Integrand size = 19, antiderivative size = 87 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=-\frac {a \log (\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \sec ^2(c+d x)}{d}-\frac {2 a \sec ^3(c+d x)}{3 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {a \sec ^5(c+d x)}{5 d} \] Output:
-a*ln(cos(d*x+c))/d+a*sec(d*x+c)/d-a*sec(d*x+c)^2/d-2/3*a*sec(d*x+c)^3/d+1 /4*a*sec(d*x+c)^4/d+1/5*a*sec(d*x+c)^5/d
Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=-\frac {a \log (\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}-\frac {a \sec ^2(c+d x)}{d}-\frac {2 a \sec ^3(c+d x)}{3 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {a \sec ^5(c+d x)}{5 d} \] Input:
Integrate[(a + a*Sec[c + d*x])*Tan[c + d*x]^5,x]
Output:
-((a*Log[Cos[c + d*x]])/d) + (a*Sec[c + d*x])/d - (a*Sec[c + d*x]^2)/d - ( 2*a*Sec[c + d*x]^3)/(3*d) + (a*Sec[c + d*x]^4)/(4*d) + (a*Sec[c + d*x]^5)/ (5*d)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, 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.
\(\displaystyle \int \tan ^5(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^5 (1-\cos (c+d x))^2 (\cos (c+d x)+1)^3 \sec ^6(c+d x)d\cos (c+d x)}{a^4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int (1-\cos (c+d x))^2 (\cos (c+d x)+1)^3 \sec ^6(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a \int \left (\sec ^6(c+d x)+\sec ^5(c+d x)-2 \sec ^4(c+d x)-2 \sec ^3(c+d x)+\sec ^2(c+d x)+\sec (c+d x)\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (-\frac {1}{5} \sec ^5(c+d x)-\frac {1}{4} \sec ^4(c+d x)+\frac {2}{3} \sec ^3(c+d x)+\sec ^2(c+d x)-\sec (c+d x)+\log (\cos (c+d x))\right )}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])*Tan[c + d*x]^5,x]
Output:
-((a*(Log[Cos[c + d*x]] - Sec[c + d*x] + Sec[c + d*x]^2 + (2*Sec[c + d*x]^ 3)/3 - Sec[c + d*x]^4/4 - Sec[c + d*x]^5/5))/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 = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sec \left (d x +c \right )^{5}}{5}+\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {2 \sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
default | \(\frac {a \left (\frac {\sec \left (d x +c \right )^{5}}{5}+\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {2 \sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a \left (\frac {\sec \left (d x +c \right )^{5}}{5}-\frac {2 \sec \left (d x +c \right )^{3}}{3}+\sec \left (d x +c \right )\right )}{d}\) | \(73\) |
risch | \(i a x +\frac {2 i a c}{d}+\frac {2 a \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}-30 \,{\mathrm e}^{8 i \left (d x +c \right )}+20 \,{\mathrm e}^{7 i \left (d x +c \right )}-60 \,{\mathrm e}^{6 i \left (d x +c \right )}+58 \,{\mathrm e}^{5 i \left (d x +c \right )}-60 \,{\mathrm e}^{4 i \left (d x +c \right )}+20 \,{\mathrm e}^{3 i \left (d x +c \right )}-30 \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(152\) |
Input:
int((a+a*sec(d*x+c))*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
a/d*(1/5*sec(d*x+c)^5+1/4*sec(d*x+c)^4-2/3*sec(d*x+c)^3-sec(d*x+c)^2+sec(d *x+c)+ln(sec(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=-\frac {60 \, a \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 60 \, a \cos \left (d x + c\right )^{4} + 60 \, a \cos \left (d x + c\right )^{3} + 40 \, a \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 12 \, a}{60 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)^5,x, algorithm="fricas")
Output:
-1/60*(60*a*cos(d*x + c)^5*log(-cos(d*x + c)) - 60*a*cos(d*x + c)^4 + 60*a *cos(d*x + c)^3 + 40*a*cos(d*x + c)^2 - 15*a*cos(d*x + c) - 12*a)/(d*cos(d *x + c)^5)
Time = 0.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.29 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{5 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {4 a \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {8 a \sec {\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sec {\left (c \right )} + a\right ) \tan ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)**5,x)
Output:
Piecewise((a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**4*sec(c + d* x)/(5*d) + a*tan(c + d*x)**4/(4*d) - 4*a*tan(c + d*x)**2*sec(c + d*x)/(15* d) - a*tan(c + d*x)**2/(2*d) + 8*a*sec(c + d*x)/(15*d), Ne(d, 0)), (x*(a*s ec(c) + a)*tan(c)**5, True))
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=-\frac {60 \, a \log \left (\cos \left (d x + c\right )\right ) - \frac {60 \, a \cos \left (d x + c\right )^{4} - 60 \, a \cos \left (d x + c\right )^{3} - 40 \, a \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 12 \, a}{\cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/60*(60*a*log(cos(d*x + c)) - (60*a*cos(d*x + c)^4 - 60*a*cos(d*x + c)^3 - 40*a*cos(d*x + c)^2 + 15*a*cos(d*x + c) + 12*a)/cos(d*x + c)^5)/d
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=-\frac {60 \, a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {60 \, a \cos \left (d x + c\right )^{4} - 60 \, a \cos \left (d x + c\right )^{3} - 40 \, a \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 12 \, a}{\cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate((a+a*sec(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/60*(60*a*log(abs(cos(d*x + c))) - (60*a*cos(d*x + c)^4 - 60*a*cos(d*x + c)^3 - 40*a*cos(d*x + c)^2 + 15*a*cos(d*x + c) + 12*a)/cos(d*x + c)^5)/d
Time = 16.71 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.74 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {62\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {22\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {16\,a}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(c + d*x)^5*(a + a/cos(c + d*x)),x)
Output:
(2*a*atanh(tan(c/2 + (d*x)/2)^2))/d - ((16*a)/15 - (22*a*tan(c/2 + (d*x)/2 )^2)/3 + (62*a*tan(c/2 + (d*x)/2)^4)/3 - 10*a*tan(c/2 + (d*x)/2)^6 + 2*a*t an(c/2 + (d*x)/2)^8)/(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 = 80, normalized size of antiderivative = 0.92 \[ \int (a+a \sec (c+d x)) \tan ^5(c+d x) \, dx=\frac {a \left (30 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+12 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{4}-16 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+32 \sec \left (d x +c \right )+15 \tan \left (d x +c \right )^{4}-30 \tan \left (d x +c \right )^{2}\right )}{60 d} \] Input:
int((a+a*sec(d*x+c))*tan(d*x+c)^5,x)
Output:
(a*(30*log(tan(c + d*x)**2 + 1) + 12*sec(c + d*x)*tan(c + d*x)**4 - 16*sec (c + d*x)*tan(c + d*x)**2 + 32*sec(c + d*x) + 15*tan(c + d*x)**4 - 30*tan( c + d*x)**2))/(60*d)