Integrand size = 33, antiderivative size = 329 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} d}-\frac {2 (2 a+b) (3 A b-(4 a+b) B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} d}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 d} \]
-2/3*(6*A*a^2*b-3*A*b^3-8*B*a^3+5*B*a*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))*(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*(2*a+b)*(3*A*b-(4 *a+b)*B)*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^2*(A*b-B*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x +c))^(1/2)+2/3*B*(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. \(3460\) vs. \(2(329)=658\).
Time = 31.60 (sec) , antiderivative size = 3460, normalized size of antiderivative = 10.52 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(-6*a^2*A*b + 3*A*b^3 + 8*a^3*B - 5*a*b^2*B)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)) + (2*(a^2*A*b*Sin[c + d*x ] - a^3*B*Sin[c + d*x]))/(b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])) + (2*B*Ta n[c + d*x])/(3*b^2)))/(d*(a + b*Sec[c + d*x])^(3/2)) - (2*(b + a*Cos[c + d *x])*((2*a^2*A)/(b*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x] ]) - (A*b)/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (5 *a*B)/(3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^ 3*B)/(3*b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2 *a*A*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3* A*Sqrt[Sec[c + d*x]])/(b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (8*a^4 *B*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (7* a^2*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (b *B*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (a*A*Co s[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (2*a^3*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b^2*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (8*a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^3* (-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (5*a^2*B*Cos[2*(c + d*x)]*Sqrt[Se c[c + d*x]])/(3*b*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]))*Sec[c + d*x]^(3/ 2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-6*a^2*A*b + 3*A*b^3 + 8*a^3*B - 5*a*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + ...
Time = 1.47 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4516, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4516 |
| \(\displaystyle -\frac {2 \int -\frac {\sec (c+d x) \left (b \left (a^2-b^2\right ) B \sec ^2(c+d x)+\left (2 a^2-b^2\right ) (A b-a B) \sec (c+d x)+a b (A b-a B)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (b \left (a^2-b^2\right ) B \sec ^2(c+d x)+\left (2 a^2-b^2\right ) (A b-a B) \sec (c+d x)+a b (A b-a B)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (a^2-b^2\right ) B \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 a^2-b^2\right ) (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+a b (A b-a B)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \sec (c+d x) b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) (2 a+b) (-4 a B+3 A b-b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (2 a+b) (-4 a B+3 A b-b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (2 a+b) (-4 a B+3 A b-b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (2 a+b) (-4 a B+3 A b-b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
(-2*a^2*(A*b - a*B)*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d* x]]) + (((-2*(a - b)*Sqrt[a + b]*(6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 5*a*b^2* 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 + b)*(3*A*b - 4*a*B - b*B)*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 + Se c[c + d*x]))/(a - b))])/d)/(3*b) + (2*(a^2 - b^2)*B*Sqrt[a + b*Sec[c + d*x ]]*Tan[c + d*x])/(3*d))/(b^2*(a^2 - b^2))
3.4.78.3.1 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[(-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. \(4227\) vs. \(2(301)=602\).
Time = 30.24 (sec) , antiderivative size = 4228, normalized size of antiderivative = 12.85
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4228\) |
| default | \(\text {Expression too large to display}\) | \(4252\) |
2*A/d/b^2/(a+b)/(a-b)*(a^2*b*sin(d*x+c)+2*EllipticE(cot(d*x+c)-csc(d*x+c), ((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b-EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/ (a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*a*b^2+2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 /2)*(cos(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^3*cos(d*x+c)^2-(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) )^(1/2)*(cos(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))*b^3*cos(d*x+c)^2-2*EllipticF(cot(d*x+c)-csc(d*x+c),(( a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d* x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b-EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a +b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c os(d*x+c)+1))^(1/2)*a*b^2+2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^ (1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*b^3*cos(d*x+c)+4*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ (1/2)*(cos(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^3*cos(d*x+c)+EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/( a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/( cos(d*x+c)+1))^(1/2)*b^3*cos(d*x+c)^2-2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x +c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-cs...
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b ^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]