Integrand size = 41, antiderivative size = 192 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (5 A b+5 a B+3 b C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 (b B+a (3 A+C)) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 (5 A b+5 a B+3 b C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b B+a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 b C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
-2/5*(5*A*b+5*B*a+3*C*b)*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+2/3*(B*b+a*(3*A+C))*cos(d*x+c)^(1/2)*InverseJacob iAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/5*(5*A*b+5*B*a+3*C*b)*sec( d*x+c)^(1/2)*sin(d*x+c)/d+2/3*(B*b+C*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5* b*C*sec(d*x+c)^(5/2)*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.07 (sec) , antiderivative size = 1140, normalized size of antiderivative = 5.94 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
Integrate[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C* Sec[c + d*x]^2),x]
Output:
(2*Sqrt[2]*A*b*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^ ((2*I)*(c + d*x))]*Cos[c + d*x]^3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^(( 2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x ]^2))/(3*d*E^(I*d*x)*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A* Cos[2*c + 2*d*x])) + (2*Sqrt[2]*a*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1 [1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*Sqrt[2]*b*C*Sqrt[E^(I*(c + d* x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^ 3*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)* c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(b + a*Cos[ c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*a*A*Cos[ c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*b*B*Cos[c + d*x]^(7/ 2)*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A...
Time = 1.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.95, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.390, Rules used = {3042, 4564, 27, 3042, 4535, 3042, 4255, 3042, 4258, 3042, 3119, 4534, 3042, 4258, 3042, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4564 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} \left (5 (b B+a C) \sec ^2(c+d x)+(5 A b+3 C b+5 a B) \sec (c+d x)+5 a A\right )dx+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sec (c+d x)} \left (5 (b B+a C) \sec ^2(c+d x)+(5 A b+3 C b+5 a B) \sec (c+d x)+5 a A\right )dx+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(5 A b+3 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a A\right )dx+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left ((5 a B+5 A b+3 b C) \int \sec ^{\frac {3}{2}}(c+d x)dx+\int \sqrt {\sec (c+d x)} \left (5 (b B+a C) \sec ^2(c+d x)+5 a A\right )dx\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left ((5 a B+5 A b+3 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} (a (3 A+C)+b B) \int \sqrt {\sec (c+d x)}dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} (a (3 A+C)+b B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} (a (3 A+C)+b B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} (a (3 A+C)+b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (a (3 A+C)+b B)}{3 d}+(5 a B+5 A b+3 b C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
Input:
Int[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(2*b*C*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + ((10*(b*B + a*(3*A + C))*S qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (1 0*(b*B + a*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (5*A*b + 5*a*B + 3* b*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]) /d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
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_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ n/(f*(n + 2))), x] + Simp[1/(n + 2) Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(714\) vs. \(2(171)=342\).
Time = 5.97 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.72
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(715\) |
| parts | \(\text {Expression too large to display}\) | \(902\) |
Input:
int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_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*a*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* (A*b+B*a)/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*(B*b+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*(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)))+2/5*C*b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*s in(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*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip ticE(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))/sin(1/2*d*x+1/ 2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.35 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, {\left (3 \, A + C\right )} a + i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, {\left (3 \, A + C\right )} a - i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a + i \, {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a - i \, {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (5 \, B a + {\left (5 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 3 \, C b + 5 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
Output:
-1/15*(5*sqrt(2)*(I*(3*A + C)*a + I*B*b)*cos(d*x + c)^2*weierstrassPInvers e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*(3*A + C)*a - I*B* b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) ) + 3*sqrt(2)*(5*I*B*a + I*(5*A + 3*C)*b)*cos(d*x + c)^2*weierstrassZeta(- 4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt( 2)*(-5*I*B*a - I*(5*A + 3*C)*b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(5*B*a + (5* A + 3*C)*b)*cos(d*x + c)^2 + 3*C*b + 5*(C*a + B*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)* *2),x)
Output:
Timed out
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(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)*sqr t(sec(d*x + c)), x)
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(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)*sqr t(sec(d*x + c)), x)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:
int((a + b/cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/co s(c + d*x)^2),x)
Output:
int((a + b/cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/co s(c + d*x)^2), x)
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a b \] Input:
int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
int(sqrt(sec(c + d*x)),x)*a**2 + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x) *b*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a*c + int(sqrt(sec(c + d* x))*sec(c + d*x)**2,x)*b**2 + 2*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*a*b