Integrand size = 33, antiderivative size = 221 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b (5 A b+7 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 b B \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d} \] Output:
-2/5*(3*b^2*B+5*a*(2*A*b+B*a))*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*(3*A*a^2+A*b^2+2*B*a*b)*cos(d*x+c)^(1/2 )*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/5*(3*b^2*B+5 *a*(2*A*b+B*a))*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/15*b*(5*A*b+7*B*a)*sec(d*x +c)^(3/2)*sin(d*x+c)/d+2/5*b*B*sec(d*x+c)^(3/2)*(a+b*sec(d*x+c))*sin(d*x+c )/d
Time = 2.91 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.77 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-12 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (3 a^2 A+A b^2+2 a b B\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (15 \left (2 a A b+a^2 B+b^2 B\right )+10 b (A b+2 a B) \cos (c+d x)+3 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x ]
Output:
(Sec[c + d*x]^(5/2)*(-12*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[c + d*x]^(5/2) *EllipticE[(c + d*x)/2, 2] + 20*(3*a^2*A + A*b^2 + 2*a*b*B)*Cos[c + d*x]^( 5/2)*EllipticF[(c + d*x)/2, 2] + 2*(15*(2*a*A*b + a^2*B + b^2*B) + 10*b*(A *b + 2*a*B)*Cos[c + d*x] + 3*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Cos[2*(c + d*x )])*Sin[c + d*x]))/(30*d)
Time = 1.27 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.93, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4514, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4514 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} \left (b (5 A b+7 a B) \sec ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \sec (c+d x)+a (5 a A+b B)\right )dx+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sec (c+d x)} \left (b (5 A b+7 a B) \sec ^2(c+d x)+\left (3 B b^2+5 a (2 A b+a B)\right ) \sec (c+d x)+a (5 a A+b B)\right )dx+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B b^2+5 a (2 A b+a B)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a (5 a A+b B)\right )dx+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \int \sec ^{\frac {3}{2}}(c+d x)dx+\int \sqrt {\sec (c+d x)} \left (b (5 A b+7 a B) \sec ^2(c+d x)+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A b+7 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (5 a A+b B)\right )dx+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \int \sqrt {\sec (c+d x)}dx+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\left (5 a (a B+2 A b)+3 b^2 B\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\) |
Input:
Int[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
Output:
(2*b*B*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((10* (3*a^2*A + A*b^2 + 2*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*S qrt[Sec[c + d*x]])/(3*d) + (2*b*(5*A*b + 7*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (3*b^2*B + 5*a*(2*A*b + a*B))*((-2*Sqrt[Cos[c + d*x]]*Ellip ticE[(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_)]*(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/(m + n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n* Simp[a^2*A*(m + n) + a*b*B*n + (a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1) )*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2, x], x] , x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && !(IGtQ[n, 1] && !IntegerQ[m])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(722\) vs. \(2(200)=400\).
Time = 6.39 (sec) , antiderivative size = 723, normalized size of antiderivative = 3.27
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(723\) |
| parts | \(\text {Expression too large to display}\) | \(914\) |
Input:
int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNV ERBOSE)
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^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*(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*cos(1/2*d*x+1/2*c)*sin(1/2*d *x+1/2*c)^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)^2-1)^(1/2))+2*b*(A*b+2*B*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*b^2*B/(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*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)*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*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)* sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/si n(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 = 0.11 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.29 \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{2} + 10 i \, A a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{2} - 10 i \, A a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, B b^{2} + 3 \, {\left (5 \, B a^{2} + 10 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(3*I*A*a^2 + 2*I*B*a*b + I*A*b^2)*cos(d*x + c)^2*weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3*I*A*a^2 - 2*I*B*a*b - I*A*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*B*a^2 + 10*I*A*a*b + 3*I*B*b^2)*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^2 - 10*I*A*a*b - 3*I*B*b^2)*cos( d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*B*b^2 + 3*(5*B*a^2 + 10*A*a*b + 3*B*b^2)*cos(d*x + c)^2 + 5*(2*B*a*b + A*b^2)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c) ))/(d*cos(d*x + c)^2)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(1/2)*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="maxima")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2),x)
Output:
int((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2), x)
\[ \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{3}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a^{2} b \] Input:
int(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
Output:
int(sqrt(sec(c + d*x)),x)*a**3 + int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x) *b**3 + 3*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*a*b**2 + 3*int(sqrt(se c(c + d*x))*sec(c + d*x),x)*a**2*b