Integrand size = 21, antiderivative size = 74 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^2 d}+\frac {a^2-b^2}{a b^2 d (a+b \sec (c+d x))} \] Output:
ln(cos(d*x+c))/a^2/d+(a^2+b^2)*ln(a+b*sec(d*x+c))/a^2/b^2/d+(a^2-b^2)/a/b^ 2/d/(a+b*sec(d*x+c))
Time = 0.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {b-\frac {b^3}{a^2}}{b+a \cos (c+d x)}+\log (\cos (c+d x))-\frac {\left (a^2+b^2\right ) \log (b+a \cos (c+d x))}{a^2}}{b^2 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
Output:
-(((b - b^3/a^2)/(b + a*Cos[c + d*x]) + Log[Cos[c + d*x]] - ((a^2 + b^2)*L og[b + a*Cos[c + d*x]])/a^2)/(b^2*d))
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 25, 4373, 522, 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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}{\left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (b^2-b^2 \sec ^2(c+d x)\right )}{b (a+b \sec (c+d x))^2}d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (\frac {-a^2-b^2}{a^2 (a+b \sec (c+d x))}+\frac {b \cos (c+d x)}{a^2}+\frac {a^2-b^2}{a (a+b \sec (c+d x))^2}\right )d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {a^2-b^2}{a (a+b \sec (c+d x))}+\frac {b^2 \log (b \sec (c+d x))}{a^2}-\frac {\left (a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2}}{b^2 d}\) |
Input:
Int[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
Output:
-(((b^2*Log[b*Sec[c + d*x]])/a^2 - ((a^2 + b^2)*Log[a + b*Sec[c + d*x]])/a ^2 - (a^2 - b^2)/(a*(a + b*Sec[c + d*x])))/(b^2*d))
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b^{2}}-\frac {a^{2}-b^{2}}{a^{2} b \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{2} a^{2}}}{d}\) | \(72\) |
default | \(\frac {-\frac {\ln \left (\cos \left (d x +c \right )\right )}{b^{2}}-\frac {a^{2}-b^{2}}{a^{2} b \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{2} a^{2}}}{d}\) | \(72\) |
risch | \(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}-\frac {2 \left (a^{2}-b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{a^{2} b d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}\) | \(163\) |
Input:
int(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^2*ln(cos(d*x+c))-(a^2-b^2)/a^2/b/(b+a*cos(d*x+c))+(a^2+b^2)/b^2/ a^2*ln(b+a*cos(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.38 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {a^{2} b - b^{3} - {\left (a^{2} b + b^{3} + {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (a^{3} \cos \left (d x + c\right ) + a^{2} b\right )} \log \left (-\cos \left (d x + c\right )\right )}{a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-(a^2*b - b^3 - (a^2*b + b^3 + (a^3 + a*b^2)*cos(d*x + c))*log(a*cos(d*x + c) + b) + (a^3*cos(d*x + c) + a^2*b)*log(-cos(d*x + c)))/(a^3*b^2*d*cos(d *x + c) + a^2*b^3*d)
\[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+b*sec(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**3/(a + b*sec(c + d*x))**2, x)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {a^{2} - b^{2}}{a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}} + \frac {\log \left (\cos \left (d x + c\right )\right )}{b^{2}} - \frac {{\left (a^{2} + b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{2}}}{d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-((a^2 - b^2)/(a^3*b*cos(d*x + c) + a^2*b^2) + log(cos(d*x + c))/b^2 - (a^ 2 + b^2)*log(a*cos(d*x + c) + b)/(a^2*b^2))/d
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{b^{2} d} + \frac {{\left (a^{2} + b^{2}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{2} b^{2} d} - \frac {a^{2} b - b^{3}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{2} b^{2} d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
-log(abs(cos(d*x + c)))/(b^2*d) + (a^2 + b^2)*log(abs(a*cos(d*x + c) + b)) /(a^2*b^2*d) - (a^2*b - b^3)/((a*cos(d*x + c) + b)*a^2*b^2*d)
Time = 10.85 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.68 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (\frac {1}{a^2}+\frac {1}{b^2}\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {2\,\left (a+b\right )}{a\,b\,d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \] Input:
int(tan(c + d*x)^3/(a + b/cos(c + d*x))^2,x)
Output:
(log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(1/a^2 + 1/b ^2))/d - log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - log(tan(c/2 + (d*x)/2)^2 - 1)/(b^2*d) - (2*(a + b))/(a*b*d*(a + b - tan(c/2 + (d*x)/2)^2*(a - b)))
Time = 0.17 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.78 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{3}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a \,b^{2}+\cos \left (d x +c \right ) a^{2} b -\cos \left (d x +c \right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) b^{3}-a^{2} b +a \,b^{2}}{a^{2} b^{2} d \left (\cos \left (d x +c \right ) a +b \right )} \] Input:
int(tan(d*x+c)^3/(a+b*sec(d*x+c))^2,x)
Output:
( - cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a*b**2 - cos(c + d*x)*log(ta n((c + d*x)/2) - 1)*a**3 - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3 + c os(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a** 3 + cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b )*a*b**2 + cos(c + d*x)*a**2*b - cos(c + d*x)*a*b**2 - log(tan((c + d*x)/2 )**2 + 1)*b**3 - log(tan((c + d*x)/2) - 1)*a**2*b - log(tan((c + d*x)/2) + 1)*a**2*b + log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a* *2*b + log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*b**3 - a **2*b + a*b**2)/(a**2*b**2*d*(cos(c + d*x)*a + b))