Integrand size = 43, antiderivative size = 342 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (15 A b^2-10 a b B+8 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 d}-\frac {2 \sqrt {a+b} \left (15 A b^2-b^2 (5 B-9 C)+8 a^2 C-2 a b (5 B+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}+\frac {2 (5 b B-4 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 C \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d} \]
-2/15*(a-b)*(15*A*b^2-10*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*(15*A*b^2-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
Leaf count is larger than twice the leaf count of optimal. \(3332\) vs. \(2(342)=684\).
Time = 29.12 (sec) , antiderivative size = 3332, normalized size of antiderivative = 9.74 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]
(Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) *((4*(15*A*b^2 - 10*a*b*B + 8*a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b^3) + (4 *Sec[c + d*x]*(5*b*B*Sin[c + d*x] - 4*a*C*Sin[c + d*x]))/(15*b^2) + (4*C*S ec[c + d*x]*Tan[c + d*x])/(5*b)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2 *c + 2*d*x])*Sqrt[a + b*Sec[c + d*x]]) + (4*((-2*A)/(Sqrt[b + a*Cos[c + d* x]]*Sqrt[Sec[c + d*x]]) + (4*a*B)/(3*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*C)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^ 2*C)/(15*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a*A*Sqrt[Se c[c + d*x]])/(b*Sqrt[b + a*Cos[c + d*x]]) + (2*B*Sqrt[Sec[c + d*x]])/(3*Sq rt[b + a*Cos[c + d*x]]) + (4*a^2*B*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*C os[c + d*x]]) - (16*a^3*C*Sqrt[Sec[c + d*x]])/(15*b^3*Sqrt[b + a*Cos[c + d *x]]) - (14*a*C*Sqrt[Sec[c + d*x]])/(15*b*Sqrt[b + a*Cos[c + d*x]]) - (2*a *A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b*Sqrt[b + a*Cos[c + d*x]]) + (4* a^2*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]] ) - (16*a^3*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*b^3*Sqrt[b + a*Cos[ c + d*x]]) - (6*a*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*b*Sqrt[b + a*C os[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-2*(a + b)*(15*A*b^2 - 10*a*b*B + 8*a^2*C + 9*b^2*C)*S qrt[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)]...
Time = 1.28 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4580, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4580 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x) \left ((5 b B-4 a C) \sec ^2(c+d x)+b (5 A+3 C) \sec (c+d x)+2 a C\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec (c+d x) \left ((5 b B-4 a C) \sec ^2(c+d x)+b (5 A+3 C) \sec (c+d x)+2 a C\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left ((5 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (b (5 b B+2 a C)+\left (8 C a^2-10 b B a+15 A b^2+9 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b (5 b B+2 a C)+\left (8 C a^2-10 b B a+15 A b^2+9 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (5 b B+2 a C)+\left (8 C a^2-10 b B a+15 A b^2+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {\frac {\left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\left (8 a^2 C-2 a b (5 B+C)+15 A b^2-b^2 (5 B-9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (8 a^2 C-2 a b (5 B+C)+15 A b^2-b^2 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {\left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (8 a^2 C-2 a b (5 B+C)+15 A b^2-b^2 (5 B-9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {\frac {-\frac {2 \sqrt {a+b} \cot (c+d x) \left (8 a^2 C-2 a b (5 B+C)+15 A b^2-b^2 (5 B-9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (8 a^2 C-10 a b B+15 A b^2+9 b^2 C\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}+\frac {2 (5 b B-4 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\) |
(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]*(15*A*b^2 - 10*a*b*B + 8*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)]*S qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b)) ])/(b^2*d) - (2*Sqrt[a + b]*(15*A*b^2 - 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*Se c[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b)
3.10.59.3.1 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_)]*((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]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3929\) vs. \(2(312)=624\).
Time = 31.89 (sec) , antiderivative size = 3930, normalized size of antiderivative = 11.49
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3930\) |
default | \(\text {Expression too large to display}\) | \(3949\) |
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
-2*A/d/b*(EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b*c os(d*x+c)^2-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a *cos(d*x+c)^2-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d* x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2) *b*cos(d*x+c)^2+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^( 1/2)*b*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*( cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c))) ^(1/2)*a*cos(d*x+c)-2*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c) ))^(1/2)*b*cos(d*x+c)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos( d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b) )^(1/2))*b-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+ cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a- (cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)) )^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b-cos(d*x+c)* sin(d*x+c)*a-sin(d*x+c)*b)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(1+cos( d*x+c))-2/3*B/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(1+cos(d*x+...
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 /2),x, algorithm="fricas")
integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)/sqrt(b*s ec(d*x + c) + a), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/sqrt(a + b*sec(c + d*x)), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*se c(d*x + c) + a), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^2/sqrt(b*se c(d*x + c) + a), x)
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]