Integrand size = 29, antiderivative size = 106 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (4 A+3 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a (4 A+3 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a (A+B) \tan ^3(c+d x)}{3 d} \] Output:
1/8*a*(4*A+3*B)*arctanh(sin(d*x+c))/d+a*(A+B)*tan(d*x+c)/d+1/8*a*(4*A+3*B) *sec(d*x+c)*tan(d*x+c)/d+1/4*a*B*sec(d*x+c)^3*tan(d*x+c)/d+1/3*a*(A+B)*tan (d*x+c)^3/d
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (3 (4 A+3 B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (12 A+9 B+8 (A+B) (2+\cos (2 (c+d x))) \sec (c+d x)+6 B \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{24 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(3*(4*A + 3*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(12*A + 9*B + 8*(A + B)*(2 + Cos[2*(c + d*x)])*Sec[c + d*x] + 6*B*Sec[c + d*x]^2)*Tan[c + d*x ]))/(24*d)
Time = 0.62 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {1}{4} \int \sec ^3(c+d x) (a (4 A+3 B)+4 a (A+B) \sec (c+d x))dx+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (4 A+3 B)+4 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{4} \left (4 a (A+B) \int \sec ^4(c+d x)dx+a (4 A+3 B) \int \sec ^3(c+d x)dx\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a (A+B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+3 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*(4*A + 3*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*a*(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 = 0.78 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(-\frac {\left (a A +B a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(119\) |
| derivativedivides | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(131\) |
| default | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(131\) |
| norman | \(\frac {-\frac {a \left (4 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {7 a \left (4 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a \left (12 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (52 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a \left (4 A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +3 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(163\) |
| parallelrisch | \(\frac {8 a \left (-\frac {3 \left (A +\frac {3 B}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (A +\frac {3 B}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (A +\frac {3 B}{4}\right ) \sin \left (3 d x +3 c \right )}{8}+\frac {\left (A +B \right ) \sin \left (4 d x +4 c \right )}{4}+\frac {3 \sin \left (d x +c \right ) \left (A +\frac {11 B}{4}\right )}{8}\right )}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(170\) |
| risch | \(-\frac {i a \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+9 B \,{\mathrm e}^{7 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+33 B \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-33 B \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 \,{\mathrm e}^{i \left (d x +c \right )} A -9 B \,{\mathrm e}^{i \left (d x +c \right )}-16 A -16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}\) | \(253\) |
Input:
int(sec(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
-(A*a+B*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*a/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)))+a*A/d*(1/2*sec( d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 6 \, B a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/48*(3*(4*A + 3*B)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*B) *a*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(A + B)*a*cos(d*x + c)^3 + 3*(4*A + 3*B)*a*cos(d*x + c)^2 + 8*(A + B)*a*cos(d*x + c) + 6*B*a)*sin(d *x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x)**4, x) + Integ ral(B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x))
Time = 0.03 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.54 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, B a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*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))*A*a + 16*(tan(d*x + c)^3 + 3*ta n(d*x + c))*B*a - 3*B*a*(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
Time = 0.15 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.77 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (4 \, A a + 3 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A a + 3 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 28 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 49 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 52 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 31 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 39 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/24*(3*(4*A*a + 3*B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3* B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 + 9*B*a*tan(1/2*d*x + 1/2*c)^7 - 28*A*a*tan(1/2*d*x + 1/2*c)^5 - 49*B*a*t an(1/2*d*x + 1/2*c)^5 + 52*A*a*tan(1/2*d*x + 1/2*c)^3 + 31*B*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x + 1/2*c) - 39*B*a*tan(1/2*d*x + 1/2*c))/( tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 14.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\left (-A\,a-\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,A\,a}{3}+\frac {49\,B\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {13\,A\,a}{3}-\frac {31\,B\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,B\right )}{4\,d} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x)))/cos(c + d*x)^3,x)
Output:
(tan(c/2 + (d*x)/2)*(3*A*a + (13*B*a)/4) - tan(c/2 + (d*x)/2)^7*(A*a + (3* B*a)/4) - tan(c/2 + (d*x)/2)^3*((13*A*a)/3 + (31*B*a)/12) + tan(c/2 + (d*x )/2)^5*((7*A*a)/3 + (49*B*a)/12))/(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)) + (a*ata nh(tan(c/2 + (d*x)/2))*(4*A + 3*B))/(4*d)
Time = 0.16 (sec) , antiderivative size = 377, normalized size of antiderivative = 3.56 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -12 \sin \left (d x +c \right )^{3} a -9 \sin \left (d x +c \right )^{3} b +12 \sin \left (d x +c \right ) a +15 \sin \left (d x +c \right ) b \right )}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(a*( - 16*cos(c + d*x)*sin(c + d*x)**3*a - 16*cos(c + d*x)*sin(c + d*x)**3 *b + 24*cos(c + d*x)*sin(c + d*x)*a + 24*cos(c + d*x)*sin(c + d*x)*b - 12* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 9*log(tan((c + d*x)/2) - 1)* sin(c + d*x)**4*b + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 18*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 12*log(tan((c + d*x)/2) - 1)*a - 9*log(tan((c + d*x)/2) - 1)*b + 12*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**4*a + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **2*b + 12*log(tan((c + d*x)/2) + 1)*a + 9*log(tan((c + d*x)/2) + 1)*b - 1 2*sin(c + d*x)**3*a - 9*sin(c + d*x)**3*b + 12*sin(c + d*x)*a + 15*sin(c + d*x)*b))/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))