Integrand size = 29, antiderivative size = 114 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(4 a A+3 b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {(4 a A+3 b B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(A b+a B) \tan ^3(c+d x)}{3 d} \] Output:
1/8*(4*A*a+3*B*b)*arctanh(sin(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/8*(4*A*a+ 3*B*b)*sec(d*x+c)*tan(d*x+c)/d+1/4*b*B*sec(d*x+c)^3*tan(d*x+c)/d+1/3*(A*b+ B*a)*tan(d*x+c)^3/d
Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 (4 a A+3 b B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (12 a A+9 b B+8 (A b+a B) (2+\cos (2 (c+d x))) \sec (c+d x)+6 b B \sec ^2(c+d x)\right ) \tan (c+d x)}{24 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(3*(4*a*A + 3*b*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(12*a*A + 9*b*B + 8*(A*b + a*B)*(2 + Cos[2*(c + d*x)])*Sec[c + d*x] + 6*b*B*Sec[c + d*x]^2)* Tan[c + d*x])/(24*d)
Time = 0.62 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4485, 3042, 4274, 3042, 4254, 2009, 4255, 3042, 4257}
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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {1}{4} \int \sec ^3(c+d x) (4 a A+3 b B+4 (A b+a B) \sec (c+d x))dx+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (4 a A+3 b B+4 (A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{4} \left (4 (a B+A b) \int \sec ^4(c+d x)dx+(4 a A+3 b B) \int \sec ^3(c+d x)dx\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 (a B+A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 (a B+A b) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 (a B+A b) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 (a B+A b) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 (a B+A b) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 (a B+A b) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(b*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*a*A + 3*b*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*(A*b + a*B)*(-Tan[ c + d*x] - Tan[c + d*x]^3/3))/d)/4
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Time = 1.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04
method | result | size |
parts | \(-\frac {\left (A b +B a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(119\) |
derivativedivides | \(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(131\) |
default | \(\frac {A a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(131\) |
parallelrisch | \(\frac {-24 \left (A a +\frac {3 B b}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 \left (A a +\frac {3 B b}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (A b +B a \right ) \sin \left (2 d x +2 c \right )+3 \left (4 A a +3 B b \right ) \sin \left (3 d x +3 c \right )+8 \left (A b +B a \right ) \sin \left (4 d x +4 c \right )+3 \sin \left (d x +c \right ) \left (4 A a +11 B b \right )}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(192\) |
norman | \(\frac {\frac {\left (4 A a -8 A b -8 B a +5 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (4 A a +8 A b +8 B a +5 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A a -40 A b -40 B a -9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (12 A a +40 A b +40 B a -9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (4 A a +3 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A a +3 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(201\) |
risch | \(-\frac {i \left (12 A a \,{\mathrm e}^{7 i \left (d x +c \right )}+9 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 A a \,{\mathrm e}^{5 i \left (d x +c \right )}+33 B b \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A a \,{\mathrm e}^{3 i \left (d x +c \right )}-33 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A a \,{\mathrm e}^{i \left (d x +c \right )}-9 B b \,{\mathrm e}^{i \left (d x +c \right )}-16 A b -16 B a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A a}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A a}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{8 d}\) | \(266\) |
Input:
int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
-(A*b+B*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*A/d*(1/2*sec(d*x+c)*tan( d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+B*b/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d *x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.19 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{2} + 6 \, B b + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/48*(3*(4*A*a + 3*B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A*a + 3*B*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(B*a + A*b)*cos(d*x + c)^3 + 3*(4*A*a + 3*B*b)*cos(d*x + c)^2 + 6*B*b + 8*(B*a + A*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*ta n(d*x + c))*A*b - 3*B*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c ) - 1)) - 12*A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (106) = 212\).
Time = 0.15 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.67 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (4 \, A a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/24*(3*(4*A*a + 3*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3* B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 + 15*B*b* tan(1/2*d*x + 1/2*c)^7 - 12*A*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*a*tan(1/2*d* x + 1/2*c)^5 + 40*A*b*tan(1/2*d*x + 1/2*c)^5 + 9*B*b*tan(1/2*d*x + 1/2*c)^ 5 - 12*A*a*tan(1/2*d*x + 1/2*c)^3 - 40*B*a*tan(1/2*d*x + 1/2*c)^3 - 40*A*b *tan(1/2*d*x + 1/2*c)^3 + 9*B*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*a*tan(1/2*d* x + 1/2*c) + 24*B*a*tan(1/2*d*x + 1/2*c) + 24*A*b*tan(1/2*d*x + 1/2*c) + 1 5*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 14.88 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.70 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\left (A\,a-2\,A\,b-2\,B\,a+\frac {5\,B\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,A\,b}{3}-A\,a+\frac {10\,B\,a}{3}+\frac {3\,B\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,B\,b}{4}-\frac {10\,A\,b}{3}-\frac {10\,B\,a}{3}-A\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+2\,B\,a+\frac {5\,B\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A\,a+\frac {3\,B\,b}{4}\right )}{d} \] Input:
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x)))/cos(c + d*x)^3,x)
Output:
(tan(c/2 + (d*x)/2)*(A*a + 2*A*b + 2*B*a + (5*B*b)/4) + tan(c/2 + (d*x)/2) ^7*(A*a - 2*A*b - 2*B*a + (5*B*b)/4) - tan(c/2 + (d*x)/2)^3*(A*a + (10*A*b )/3 + (10*B*a)/3 - (3*B*b)/4) + tan(c/2 + (d*x)/2)^5*((10*A*b)/3 - A*a + ( 10*B*a)/3 + (3*B*b)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^ 2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(A*a + (3*B*b)/4))/d
Time = 0.17 (sec) , antiderivative size = 378, normalized size of antiderivative = 3.32 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +48 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{2}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{2}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-12 \sin \left (d x +c \right )^{3} a^{2}-9 \sin \left (d x +c \right )^{3} b^{2}+12 \sin \left (d x +c \right ) a^{2}+15 \sin \left (d x +c \right ) b^{2}}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
( - 32*cos(c + d*x)*sin(c + d*x)**3*a*b + 48*cos(c + d*x)*sin(c + d*x)*a*b - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 9*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**4*b**2 + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *2*a**2 + 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 - 12*log(tan(( c + d*x)/2) - 1)*a**2 - 9*log(tan((c + d*x)/2) - 1)*b**2 + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d *x)**4*b**2 - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 18*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 + 12*log(tan((c + d*x)/2) + 1)*a **2 + 9*log(tan((c + d*x)/2) + 1)*b**2 - 12*sin(c + d*x)**3*a**2 - 9*sin(c + d*x)**3*b**2 + 12*sin(c + d*x)*a**2 + 15*sin(c + d*x)*b**2)/(24*d*(sin( c + d*x)**4 - 2*sin(c + d*x)**2 + 1))