Integrand size = 21, antiderivative size = 135 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {3 a b \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (5 a^2+4 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {b^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (5 a^2+4 b^2\right ) \tan ^3(c+d x)}{15 d} \] Output:
3/4*a*b*arctanh(sin(d*x+c))/d+1/5*(5*a^2+4*b^2)*tan(d*x+c)/d+3/4*a*b*sec(d *x+c)*tan(d*x+c)/d+1/2*a*b*sec(d*x+c)^3*tan(d*x+c)/d+1/5*b^2*sec(d*x+c)^4* tan(d*x+c)/d+1/15*(5*a^2+4*b^2)*tan(d*x+c)^3/d
Time = 0.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {45 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (60 \left (a^2+b^2\right )+45 a b \sec (c+d x)+30 a b \sec ^3(c+d x)+20 \left (a^2+2 b^2\right ) \tan ^2(c+d x)+12 b^2 \tan ^4(c+d x)\right )}{60 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]
Output:
(45*a*b*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(60*(a^2 + b^2) + 45*a*b*Sec[ c + d*x] + 30*a*b*Sec[c + d*x]^3 + 20*(a^2 + 2*b^2)*Tan[c + d*x]^2 + 12*b^ 2*Tan[c + d*x]^4))/(60*d)
Time = 0.76 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4275, 3042, 4255, 3042, 4255, 3042, 4257, 4534, 3042, 4254, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle \int \sec ^4(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right )dx+2 a b \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (5 a^2+4 b^2\right ) \int \sec ^4(c+d x)dx+2 a b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {b^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+2 a b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {b^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {\left (5 a^2+4 b^2\right ) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 d}+2 a b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {b^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (5 a^2+4 b^2\right ) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 d}+2 a b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {b^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]
Output:
(b^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) - ((5*a^2 + 4*b^2)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/(5*d) + 2*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)
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_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 2.00 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
| default | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
| parts | \(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {b^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(115\) |
| risch | \(-\frac {i \left (45 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+210 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-280 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-320 \,{\mathrm e}^{4 i \left (d x +c \right )} b^{2}-210 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-200 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-160 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}-45 a b \,{\mathrm e}^{i \left (d x +c \right )}-40 a^{2}-32 b^{2}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\) | \(194\) |
| norman | \(\frac {-\frac {4 \left (25 a^{2}+29 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (4 a^{2}-5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}-\frac {\left (4 a^{2}+5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (16 a^{2}-3 a b +8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {\left (16 a^{2}+3 a b +8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {3 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {3 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(206\) |
| parallelrisch | \(\frac {-450 a \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 a \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (200 a^{2}+160 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (40 a^{2}+32 b^{2}\right ) \sin \left (5 d x +5 c \right )+420 a b \sin \left (2 d x +2 c \right )+90 a b \sin \left (4 d x +4 c \right )+160 \sin \left (d x +c \right ) \left (a^{2}+2 b^{2}\right )}{60 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(213\) |
Input:
int(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+2*a*b*(-(-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^2*(-8/15-1/5*sec (d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (45 \, a b \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right ) + 4 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 12 \, b^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/120*(45*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*b*cos(d*x + c)^5 *log(-sin(d*x + c) + 1) + 2*(45*a*b*cos(d*x + c)^3 + 8*(5*a^2 + 4*b^2)*cos (d*x + c)^4 + 30*a*b*cos(d*x + c) + 4*(5*a^2 + 4*b^2)*cos(d*x + c)^2 + 12* b^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*sec(d*x+c))**2,x)
Output:
Integral((a + b*sec(c + d*x))**2*sec(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + 8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} b^{2} - 15 \, a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + 8*(3*tan(d*x + c)^5 + 10 *tan(d*x + c)^3 + 15*tan(d*x + c))*b^2 - 15*a*b*(2*(3*sin(d*x + c)^3 - 5*s in(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
Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (123) = 246\).
Time = 0.16 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.01 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 232 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/60*(45*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a*b*log(abs(tan(1/2*d *x + 1/2*c) - 1)) - 2*(60*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*b^2*tan(1/2*d*x + 1/2*c)^9 - 160*a^2*tan(1/2*d*x + 1/2*c)^ 7 + 30*a*b*tan(1/2*d*x + 1/2*c)^7 - 80*b^2*tan(1/2*d*x + 1/2*c)^7 + 200*a^ 2*tan(1/2*d*x + 1/2*c)^5 + 232*b^2*tan(1/2*d*x + 1/2*c)^5 - 160*a^2*tan(1/ 2*d*x + 1/2*c)^3 - 30*a*b*tan(1/2*d*x + 1/2*c)^3 - 80*b^2*tan(1/2*d*x + 1/ 2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 75*a*b*tan(1/2*d*x + 1/2*c) + 60*b^ 2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 13.33 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.64 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {3\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\left (2\,a^2-\frac {5\,a\,b}{2}+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {16\,a^2}{3}+a\,b-\frac {8\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,a^2}{3}+\frac {116\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,a^2}{3}-a\,b-\frac {8\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {5\,a\,b}{2}+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b/cos(c + d*x))^2/cos(c + d*x)^4,x)
Output:
(3*a*b*atanh(tan(c/2 + (d*x)/2)))/(2*d) - (tan(c/2 + (d*x)/2)^5*((20*a^2)/ 3 + (116*b^2)/15) + tan(c/2 + (d*x)/2)^9*(2*a^2 - (5*a*b)/2 + 2*b^2) - tan (c/2 + (d*x)/2)^3*(a*b + (16*a^2)/3 + (8*b^2)/3) - tan(c/2 + (d*x)/2)^7*(( 16*a^2)/3 - a*b + (8*b^2)/3) + tan(c/2 + (d*x)/2)*((5*a*b)/2 + 2*a^2 + 2*b ^2))/(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 = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.28 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a b +90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a b -45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b +45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a b -90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a b +45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b -45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +40 \sin \left (d x +c \right )^{5} a^{2}+32 \sin \left (d x +c \right )^{5} b^{2}-100 \sin \left (d x +c \right )^{3} a^{2}-80 \sin \left (d x +c \right )^{3} b^{2}+60 \sin \left (d x +c \right ) a^{2}+60 \sin \left (d x +c \right ) b^{2}}{60 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*sec(d*x+c))^2,x)
Output:
( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b + 90*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b - 45*cos(c + d*x)* log(tan((c + d*x)/2) - 1)*a*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**4*a*b - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**2*a*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b - 45*cos(c + d*x )*sin(c + d*x)**3*a*b + 75*cos(c + d*x)*sin(c + d*x)*a*b + 40*sin(c + d*x) **5*a**2 + 32*sin(c + d*x)**5*b**2 - 100*sin(c + d*x)**3*a**2 - 80*sin(c + d*x)**3*b**2 + 60*sin(c + d*x)*a**2 + 60*sin(c + d*x)*b**2)/(60*cos(c + d *x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))