Integrand size = 19, antiderivative size = 66 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=-\frac {a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {3 a^3 \sec ^2(c+d x)}{2 d}+\frac {a^3 \sec ^3(c+d x)}{3 d} \] Output:
-a^3*ln(cos(d*x+c))/d+3*a^3*sec(d*x+c)/d+3/2*a^3*sec(d*x+c)^2/d+1/3*a^3*se c(d*x+c)^3/d
Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=-\frac {a^3 (-22-18 \cos (2 (c+d x))+9 \cos (c+d x) (-2+\log (\cos (c+d x)))+3 \cos (3 (c+d x)) \log (\cos (c+d x))) \sec ^3(c+d x)}{12 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x],x]
Output:
-1/12*(a^3*(-22 - 18*Cos[2*(c + d*x)] + 9*Cos[c + d*x]*(-2 + Log[Cos[c + d *x]]) + 3*Cos[3*(c + d*x)]*Log[Cos[c + d*x]])*Sec[c + d*x]^3)/d
Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73, 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, 49, 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.
\(\displaystyle \int \tan (c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^3 (\cos (c+d x)+1)^3 \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \int (\cos (c+d x)+1)^3 \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {a^3 \int \left (\sec ^4(c+d x)+3 \sec ^3(c+d x)+3 \sec ^2(c+d x)+\sec (c+d x)\right )d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (-\frac {1}{3} \sec ^3(c+d x)-\frac {3}{2} \sec ^2(c+d x)-3 \sec (c+d x)+\log (\cos (c+d x))\right )}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x],x]
Output:
-((a^3*(Log[Cos[c + d*x]] - 3*Sec[c + d*x] - (3*Sec[c + d*x]^2)/2 - Sec[c + d*x]^3/3))/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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 44, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right )^{3}}{3}+\frac {3 \sec \left (d x +c \right )^{2}}{2}+3 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(44\) |
default | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right )^{3}}{3}+\frac {3 \sec \left (d x +c \right )^{2}}{2}+3 \sec \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\) | \(44\) |
parts | \(\frac {a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {a^{3} \sec \left (d x +c \right )^{3}}{3 d}+\frac {3 a^{3} \sec \left (d x +c \right )}{d}+\frac {3 a^{3} \sec \left (d x +c \right )^{2}}{2 d}\) | \(67\) |
risch | \(i a^{3} x +\frac {2 i a^{3} c}{d}+\frac {2 a^{3} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}+9 \,{\mathrm e}^{4 i \left (d x +c \right )}+22 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(116\) |
Input:
int((a+a*sec(d*x+c))^3*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*a^3*(1/3*sec(d*x+c)^3+3/2*sec(d*x+c)^2+3*sec(d*x+c)+ln(sec(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=-\frac {6 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 18 \, a^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+a*sec(d*x+c))^3*tan(d*x+c),x, algorithm="fricas")
Output:
-1/6*(6*a^3*cos(d*x + c)^3*log(-cos(d*x + c)) - 18*a^3*cos(d*x + c)^2 - 9* a^3*cos(d*x + c) - 2*a^3)/(d*cos(d*x + c)^3)
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=\begin {cases} \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{3} \sec ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 a^{3} \sec ^{2}{\left (c + d x \right )}}{2 d} + \frac {3 a^{3} \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sec {\left (c \right )} + a\right )^{3} \tan {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+a*sec(d*x+c))**3*tan(d*x+c),x)
Output:
Piecewise((a**3*log(tan(c + d*x)**2 + 1)/(2*d) + a**3*sec(c + d*x)**3/(3*d ) + 3*a**3*sec(c + d*x)**2/(2*d) + 3*a**3*sec(c + d*x)/d, Ne(d, 0)), (x*(a *sec(c) + a)**3*tan(c), True))
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=-\frac {6 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac {18 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+a*sec(d*x+c))^3*tan(d*x+c),x, algorithm="maxima")
Output:
-1/6*(6*a^3*log(cos(d*x + c)) - (18*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c ) + 2*a^3)/cos(d*x + c)^3)/d
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=-\frac {6 \, a^{3} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {18 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate((a+a*sec(d*x+c))^3*tan(d*x+c),x, algorithm="giac")
Output:
-1/6*(6*a^3*log(abs(cos(d*x + c))) - (18*a^3*cos(d*x + c)^2 + 9*a^3*cos(d* x + c) + 2*a^3)/cos(d*x + c)^3)/d
Time = 13.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.59 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=\frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {20\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(c + d*x)*(a + a/cos(c + d*x))^3,x)
Output:
(2*a^3*atanh(tan(c/2 + (d*x)/2)^2))/d - (2*a^3*tan(c/2 + (d*x)/2)^4 - 6*a^ 3*tan(c/2 + (d*x)/2)^2 + (20*a^3)/3)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/ 2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int (a+a \sec (c+d x))^3 \tan (c+d x) \, dx=\frac {a^{3} \left (3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+2 \sec \left (d x +c \right )^{3}+9 \sec \left (d x +c \right )^{2}+18 \sec \left (d x +c \right )\right )}{6 d} \] Input:
int((a+a*sec(d*x+c))^3*tan(d*x+c),x)
Output:
(a**3*(3*log(tan(c + d*x)**2 + 1) + 2*sec(c + d*x)**3 + 9*sec(c + d*x)**2 + 18*sec(c + d*x)))/(6*d)