Integrand size = 33, antiderivative size = 417 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 b (A-3 B)-3 b^3 (A-B)-8 a^3 B+3 a b^2 (A+3 B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
2/3*(2*A*a^3*b-6*A*a*b^3-8*B*a^4+15*B*a^2*b^2-3*B*b^4)*cot(d*x+c)*Elliptic E((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)/(a-b)/b^4/(a+b)^(3/2)/d+2/3 *(2*a^2*b*(A-3*B)-3*b^3*(A-B)-8*B*a^3+3*a*b^2*(A+3*B))*cot(d*x+c)*Elliptic F((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/(a^2-b^2)/d/(a+b)^(1/2) -2/3*a^2*(A*b-B*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*a *(2*A*a^2*b-6*A*b^3-5*B*a^3+9*B*a*b^2)*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*s ec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3920\) vs. \(2(417)=834\).
Time = 28.44 (sec) , antiderivative size = 3920, normalized size of antiderivative = 9.40 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(-2*a^3*A*b + 6*a*A*b^3 + 8*a^4 *B - 15*a^2*b^2*B + 3*b^4*B)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) - (2*(a* A*b*Sin[c + d*x] - a^2*B*Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(b + a*Cos[c + d *x])^2) - (2*(-(a^3*A*b*Sin[c + d*x]) + 5*a*A*b^3*Sin[c + d*x] + 4*a^4*B*S in[c + d*x] - 8*a^2*b^2*B*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[ c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) - (2*(b + a*Cos[c + d*x])^2*(( 2*a^3*A)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a*A*b)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (5*a^2*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^4*B)/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] ]) - (b^2*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (5*a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*a^4*A*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d *x]]) + (A*b^2*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x] ]) - (8*a^5*B*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (17*a^3*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos [c + d*x]]) - (3*a*b*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[ c + d*x]]) - (2*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2 *Sqrt[b + a*Cos[c + d*x]]) + (2*a^4*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]]) /(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^5*B*Cos[2*(c + ...
Time = 1.59 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4516, 27, 3042, 4568, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4516 |
| \(\displaystyle -\frac {2 \int -\frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) B \sec ^2(c+d x)+\left (2 a^2-3 b^2\right ) (A b-a B) \sec (c+d x)+3 a b (A b-a B)\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) B \sec ^2(c+d x)+\left (2 a^2-3 b^2\right ) (A b-a B) \sec (c+d x)+3 a b (A b-a B)\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b \left (a^2-b^2\right ) B \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 a^2-3 b^2\right ) (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a b (A b-a B)\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle \frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {\sec (c+d x) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {b (a-b) \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b (a-b) \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
(-2*a^2*(A*b - a*B)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x] )^(3/2)) + (((2*(a - b)*Sqrt[a + b]*(2*a^3*A*b - 6*a*A*b^3 - 8*a^4*B + 15* a^2*b^2*B - 3*b^4*B)*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]*(2*a^2 *b*(A - 3*B) - 3*b^3*(A - B) - 8*a^3*B + 3*a*b^2*(A + 3*B))*Cot[c + d*x]*E llipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))] )/d)/(b*(a^2 - b^2)) + (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^3*B + 9*a*b^2*B)*Ta n[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*b^2*(a^2 - b^2))
3.4.86.3.1 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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B) *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[a*b*(A*b - a*B)*(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Cs c[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 ]
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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] )^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(7754\) vs. \(2(387)=774\).
Time = 27.55 (sec) , antiderivative size = 7755, normalized size of antiderivative = 18.60
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(7755\) |
| default | \(\text {Expression too large to display}\) | \(7823\) |
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b ^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]