Integrand size = 21, antiderivative size = 120 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \sec ^2(c+d x)}{2 d}-\frac {4 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \sec ^4(c+d x)}{4 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \sec ^6(c+d x)}{6 d} \] Output:
-a^2*ln(cos(d*x+c))/d+2*a^2*sec(d*x+c)/d-1/2*a^2*sec(d*x+c)^2/d-4/3*a^2*se c(d*x+c)^3/d-1/4*a^2*sec(d*x+c)^4/d+2/5*a^2*sec(d*x+c)^5/d+1/6*a^2*sec(d*x +c)^6/d
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {a^2 (312 \cos (c+d x)-5 (14-28 \cos (3 (c+d x))+6 \cos (4 (c+d x))-12 \cos (5 (c+d x))+30 \log (\cos (c+d x))+18 \cos (4 (c+d x)) \log (\cos (c+d x))+3 \cos (6 (c+d x)) \log (\cos (c+d x))+9 \cos (2 (c+d x)) (4+5 \log (\cos (c+d x))))) \sec ^6(c+d x)}{480 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^5,x]
Output:
(a^2*(312*Cos[c + d*x] - 5*(14 - 28*Cos[3*(c + d*x)] + 6*Cos[4*(c + d*x)] - 12*Cos[5*(c + d*x)] + 30*Log[Cos[c + d*x]] + 18*Cos[4*(c + d*x)]*Log[Cos [c + d*x]] + 3*Cos[6*(c + d*x)]*Log[Cos[c + d*x]] + 9*Cos[2*(c + d*x)]*(4 + 5*Log[Cos[c + d*x]])))*Sec[c + d*x]^6)/(480*d)
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4367, 27, 99, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2dx\) |
| \(\Big \downarrow \) 4367 |
| \(\displaystyle -\frac {\int a^6 (1-\cos (c+d x))^2 (\cos (c+d x)+1)^4 \sec ^7(c+d x)d\cos (c+d x)}{a^4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \int (1-\cos (c+d x))^2 (\cos (c+d x)+1)^4 \sec ^7(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^2 \int \left (\sec ^7(c+d x)+2 \sec ^6(c+d x)-\sec ^5(c+d x)-4 \sec ^4(c+d x)-\sec ^3(c+d x)+2 \sec ^2(c+d x)+\sec (c+d x)\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \left (-\frac {1}{6} \sec ^6(c+d x)-\frac {2}{5} \sec ^5(c+d x)+\frac {1}{4} \sec ^4(c+d x)+\frac {4}{3} \sec ^3(c+d x)+\frac {1}{2} \sec ^2(c+d x)-2 \sec (c+d x)+\log (\cos (c+d x))\right )}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^5,x]
Output:
-((a^2*(Log[Cos[c + d*x]] - 2*Sec[c + d*x] + Sec[c + d*x]^2/2 + (4*Sec[c + d*x]^3)/3 + Sec[c + d*x]^4/4 - (2*Sec[c + d*x]^5)/5 - Sec[c + d*x]^6/6))/ d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right )^{6}}{6}+\frac {2 \sec \left (d x +c \right )^{5}}{5}-\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {4 \sec \left (d x +c \right )^{3}}{3}-\frac {\sec \left (d x +c \right )^{2}}{2}+2 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(74\) |
| default | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right )^{6}}{6}+\frac {2 \sec \left (d x +c \right )^{5}}{5}-\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {4 \sec \left (d x +c \right )^{3}}{3}-\frac {\sec \left (d x +c \right )^{2}}{2}+2 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(74\) |
| parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{2} \tan \left (d x +c \right )^{6}}{6 d}+\frac {2 a^{2} \left (\frac {\sec \left (d x +c \right )^{5}}{5}-\frac {2 \sec \left (d x +c \right )^{3}}{3}+\sec \left (d x +c \right )\right )}{d}\) | \(94\) |
| risch | \(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {2 a^{2} \left (30 \,{\mathrm e}^{11 i \left (d x +c \right )}-15 \,{\mathrm e}^{10 i \left (d x +c \right )}+70 \,{\mathrm e}^{9 i \left (d x +c \right )}-90 \,{\mathrm e}^{8 i \left (d x +c \right )}+156 \,{\mathrm e}^{7 i \left (d x +c \right )}-70 \,{\mathrm e}^{6 i \left (d x +c \right )}+156 \,{\mathrm e}^{5 i \left (d x +c \right )}-90 \,{\mathrm e}^{4 i \left (d x +c \right )}+70 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{2 i \left (d x +c \right )}+30 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(182\) |
Input:
int((a+a*sec(d*x+c))^2*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
a^2/d*(1/6*sec(d*x+c)^6+2/5*sec(d*x+c)^5-1/4*sec(d*x+c)^4-4/3*sec(d*x+c)^3 -1/2*sec(d*x+c)^2+2*sec(d*x+c)+ln(sec(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {60 \, a^{2} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 120 \, a^{2} \cos \left (d x + c\right )^{5} + 30 \, a^{2} \cos \left (d x + c\right )^{4} + 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )^{2} - 24 \, a^{2} \cos \left (d x + c\right ) - 10 \, a^{2}}{60 \, d \cos \left (d x + c\right )^{6}} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^5,x, algorithm="fricas")
Output:
-1/60*(60*a^2*cos(d*x + c)^6*log(-cos(d*x + c)) - 120*a^2*cos(d*x + c)^5 + 30*a^2*cos(d*x + c)^4 + 80*a^2*cos(d*x + c)^3 + 15*a^2*cos(d*x + c)^2 - 2 4*a^2*cos(d*x + c) - 10*a^2)/(d*cos(d*x + c)^6)
Time = 0.64 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.58 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac {2 a^{2} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} + \frac {a^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{6 d} - \frac {8 a^{2} \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} - \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {a^{2} \sec ^{2}{\left (c + d x \right )}}{6 d} + \frac {16 a^{2} \sec {\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\left (c \right )} + a\right )^{2} \tan ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+a*sec(d*x+c))**2*tan(d*x+c)**5,x)
Output:
Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**4*sec( c + d*x)**2/(6*d) + 2*a**2*tan(c + d*x)**4*sec(c + d*x)/(5*d) + a**2*tan(c + d*x)**4/(4*d) - a**2*tan(c + d*x)**2*sec(c + d*x)**2/(6*d) - 8*a**2*tan (c + d*x)**2*sec(c + d*x)/(15*d) - a**2*tan(c + d*x)**2/(2*d) + a**2*sec(c + d*x)**2/(6*d) + 16*a**2*sec(c + d*x)/(15*d), Ne(d, 0)), (x*(a*sec(c) + a)**2*tan(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {120 \, a^{2} \cos \left (d x + c\right )^{5} - 30 \, a^{2} \cos \left (d x + c\right )^{4} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )^{2} + 24 \, a^{2} \cos \left (d x + c\right ) + 10 \, a^{2}}{\cos \left (d x + c\right )^{6}}}{60 \, d} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/60*(60*a^2*log(cos(d*x + c)) - (120*a^2*cos(d*x + c)^5 - 30*a^2*cos(d*x + c)^4 - 80*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c)^2 + 24*a^2*cos(d*x + c) + 10*a^2)/cos(d*x + c)^6)/d
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {60 \, a^{2} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {120 \, a^{2} \cos \left (d x + c\right )^{5} - 30 \, a^{2} \cos \left (d x + c\right )^{4} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )^{2} + 24 \, a^{2} \cos \left (d x + c\right ) + 10 \, a^{2}}{\cos \left (d x + c\right )^{6}}}{60 \, d} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/60*(60*a^2*log(abs(cos(d*x + c))) - (120*a^2*cos(d*x + c)^5 - 30*a^2*co s(d*x + c)^4 - 80*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c)^2 + 24*a^2*cos( d*x + c) + 10*a^2)/cos(d*x + c)^6)/d
Time = 16.08 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.60 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {92\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-44\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {74\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {32\,a^2}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int(tan(c + d*x)^5*(a + a/cos(c + d*x))^2,x)
Output:
(2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d - ((74*a^2*tan(c/2 + (d*x)/2)^2)/5 - 44*a^2*tan(c/2 + (d*x)/2)^4 + (92*a^2*tan(c/2 + (d*x)/2)^6)/3 - 12*a^2*ta n(c/2 + (d*x)/2)^8 + 2*a^2*tan(c/2 + (d*x)/2)^10 - (32*a^2)/15)/(d*(15*tan (c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15* tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1 ))
Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07 \[ \int (a+a \sec (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {a^{2} \left (30 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+10 \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{4}-10 \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}+10 \sec \left (d x +c \right )^{2}+24 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{4}-32 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+64 \sec \left (d x +c \right )+15 \tan \left (d x +c \right )^{4}-30 \tan \left (d x +c \right )^{2}\right )}{60 d} \] Input:
int((a+a*sec(d*x+c))^2*tan(d*x+c)^5,x)
Output:
(a**2*(30*log(tan(c + d*x)**2 + 1) + 10*sec(c + d*x)**2*tan(c + d*x)**4 - 10*sec(c + d*x)**2*tan(c + d*x)**2 + 10*sec(c + d*x)**2 + 24*sec(c + d*x)* tan(c + d*x)**4 - 32*sec(c + d*x)*tan(c + d*x)**2 + 64*sec(c + d*x) + 15*t an(c + d*x)**4 - 30*tan(c + d*x)**2))/(60*d)