Integrand size = 19, antiderivative size = 56 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {1}{2 a^3 d (1+\cos (c+d x))^2}-\frac {2}{a^3 d (1+\cos (c+d x))}-\frac {\log (1+\cos (c+d x))}{a^3 d} \] Output:
1/2/a^3/d/(1+cos(d*x+c))^2-2/a^3/d/(1+cos(d*x+c))-ln(1+cos(d*x+c))/a^3/d
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.41 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-1+8 \cos ^2\left (\frac {1}{2} (c+d x)\right )+16 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^3(c+d x)}{a^3 d (1+\sec (c+d x))^3} \] Input:
Integrate[Tan[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
-((Cos[(c + d*x)/2]^2*(-1 + 8*Cos[(c + d*x)/2]^2 + 16*Cos[(c + d*x)/2]^4*L og[Cos[(c + d*x)/2]])*Sec[c + d*x]^3)/(a^3*d*(1 + Sec[c + d*x])^3))
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x)}{a^3 (\cos (c+d x)+1)^3}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x)}{(\cos (c+d x)+1)^3}d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (\frac {1}{\cos (c+d x)+1}-\frac {2}{(\cos (c+d x)+1)^2}+\frac {1}{(\cos (c+d x)+1)^3}\right )d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {2}{\cos (c+d x)+1}-\frac {1}{2 (\cos (c+d x)+1)^2}+\log (\cos (c+d x)+1)}{a^3 d}\) |
Input:
Int[Tan[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
-((-1/2*1/(1 + Cos[c + d*x])^2 + 2/(1 + Cos[c + d*x]) + Log[1 + Cos[c + d* x]])/(a^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 = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\ln \left (\sec \left (d x +c \right )\right )+\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}+\frac {1}{1+\sec \left (d x +c \right )}-\ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{3}}\) | \(49\) |
default | \(\frac {\ln \left (\sec \left (d x +c \right )\right )+\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}+\frac {1}{1+\sec \left (d x +c \right )}-\ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{3}}\) | \(49\) |
risch | \(\frac {i x}{a^{3}}+\frac {2 i c}{d \,a^{3}}-\frac {2 \left (2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(94\) |
Input:
int(tan(d*x+c)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(ln(sec(d*x+c))+1/2/(1+sec(d*x+c))^2+1/(1+sec(d*x+c))-ln(1+sec(d*x +c)))
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.36 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, \cos \left (d x + c\right ) + 3}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
-1/2*(2*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 4*cos(d*x + c) + 3)/(a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + a^3*d )
Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (48) = 96\).
Time = 17.36 (sec) , antiderivative size = 411, normalized size of antiderivative = 7.34 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\begin {cases} \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec ^{2}{\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\sec {\left (c + d x \right )} + 1 \right )} \sec ^{2}{\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {4 \log {\left (\sec {\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} - \frac {2 \log {\left (\sec {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \sec {\left (c + d x \right )}}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} + \frac {3}{2 a^{3} d \sec ^{2}{\left (c + d x \right )} + 4 a^{3} d \sec {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \tan {\left (c \right )}}{\left (a \sec {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))**3,x)
Output:
Piecewise((log(tan(c + d*x)**2 + 1)*sec(c + d*x)**2/(2*a**3*d*sec(c + d*x) **2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + 2*log(tan(c + d*x)**2 + 1)*sec(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + lo g(tan(c + d*x)**2 + 1)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) - 2*log(sec(c + d*x) + 1)*sec(c + d*x)**2/(2*a**3*d*sec(c + d*x )**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) - 4*log(sec(c + d*x) + 1)*sec(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) - 2*lo g(sec(c + d*x) + 1)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2* a**3*d) + 2*sec(c + d*x)/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a**3*d) + 3/(2*a**3*d*sec(c + d*x)**2 + 4*a**3*d*sec(c + d*x) + 2*a** 3*d), Ne(d, 0)), (x*tan(c)/(a*sec(c) + a)**3, True))
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.07 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {4 \, \cos \left (d x + c\right ) + 3}{a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {2 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-1/2*((4*cos(d*x + c) + 3)/(a^3*cos(d*x + c)^2 + 2*a^3*cos(d*x + c) + a^3) + 2*log(cos(d*x + c) + 1)/a^3)/d
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} - \frac {4 \, \cos \left (d x + c\right ) + 3}{2 \, a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
-log(abs(cos(d*x + c) + 1))/(a^3*d) - 1/2*(4*cos(d*x + c) + 3)/(a^3*d*(cos (d*x + c) + 1)^2)
Time = 12.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}}{a^3\,d} \] Input:
int(tan(c + d*x)/(a + a/cos(c + d*x))^3,x)
Output:
(log(tan(c/2 + (d*x)/2)^2 + 1) - (3*tan(c/2 + (d*x)/2)^2)/4 + tan(c/2 + (d *x)/2)^4/8)/(a^3*d)
Time = 0.15 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.57 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sec \left (d x +c \right )^{2}+2 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sec \left (d x +c \right )+\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-2 \,\mathrm {log}\left (\sec \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\sec \left (d x +c \right )+1\right ) \sec \left (d x +c \right )-2 \,\mathrm {log}\left (\sec \left (d x +c \right )+1\right )-3 \sec \left (d x +c \right )^{2}-4 \sec \left (d x +c \right )}{2 a^{3} d \left (\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1\right )} \] Input:
int(tan(d*x+c)/(a+a*sec(d*x+c))^3,x)
Output:
(log(tan(c + d*x)**2 + 1)*sec(c + d*x)**2 + 2*log(tan(c + d*x)**2 + 1)*sec (c + d*x) + log(tan(c + d*x)**2 + 1) - 2*log(sec(c + d*x) + 1)*sec(c + d*x )**2 - 4*log(sec(c + d*x) + 1)*sec(c + d*x) - 2*log(sec(c + d*x) + 1) - 3* sec(c + d*x)**2 - 4*sec(c + d*x))/(2*a**3*d*(sec(c + d*x)**2 + 2*sec(c + d *x) + 1))