C \cos (3 (c+d x))+40 (3 A+2 B+C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 i (5 B+4 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+15 A \sin (c+d x)+30 B \sin (c+d x)+36 C \sin (c+d x)+10 B \sin (2 (c+d x))+20 C \sin (2 (c+d x))+15 A \sin (3 (c+d x))+30 B \sin (3 (c+d x))+24 C \sin (3 (c+d x))\right )}{30 d} \]
Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) /Sqrt[Sec[c + d*x]],x]
(a^2*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((-90*I)*B*Cos[c + d*x] - (72*I)*C*Cos[c + d*x] - (30*I)*B*Cos[3*(c + d*x)] - (24*I)*C*Cos[3*(c + d* x)] + 40*(3*A + 2*B + C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + (( 2*I)*(5*B + 4*C)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/ 4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 15*A*Sin[c + d*x] + 30*B* Sin[c + d*x] + 36*C*Sin[c + d*x] + 10*B*Sin[2*(c + d*x)] + 20*C*Sin[2*(c + d*x)] + 15*A*Sin[3*(c + d*x)] + 30*B*Sin[3*(c + d*x)] + 24*C*Sin[3*(c + d *x)]))/(30*d*E^(I*d*x))
Time = 1.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4576, 27, 3042, 4506, 27, 3042, 4485, 25, 3042, 4274, 3042, 4258, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^2 (a (5 A-C)+a (5 B+4 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^2 (a (5 A-C)+a (5 B+4 C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (5 A-C)+a (5 B+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left ((15 A-5 B-7 C) a^2+(15 A+25 B+17 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(\sec (c+d x) a+a) \left ((15 A-5 B-7 C) a^2+(15 A+25 B+17 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((15 A-5 B-7 C) a^2+(15 A+25 B+17 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \int -\frac {3 a^3 (5 B+4 C)-5 a^3 (3 A+2 B+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \int \frac {3 a^3 (5 B+4 C)-5 a^3 (3 A+2 B+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \int \frac {3 a^3 (5 B+4 C)-5 a^3 (3 A+2 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (3 a^3 (5 B+4 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (3 A+2 B+C) \int \sqrt {\sec (c+d x)}dx\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (3 a^3 (5 B+4 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (3 A+2 B+C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (3 a^3 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 a^3 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (3 a^3 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a^3 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (\frac {6 a^3 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a^3 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (\frac {6 a^3 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 a^3 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )\right )+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
(2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2*( 5*B + 4*C)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3*d) + (-2*((6*a^3*(5*B + 4*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq rt[Sec[c + d*x]])/d - (10*a^3*(3*A + 2*B + C)*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d) + (2*a^3*(15*A + 25*B + 17*C)*Sqrt [Sec[c + d*x]]*Sin[c + d*x])/d)/3)/(5*a)
3.6.43.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(880\) vs. \(2(242)=484\).
Time = 5.37 (sec) , antiderivative size = 881, normalized size of antiderivative = 4.12
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(881\) |
| parts | \(\text {Expression too large to display}\) | \(1065\) |
int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, method=_RETURNVERBOSE)
-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(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*B*(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*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*C/sin(1/2* d*x+1/2*c)^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)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c) ^2-1)^(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2 *sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*s in(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1 /2*c),2^(1/2))*(2*sin(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)+8*(1/4*B+1/2*C)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*s in(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*(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+...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, {\left (5 \, A + 10 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1 /2),x, algorithm="fricas")
-2/15*(5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)^2*weierstrassPInverse( -4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 2*B + C)*a^2*co s(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3 *I*sqrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstra ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(5*B + 4*C )*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos (d*x + c) - I*sin(d*x + c))) - (3*(5*A + 10*B + 8*C)*a^2*cos(d*x + c)^2 + 5*(B + 2*C)*a^2*cos(d*x + c) + 3*C*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/( d*cos(d*x + c)^2)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/s qrt(sec(d*x + c)), x)
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/s qrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
int(((a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/co s(c + d*x))^(1/2),x)