Integrand size = 29, antiderivative size = 134 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 a (A+B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 B) \tan (c+d x)}{5 d}+\frac {3 a (A+B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a B \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 B) \tan ^3(c+d x)}{15 d} \] Output:
3/8*a*(A+B)*arctanh(sin(d*x+c))/d+1/5*a*(5*A+4*B)*tan(d*x+c)/d+3/8*a*(A+B) *sec(d*x+c)*tan(d*x+c)/d+1/4*a*(A+B)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*B*sec (d*x+c)^4*tan(d*x+c)/d+1/15*a*(5*A+4*B)*tan(d*x+c)^3/d
Time = 0.54 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (45 (A+B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 (A+B) \sec (c+d x)+30 (A+B) \sec ^3(c+d x)+8 \left (15 (A+B)+5 (A+2 B) \tan ^2(c+d x)+3 B \tan ^4(c+d x)\right )\right )\right )}{120 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*(45*(A + B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*(A + B)*Sec[c + d* x] + 30*(A + B)*Sec[c + d*x]^3 + 8*(15*(A + B) + 5*(A + 2*B)*Tan[c + d*x]^ 2 + 3*B*Tan[c + d*x]^4))))/(120*d)
Time = 0.74 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4485, 3042, 4274, 3042, 4254, 2009, 4255, 3042, 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.
\(\displaystyle \int \sec ^4(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {1}{5} \int \sec ^4(c+d x) (a (5 A+4 B)+5 a (A+B) \sec (c+d x))dx+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (5 A+4 B)+5 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \sec ^5(c+d x)dx+a (5 A+4 B) \int \sec ^4(c+d x)dx\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (a (5 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+5 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a (5 A+4 B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a (5 A+4 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a B \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(a*B*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (-((a*(5*A + 4*B)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d) + 5*a*(A + B)*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d ) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/ 4))/5
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 = 0.94 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.87
method | result | size |
parts | \(\frac {\left (a A +B a \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(117\) |
derivativedivides | \(\frac {-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
default | \(\frac {-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
norman | \(\frac {-\frac {13 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {4 a \left (25 A +29 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a \left (29 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {a \left (35 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(171\) |
parallelrisch | \(\frac {8 \left (-\frac {45 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{2}+\frac {\cos \left (5 d x +5 c \right )}{10}\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {45 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{2}+\frac {\cos \left (5 d x +5 c \right )}{10}\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\frac {21 \left (A +B \right ) \sin \left (2 d x +2 c \right )}{16}+\left (\frac {5 A}{4}+B \right ) \sin \left (3 d x +3 c \right )+\frac {9 \left (A +B \right ) \sin \left (4 d x +4 c \right )}{32}+\left (\frac {A}{4}+\frac {B}{5}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +2 B \right )\right ) a}{3 d \left (10 \cos \left (d x +c \right )+5 \cos \left (3 d x +3 c \right )+\cos \left (5 d x +5 c \right )\right )}\) | \(203\) |
risch | \(-\frac {i a \left (45 A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 B \,{\mathrm e}^{9 i \left (d x +c \right )}+210 A \,{\mathrm e}^{7 i \left (d x +c \right )}+210 B \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 B \,{\mathrm e}^{4 i \left (d x +c \right )}-210 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 B \,{\mathrm e}^{2 i \left (d x +c \right )}-45 \,{\mathrm e}^{i \left (d x +c \right )} A -45 B \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\) | \(265\) |
Input:
int(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
(A*a+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)))-B*a/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+ c)-a*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} + 45 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 24 \, B a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/240*(45*(A + B)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*(A + B)*a*co s(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*A + 4*B)*a*cos(d*x + c)^4 + 45*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^2 + 30*(A + B) *a*cos(d*x + c) + 24*B*a)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
a*(Integral(A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Integ ral(B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x))
Time = 0.03 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a - 15 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(3*tan(d*x + c)^5 + 1 0*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a - 15*A*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)) - 15*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)))/d
Time = 0.15 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.60 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {45 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 290 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 130 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 350 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/120*(45*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*(A*a + B*a)* log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 45 *B*a*tan(1/2*d*x + 1/2*c)^9 - 290*A*a*tan(1/2*d*x + 1/2*c)^7 - 130*B*a*tan (1/2*d*x + 1/2*c)^7 + 400*A*a*tan(1/2*d*x + 1/2*c)^5 + 464*B*a*tan(1/2*d*x + 1/2*c)^5 - 350*A*a*tan(1/2*d*x + 1/2*c)^3 - 190*B*a*tan(1/2*d*x + 1/2*c )^3 + 195*A*a*tan(1/2*d*x + 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 13.73 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+B\right )}{4\,d}-\frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {29\,A\,a}{6}-\frac {13\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,B\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {35\,A\,a}{6}-\frac {19\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x)))/cos(c + d*x)^4,x)
Output:
(3*a*atanh(tan(c/2 + (d*x)/2))*(A + B))/(4*d) - (tan(c/2 + (d*x)/2)*((13*A *a)/4 + (13*B*a)/4) + tan(c/2 + (d*x)/2)^9*((3*A*a)/4 + (3*B*a)/4) - tan(c /2 + (d*x)/2)^7*((29*A*a)/6 + (13*B*a)/6) - tan(c/2 + (d*x)/2)^3*((35*A*a) /6 + (19*B*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A*a)/3 + (116*B*a)/15))/(d*(5 *tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.15 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.57 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
(a*( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 45*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 90*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 90*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1) *a - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 45*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 45*cos(c + d*x)*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**4*b - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 45*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*b - 45*cos(c + d*x)*sin(c + d*x)**3*a - 45*cos(c + d*x)*s in(c + d*x)**3*b + 75*cos(c + d*x)*sin(c + d*x)*a + 75*cos(c + d*x)*sin(c + d*x)*b + 80*sin(c + d*x)**5*a + 64*sin(c + d*x)**5*b - 200*sin(c + d*x)* *3*a - 160*sin(c + d*x)**3*b + 120*sin(c + d*x)*a + 120*sin(c + d*x)*b))/( 120*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))