Integrand size = 35, antiderivative size = 196 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {16 a^2 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (15 A+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {8 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
-16/5*a^2*C*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x +c)^(1/2)/d+4/3*a^2*(3*A+C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c ,2^(1/2))*sec(d*x+c)^(1/2)/d+2/15*a^2*(15*A+17*C)*sec(d*x+c)^(1/2)*sin(d*x +c)/d+2/5*C*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+8/15*C*sec(d* x+c)^(1/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.71 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \left (12 C \left (1+e^{2 i (c+d x)}\right )+12 C \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 (3 A+C) e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {-3 (-5 A-16 C+5 A \cos (2 c)) \cos (d x) \csc (c)+30 A \cos (c) \sin (d x)+2 C (10+3 \sec (c+d x)) \tan (c+d x)}{\sec ^{\frac {7}{2}}(c+d x)}\right )}{30 d (A+2 C+A \cos (2 (c+d x)))} \] Input:
Integrate[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x ]],x]
Output:
(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*(((-4* I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Cos[c + d*x]^4* (12*C*(1 + E^((2*I)*(c + d*x))) + 12*C*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I )*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5* (3*A + C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))] *Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x)) *(-1 + E^((2*I)*c))) + (-3*(-5*A - 16*C + 5*A*Cos[2*c])*Cos[d*x]*Csc[c] + 30*A*Cos[c]*Sin[d*x] + 2*C*(10 + 3*Sec[c + d*x])*Tan[c + d*x])/Sec[c + d*x ]^(7/2)))/(30*d*(A + 2*C + A*Cos[2*(c + d*x)]))
Time = 1.20 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4577, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4577 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^2 (a (5 A-C)+4 a C \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^2 (a (5 A-C)+4 a C \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (5 A-C)+4 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left ((15 A-7 C) a^2+(15 A+17 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(\sec (c+d x) a+a) \left ((15 A-7 C) a^2+(15 A+17 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x)}}dx+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((15 A-7 C) a^2+(15 A+17 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \int -\frac {12 a^3 C-5 a^3 (3 A+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \int \frac {12 a^3 C-5 a^3 (3 A+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \int \frac {12 a^3 C-5 a^3 (3 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (12 a^3 C \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (3 A+C) \int \sqrt {\sec (c+d x)}dx\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (12 a^3 C \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (3 A+C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (12 a^3 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 a^3 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (12 a^3 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a^3 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (\frac {24 a^3 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a^3 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {2 a^3 (15 A+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-2 \left (\frac {24 a^3 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 a^3 (3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )\right )+\frac {8 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{3 d}}{5 a}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d}\) |
Input:
Int[((a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]
Output:
(2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((8*C *Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3*d) + (-2*((2 4*a^3*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (10*a^3*(3*A + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec [c + d*x]])/d) + (2*a^3*(15*A + 17*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/ 3)/(5*a)
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_)]^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*Csc[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n *Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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. \(755\) vs. \(2(175)=350\).
Time = 3.16 (sec) , antiderivative size = 756, normalized size of antiderivative = 3.86
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(756\) |
| parts | \(\text {Expression too large to display}\) | \(1043\) |
Input:
int((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETUR NVERBOSE)
Output:
-4/15*a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(60*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(2*sin(1 /2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+96*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x +1/2*c)^6-20*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-48*C*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d *x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-60*A*cos(1/2*d*x+1/2*c)*sin(1/2*d* x+1/2*c)^4+60*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1 /2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-116*C*cos(1 /2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*( sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2* d*x+1/2*c)^2+48*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^ (1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+15*sin(1/ 2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+37*sin( 1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C-5*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*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(...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.14 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {2} C a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {2} C a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, {\left (5 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \, C a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(3*A + 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 + C)*a^2*cos(d*x + c)^2 *weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(2)* C*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos (d*x + c) + I*sin(d*x + c))) - 12*I*sqrt(2)*C*a^2*cos(d*x + c)^2*weierstra ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (3*(5*A + 8*C)*a^2*cos(d*x + c)^2 + 10*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. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/sqrt(sec(d*x + c)) , x)
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algori thm="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2/sqrt(sec(d*x + c)) , x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(1/2) ,x)
Output:
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(1/2) , x)
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=a^{2} \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) c +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) c \right ) \] Input:
int((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x)
Output:
a**2*(int(sqrt(sec(c + d*x))/sec(c + d*x),x)*a + 2*int(sqrt(sec(c + d*x)), x)*a + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x)*c + 2*int(sqrt(sec(c + d* x))*sec(c + d*x)**2,x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a + int( sqrt(sec(c + d*x))*sec(c + d*x),x)*c)