Integrand size = 43, antiderivative size = 352 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-5 C)-8 a^3 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} d}+\frac {2 \left (3 A b^2-(2 a+b) (b (3 B-C)-4 a C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} d}+\frac {2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 d} \]
-2/3*(6*B*a^2*b-3*B*b^3-a*b^2*(3*A-5*C)-8*a^3*C)*cot(d*x+c)*EllipticE((a+b *sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(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/(a+b)^(1/2)+2/3*(3*A*b^2-(2 *a+b)*(b*(3*B-C)-4*C*a))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b) ^(1/2),((a+b)/(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/(a+b)^(1/2)+2*a*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b^2/ (a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/3*C*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c) /b^2/d
Leaf count is larger than twice the leaf count of optimal. \(3856\) vs. \(2(352)=704\).
Time = 27.24 (sec) , antiderivative size = 3856, normalized size of antiderivative = 10.95 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x])^(3/2),x]
((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(3*a*A *b^2 - 6*a^2*b*B + 3*b^3*B + 8*a^3*C - 5*a*b^2*C)*Sin[c + d*x])/(3*b^3*(-a ^2 + b^2)) - (4*(a*A*b^2*Sin[c + d*x] - a^2*b*B*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (4*C*Tan[c + d*x])/(3* b^2)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(3/2)) - (4*(b + a*Cos[c + d*x])*((-2*a*A)/((-a^2 + b^2)*Sqrt[b + a *Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^2*B)/(b*(-a^2 + b^2)*Sqrt[b + a* Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*b*B)/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (10*a*C)/(3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^3*C)/(3*b^2*(-a^2 + b^2)*Sqrt[b + a*Cos [c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + d*x]])/(b*(-a^2 + b ^2)*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*S qrt[b + a*Cos[c + d*x]]) - (4*a*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*B*Sqrt[Sec[c + d*x]])/(b^2*(-a^2 + b^2)*Sqrt[ b + a*Cos[c + d*x]]) - (16*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)*S qrt[b + a*Cos[c + d*x]]) + (14*a^2*C*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2) *Sqrt[b + a*Cos[c + d*x]]) + (2*b*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)*Sq rt[b + a*Cos[c + d*x]]) - (2*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b *(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (2*a*B*Cos[2*(c + d*x)]*Sqrt[Sec [c + d*x]])/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*B*Cos[2*(c...
Time = 1.60 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4578, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4578 |
\(\displaystyle \frac {2 \int -\frac {\sec (c+d x) \left (-b \left (a^2-b^2\right ) C \sec ^2(c+d x)-\left (-2 C a^3+2 b B a^2-b^2 (A-C) a-b^3 B\right ) \sec (c+d x)+b \left (A b^2-a (b B-a C)\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}+\frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\sec (c+d x) \left (-b \left (a^2-b^2\right ) C \sec ^2(c+d x)-\left (-2 C a^3+2 b B a^2-b^2 (A-C) a-b^3 B\right ) \sec (c+d x)+b \left (A b^2-a (b B-a C)\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-b \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 C a^3-2 b B a^2+b^2 (A-C) a+b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left (A b^2-a (b B-a C)\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 \int \frac {\sec (c+d x) \left (b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\sec (c+d x) \left (b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b^2 \left (2 C a^2-3 b B a+3 A b^2+b^2 C\right )-b \left (-8 C a^3+6 b B a^2-b^2 (3 A-5 C) a-3 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\left (b (a-b) \left (3 A b^2-(2 a+b) (b (3 B-C)-4 a C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (b (a-b) \left (3 A b^2-(2 a+b) (b (3 B-C)-4 a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-b \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 A b^2-(2 a+b) (b (3 B-C)-4 a C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {2 a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-8 a^3 C+6 a^2 b B-a b^2 (3 A-5 C)-3 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 A b^2-(2 a+b) (b (3 B-C)-4 a C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}-\frac {2 C \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}\) |
(2*a*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*S ec[c + d*x]]) - (((2*(a - b)*Sqrt[a + b]*(6*a^2*b*B - 3*b^3*B - a*b^2*(3*A - 5*C) - 8*a^3*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*d) - (2*(a - b)*Sqrt[a + b]*(3*A*b^2 - (2*a + b)*(b*(3*B - C) - 4*a*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))])/d)/(3*b) - (2*(a^2 - b ^2)*C*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(b^2*(a^2 - b^2))
3.10.65.3.1 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[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x ])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b *B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 ]
Leaf count of result is larger than twice the leaf count of optimal. \(5332\) vs. \(2(324)=648\).
Time = 28.51 (sec) , antiderivative size = 5333, normalized size of antiderivative = 15.15
method | result | size |
parts | \(\text {Expression too large to display}\) | \(5333\) |
default | \(\text {Expression too large to display}\) | \(5348\) |
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, method=_RETURNVERBOSE)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /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)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a + b*s ec(c + d*x))**(3/2), x)
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="maxima")
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^2/(b*sec(d* x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]