Integrand size = 29, antiderivative size = 159 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7 a^4 (5 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {8 a^4 (5 A+4 B) \tan (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {4 a^4 (5 A+4 B) \tan ^3(c+d x)}{15 d} \] Output:
7/8*a^4*(5*A+4*B)*arctanh(sin(d*x+c))/d+8/5*a^4*(5*A+4*B)*tan(d*x+c)/d+27/ 40*a^4*(5*A+4*B)*sec(d*x+c)*tan(d*x+c)/d+1/20*a^4*(5*A+4*B)*sec(d*x+c)^3*t an(d*x+c)/d+1/5*B*(a+a*sec(d*x+c))^4*tan(d*x+c)/d+4/15*a^4*(5*A+4*B)*tan(d *x+c)^3/d
Time = 4.73 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.46 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 A \coth ^{-1}(\sin (c+d x))}{d}+\frac {27 a^4 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {7 a^4 B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {8 a^4 A \tan (c+d x)}{d}+\frac {8 a^4 B \tan (c+d x)}{d}+\frac {27 a^4 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {7 a^4 B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^4 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^4 B \sec ^3(c+d x) \tan (c+d x)}{d}+\frac {4 a^4 A \tan ^3(c+d x)}{3 d}+\frac {8 a^4 B \tan ^3(c+d x)}{3 d}+\frac {a^4 B \tan ^5(c+d x)}{5 d} \] Input:
Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(a^4*A*ArcCoth[Sin[c + d*x]])/d + (27*a^4*A*ArcTanh[Sin[c + d*x]])/(8*d) + (7*a^4*B*ArcTanh[Sin[c + d*x]])/(2*d) + (8*a^4*A*Tan[c + d*x])/d + (8*a^4 *B*Tan[c + d*x])/d + (27*a^4*A*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (7*a^4*B *Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^4*A*Sec[c + d*x]^3*Tan[c + d*x])/(4 *d) + (a^4*B*Sec[c + d*x]^3*Tan[c + d*x])/d + (4*a^4*A*Tan[c + d*x]^3)/(3* d) + (8*a^4*B*Tan[c + d*x]^3)/(3*d) + (a^4*B*Tan[c + d*x]^5)/(5*d)
Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 4489, 3042, 4278, 2009}
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.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^4 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {1}{5} (5 A+4 B) \int \sec (c+d x) (\sec (c+d x) a+a)^4dx+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4dx+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \frac {1}{5} (5 A+4 B) \int \left (a^4 \sec ^5(c+d x)+4 a^4 \sec ^4(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^2(c+d x)+a^4 \sec (c+d x)\right )dx+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} (5 A+4 B) \left (\frac {35 a^4 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {4 a^4 \tan ^3(c+d x)}{3 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {a^4 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {27 a^4 \tan (c+d x) \sec (c+d x)}{8 d}\right )+\frac {B \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
Input:
Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(B*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(5*d) + ((5*A + 4*B)*((35*a^4*ArcT anh[Sin[c + d*x]])/(8*d) + (8*a^4*Tan[c + d*x])/d + (27*a^4*Sec[c + d*x]*T an[c + d*x])/(8*d) + (a^4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (4*a^4*Tan[ c + d*x]^3)/(3*d)))/5
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
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[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Time = 1.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.26
| method | result | size |
| norman | \(\frac {\frac {79 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {224 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {49 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {7 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{4} \left (93 A +100 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {7 a^{4} \left (5 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {7 a^{4} \left (5 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(201\) |
| parallelrisch | \(\frac {56 \left (-\frac {75 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{2}+\frac {\cos \left (5 d x +5 c \right )}{10}\right ) \left (A +\frac {4 B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {75 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{2}+\frac {\cos \left (5 d x +5 c \right )}{10}\right ) \left (A +\frac {4 B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\left (\frac {93 A}{112}+\frac {33 B}{28}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {19 A}{14}+\frac {11 B}{8}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {81 A}{224}+\frac {3 B}{8}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {5 A}{14}+\frac {83 B}{280}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {5 B}{4}\right )\right ) a^{4}}{3 d \left (10 \cos \left (d x +c \right )+5 \cos \left (3 d x +3 c \right )+\cos \left (5 d x +5 c \right )\right )}\) | \(217\) |
| parts | \(\frac {\left (a^{4} A +4 B \,a^{4}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (4 a^{4} A +B \,a^{4}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (4 a^{4} A +6 B \,a^{4}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{4}}{d}\) | \(227\) |
| derivativedivides | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(303\) |
| default | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-4 a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{4} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(303\) |
| risch | \(-\frac {i a^{4} \left (405 A \,{\mathrm e}^{9 i \left (d x +c \right )}+420 B \,{\mathrm e}^{9 i \left (d x +c \right )}-480 A \,{\mathrm e}^{8 i \left (d x +c \right )}-120 B \,{\mathrm e}^{8 i \left (d x +c \right )}+930 A \,{\mathrm e}^{7 i \left (d x +c \right )}+1320 B \,{\mathrm e}^{7 i \left (d x +c \right )}-2880 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1920 B \,{\mathrm e}^{6 i \left (d x +c \right )}-5120 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4720 B \,{\mathrm e}^{4 i \left (d x +c \right )}-930 A \,{\mathrm e}^{3 i \left (d x +c \right )}-1320 B \,{\mathrm e}^{3 i \left (d x +c \right )}-3520 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3200 B \,{\mathrm e}^{2 i \left (d x +c \right )}-405 \,{\mathrm e}^{i \left (d x +c \right )} A -420 B \,{\mathrm e}^{i \left (d x +c \right )}-800 A -664 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\) | \(311\) |
Input:
int(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
(79/6*a^4*(5*A+4*B)/d*tan(1/2*d*x+1/2*c)^3-224/15*a^4*(5*A+4*B)/d*tan(1/2* d*x+1/2*c)^5+49/6*a^4*(5*A+4*B)/d*tan(1/2*d*x+1/2*c)^7-7/4*a^4*(5*A+4*B)/d *tan(1/2*d*x+1/2*c)^9-1/4*a^4*(93*A+100*B)/d*tan(1/2*d*x+1/2*c))/(tan(1/2* d*x+1/2*c)^2-1)^5-7/8*a^4*(5*A+4*B)/d*ln(tan(1/2*d*x+1/2*c)-1)+7/8*a^4*(5* A+4*B)/d*ln(tan(1/2*d*x+1/2*c)+1)
Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (5 \, A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (100 \, A + 83 \, B\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (27 \, A + 28 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (10 \, A + 17 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 24 \, B a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/240*(105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 105*(5*A + 4*B)*a^4*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(100*A + 83*B)*a^ 4*cos(d*x + c)^4 + 15*(27*A + 28*B)*a^4*cos(d*x + c)^3 + 16*(10*A + 17*B)* a^4*cos(d*x + c)^2 + 30*(A + 4*B)*a^4*cos(d*x + c) + 24*B*a^4)*sin(d*x + c ))/(d*cos(d*x + c)^5)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^{4} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 4 A \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 B \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sec ^{6}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)
Output:
a**4*(Integral(A*sec(c + d*x), x) + Integral(4*A*sec(c + d*x)**2, x) + Int egral(6*A*sec(c + d*x)**3, x) + Integral(4*A*sec(c + d*x)**4, x) + Integra l(A*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**2, x) + Integral(4*B*se c(c + d*x)**3, x) + Integral(6*B*sec(c + d*x)**4, x) + Integral(4*B*sec(c + d*x)**5, x) + Integral(B*sec(c + d*x)**6, x))
Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (147) = 294\).
Time = 0.05 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.32 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} - 15 \, A a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 960 \, A a^{4} \tan \left (d x + c\right ) + 240 \, B a^{4} \tan \left (d x + c\right )}{240 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/240*(320*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^4 + 480*(tan(d*x + c)^3 + 3*tan (d*x + c))*B*a^4 - 15*A*a^4*(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)) - 60*B*a^4*(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)) - 360*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 240*B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*A*a^4*log(sec (d*x + c) + tan(d*x + c)) + 960*A*a^4*tan(d*x + c) + 240*B*a^4*tan(d*x + c ))/d
Time = 0.20 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.55 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (525 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2450 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1960 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4480 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3584 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3950 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3160 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1395 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1500 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/120*(105*(5*A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(5 *A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(525*A*a^4*tan(1/ 2*d*x + 1/2*c)^9 + 420*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 2450*A*a^4*tan(1/2*d *x + 1/2*c)^7 - 1960*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 4480*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3584*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 3950*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 3160*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1395*A*a^4*tan(1/2*d*x + 1 /2*c) + 1500*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 13.93 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.41 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (5\,A+4\,B\right )}{4\,d}-\frac {\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {245\,A\,a^4}{6}-\frac {98\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {224\,A\,a^4}{3}+\frac {896\,B\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {395\,A\,a^4}{6}-\frac {158\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+25\,B\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4)/cos(c + d*x),x)
Output:
(7*a^4*atanh(tan(c/2 + (d*x)/2))*(5*A + 4*B))/(4*d) - (tan(c/2 + (d*x)/2)* ((93*A*a^4)/4 + 25*B*a^4) + tan(c/2 + (d*x)/2)^9*((35*A*a^4)/4 + 7*B*a^4) - tan(c/2 + (d*x)/2)^7*((245*A*a^4)/6 + (98*B*a^4)/3) - tan(c/2 + (d*x)/2) ^3*((395*A*a^4)/6 + (158*B*a^4)/3) + tan(c/2 + (d*x)/2)^5*((224*A*a^4)/3 + (896*B*a^4)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 1 0*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.16 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.03 \[ \int \sec (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)
Output:
(a**4*( - 525*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 4 20*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 1050*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 840*cos(c + d*x)*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 525*cos(c + d*x)*log(tan((c + d*x )/2) - 1)*a - 420*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 525*cos(c + d *x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 420*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 1050*cos(c + d*x)*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*a - 840*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*b + 525*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 420*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 405*cos(c + d*x)*sin(c + d*x)**3*a - 420*cos(c + d*x)*sin(c + d*x)**3*b + 435*cos(c + d*x)*sin(c + d*x)*a + 54 0*cos(c + d*x)*sin(c + d*x)*b + 800*sin(c + d*x)**5*a + 664*sin(c + d*x)** 5*b - 1760*sin(c + d*x)**3*a - 1600*sin(c + d*x)**3*b + 960*sin(c + d*x)*a + 960*sin(c + d*x)*b))/(120*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d *x)**2 + 1))