Integrand size = 40, antiderivative size = 314 \[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (5 a b B-2 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}-\frac {2 (a-b) \sqrt {a+b} (5 b B-2 a C-9 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}+\frac {2 (5 b B-2 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac {2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d} \]
-2/15*(a-b)*(5*B*a*b-2*C*a^2+9*C*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c) )^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+ b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d-2/15*(a-b)*(5*B*b-2*C*a-9* C*b)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b)) ^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b ))^(1/2)/b^2/d+2/5*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/15*(5*B*b-2*C *a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
Time = 16.53 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.38 \[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {a+b \sec (c+d x)} \left (2 (a+b) \left (-5 a b B+2 a^2 C-9 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (5 b B-2 a C+9 b C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-5 a b B+2 a^2 C-9 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b^2 d (b+a \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 \left (5 a b B-2 a^2 C+9 b^2 C\right ) \sin (c+d x)}{15 b^2}+\frac {2 \sec (c+d x) (5 b B \sin (c+d x)+a C \sin (c+d x))}{15 b}+\frac {2}{5} C \sec (c+d x) \tan (c+d x)\right )}{d} \]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(2*(a + b)*(-5*a*b*B + 2*a^2*C - 9*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sq rt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan [(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(5*b*B - 2*a*C + 9*b*C)*Sqr t[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (- 5*a*b*B + 2*a^2*C - 9*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d* x)/2]^2*Tan[(c + d*x)/2]))/(15*b^2*d*(b + a*Cos[c + d*x])*Sqrt[Sec[(c + d* x)/2]^2]*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Sec[c + d*x]]*((2*(5*a*b*B - 2* a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b^2) + (2*Sec[c + d*x]*(5*b*B*Sin[c + d *x] + a*C*Sin[c + d*x]))/(15*b) + (2*C*Sec[c + d*x]*Tan[c + d*x])/5))/d
Time = 1.26 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 4560, 3042, 4498, 27, 3042, 4490, 27, 3042, 4493, 3042, 4319, 4492}
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.
| \(\displaystyle \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} (3 b C+(5 b B-2 a C) \sec (c+d x))dx}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) \sqrt {a+b \sec (c+d x)} (3 b C+(5 b B-2 a C) \sec (c+d x))dx}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (3 b C+(5 b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {\sec (c+d x) \left (b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\sec (c+d x) \left (b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (5 b B+7 a C)+\left (-2 C a^2+5 b B a+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {1}{3} \left (\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) (-2 a C+5 b B-9 b C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (-2 a C+5 b B-9 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {1}{3} \left (\left (-2 a^2 C+5 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (-2 a C+5 b B-9 b C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \left (-2 a^2 C+5 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} (-2 a C+5 b B-9 b C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\) |
(2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d) + (((-2*(a - b)*Sqrt [a + b]*(5*a*b*B - 2*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x ]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b) *Sqrt[a + b]*(5*b*B - 2*a*C - 9*b*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x] ))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 + (2*(5*b*B - 2*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*b)
3.9.16.3.1 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[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(3106\) vs. \(2(284)=568\).
Time = 17.96 (sec) , antiderivative size = 3107, normalized size of antiderivative = 9.89
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3107\) |
| default | \(\text {Expression too large to display}\) | \(3136\) |
2/3*B/d/b*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(Elliptic E(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*cos(d*x+c)^2+Elli pticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b*cos(d*x+c)^2- EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b*cos(d*x+c )^2-EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos( d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^2*cos(d *x+c)^2+2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2 *cos(d*x+c)+2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d* x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) *a*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 1/2)*a*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 ))^(1/2)*b^2*cos(d*x+c)+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ b))^(1/2))*a^2+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+...
\[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2), x, algorithm="fricas")
\[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2), x, algorithm="maxima")
\[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2), x, algorithm="giac")
Timed out. \[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{\cos \left (c+d\,x\right )} \,d x \]