Integrand size = 42, antiderivative size = 329 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (10 a b B-8 a^2 C-9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 d}+\frac {2 \sqrt {a+b} \left (b^2 (5 B-9 C)-8 a^2 C+2 a b (5 B+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}+\frac {2 (5 b B-4 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 C \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d} \]
2/15*(a-b)*(10*B*a*b-8*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^4/d+2/15*(b^2*(5*B-9*C)-8*C*a^ 2+2*a*b*(5*B+C))*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^3/d+2/15*(5*B*b-4*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d *x+c)/b^2/d+2/5*C*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
Time = 17.85 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \sqrt {\sec (c+d x)} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-10 a b B+8 a^2 C+9 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (8 a^2 C+2 a b (-5 B+C)+b^2 (5 B+9 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-10 a b B+8 a^2 C+9 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b^3 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \sec (c+d x)}}+\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (\frac {2 \left (-10 a b B+8 a^2 C+9 b^2 C\right ) \sin (c+d x)}{15 b^3}+\frac {2 \sec (c+d x) (5 b B \sin (c+d x)-4 a C \sin (c+d x))}{15 b^2}+\frac {2 C \sec (c+d x) \tan (c+d x)}{5 b}\right )}{d \sqrt {a+b \sec (c+d x)}} \]
(-2*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(- 10*a*b*B + 8*a^2*C + 9*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[( 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*(8*a^2*C + 2*a*b*(-5*B + C) + b^2*(5*B + 9*C))*Sqrt[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)] + (-10*a*b*B + 8*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^3*d*Sqrt[Sec[(c + d*x)/2]^2]* Sqrt[a + b*Sec[c + d*x]]) + ((b + a*Cos[c + d*x])*Sec[c + d*x]*((2*(-10*a* b*B + 8*a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b^3) + (2*Sec[c + d*x]*(5*b*B*S in[c + d*x] - 4*a*C*Sin[c + d*x]))/(15*b^2) + (2*C*Sec[c + d*x]*Tan[c + d* x])/(5*b)))/(d*Sqrt[a + b*Sec[c + d*x]])
Time = 1.37 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4560, 3042, 4521, 27, 3042, 4570, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4521 |
| \(\displaystyle \frac {2 \int \frac {\sec (c+d x) \left ((5 b B-4 a C) \sec ^2(c+d x)+3 b C \sec (c+d x)+2 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left ((5 b B-4 a C) \sec ^2(c+d x)+3 b C \sec (c+d x)+2 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left ((5 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b C \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (b (5 b B+2 a C)-\left (-8 C a^2+10 b B a-9 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b (5 b B+2 a C)-\left (-8 C a^2+10 b B a-9 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (5 b B+2 a C)+\left (8 C a^2-10 b B a+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {\left (-8 a^2 C+2 a b (5 B+C)+b^2 (5 B-9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-8 a^2 C+10 a b B-9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (-8 a^2 C+2 a b (5 B+C)+b^2 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^2 C+10 a b B-9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {a+b} \left (-8 a^2 C+2 a b (5 B+C)+b^2 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (-8 a^2 C+10 a b B-9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {\frac {2 \sqrt {a+b} \left (-8 a^2 C+2 a b (5 B+C)+b^2 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 C+10 a b B-9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
(2*C*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) + (((2*(a - b)*Sqrt[a + b]*(10*a*b*B - 8*a^2*C - 9*b^2*C)*Cot[c + d*x]*EllipticE[Ar cSin[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*Sqrt[a + b]*(b^2*(5*B - 9*C) - 8*a^2*C + 2*a*b*(5*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*b) + (2*(5*b*B - 4*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x ])/(3*b*d))/(5*b)
3.9.38.3.1 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_)))/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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* (m + n))), x] + Simp[d^2/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B , m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] && !IGtQ[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]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*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*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3110\) vs. \(2(299)=598\).
Time = 22.52 (sec) , antiderivative size = 3111, normalized size of antiderivative = 9.46
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3111\) |
| default | \(\text {Expression too large to display}\) | \(3135\) |
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
2/3*B/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(2*Elli pticF(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)/(co s(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)^2-2*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+4*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)-4*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)-4*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*(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)*EllipticF(cot(d*x+c)-csc(d*x+c), ((a-b)/(a+b))^(1/2))*a*b-(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+...
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 ),x, algorithm="fricas")
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 ),x, algorithm="maxima")
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]