Integrand size = 33, antiderivative size = 374 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 a (a-b) \sqrt {a+b} \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (105 a A b-35 A b^2+6 a^2 C+57 a b C-25 b^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]
-4/105*a*(a-b)*(70*A*b^2-3*C*a^2+41*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^3/d+2/105*(a-b)*(105*A*a* b-35*A*b^2+6*C*a^2+57*C*a*b-25*C*b^2)*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^2/d-4/35*a*C*(a+b*sec(d*x+c))^ (3/2)*tan(d*x+c)/b/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d-2/105*(6* C*a^2-5*b^2*(7*A+5*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. \(3214\) vs. \(2(374)=748\).
Time = 25.99 (sec) , antiderivative size = 3214, normalized size of antiderivative = 8.59 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((-8*a*( -70*A*b^2 + 3*a^2*C - 41*b^2*C)*Sin[c + d*x])/(105*b^2) + (4*Sec[c + d*x]* (35*A*b^2*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(1 05*b) + (32*a*C*Sec[c + d*x]*Tan[c + d*x])/35 + (4*b*C*Sec[c + d*x]^2*Tan[ c + d*x])/7))/(d*(b + a*Cos[c + d*x])*(A + 2*C + A*Cos[2*c + 2*d*x])) + (8 *((-8*a*A*b)/(3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^3*C)/( 35*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (164*a*b*C)/(105*Sqrt[ b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + d*x]])/(3* Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos [c + d*x]]) - (62*a^2*C*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*C*Sqrt[Sec[c + d*x]])/(35*b^2*Sqrt[b + a*Cos[c + d*x]]) + (10*b^ 2*C*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (8*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) - (164*a^2*C*Cos[ 2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*C *Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(35*b^2*Sqrt[b + a*Cos[c + d*x]]))*S qrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + C*Sec [c + d*x]^2)*(2*a*(a + b)*(-70*A*b^2 + 3*a^2*C - 41*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)] + b*(a + b)*(-6* a^2*C + 5*b^2*(7*A + 5*C) + 3*a*b*(35*A + 19*C))*Sqrt[Cos[c + d*x]/(1 +...
Time = 1.52 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4571, 27, 3042, 4490, 27, 3042, 4490, 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.
\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4571 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} (b (7 A+5 C)-2 a C \sec (c+d x))dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (b (7 A+5 C)-2 a C \sec (c+d x))dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (b (7 A+5 C)-2 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (a b (35 A+19 C)-\left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (a b (35 A+19 C)-\left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (a b (35 A+19 C)+\left (5 b^2 (7 A+5 C)-6 a^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 4490 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {4 a (a-b) \sqrt {a+b} \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\) |
(2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-4*a*C*(a + b*Se c[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-4*a*(a - b)*Sqrt[a + b]*(70*A* b^2 - 3*a^2*C + 41*b^2*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^2*d) + (2*(a - b)*Sqrt[a + b] *(105*a*A*b - 35*A*b^2 + 6*a^2*C + 57*a*b*C - 25*b^2*C)*Cot[c + d*x]*Ellip ticF[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 - (2*(6*a^2*C - 5*b^2*(7*A + 5*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5)/(7*b)
3.8.19.3.1 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_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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_)]^2*(C_.))*(csc[ (e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] *((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) In t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3726\) vs. \(2(340)=680\).
Time = 28.33 (sec) , antiderivative size = 3727, normalized size of antiderivative = 9.97
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3727\) |
default | \(\text {Expression too large to display}\) | \(3782\) |
2/3*A/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(4*Elliptic E(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)^2+4*El lipticE(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)^ 2-3*(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^2*cos(d *x+c)^2-4*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)*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+8*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)+8*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)-6*(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^2*cos(d*x+c)-8*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))/(co s(d*x+c)+1))^(1/2)*a*b*cos(d*x+c)-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a...
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integral((C*b*sec(d*x + c)^4 + C*a*sec(d*x + c)^3 + A*b*sec(d*x + c)^2 + A *a*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
\[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \]