Integrand size = 21, antiderivative size = 59 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\log (\cos (c+d x))}{a d}-\frac {\left (a^2-b^2\right ) \log (a+b \sec (c+d x))}{a b^2 d}+\frac {\sec (c+d x)}{b d} \] Output:
ln(cos(d*x+c))/a/d-(a^2-b^2)*ln(a+b*sec(d*x+c))/a/b^2/d+sec(d*x+c)/b/d
Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {b^2 \log (\cos (c+d x))+\left (-a^2+b^2\right ) \log (a+b \sec (c+d x))+a b \sec (c+d x)}{a b^2 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + b*Sec[c + d*x]),x]
Output:
(b^2*Log[Cos[c + d*x]] + (-a^2 + b^2)*Log[a + b*Sec[c + d*x]] + a*b*Sec[c + d*x])/(a*b^2*d)
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^3}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}{a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (b^2-b^2 \sec ^2(c+d x)\right )}{b (a+b \sec (c+d x))}d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (\frac {a^2-b^2}{a (a+b \sec (c+d x))}+\frac {b \cos (c+d x)}{a}-1\right )d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\left (a^2-b^2\right ) \log (a+b \sec (c+d x))}{a}+\frac {b^2 \log (b \sec (c+d x))}{a}-b \sec (c+d x)}{b^2 d}\) |
Input:
Int[Tan[c + d*x]^3/(a + b*Sec[c + d*x]),x]
Output:
-(((b^2*Log[b*Sec[c + d*x]])/a + ((a^2 - b^2)*Log[a + b*Sec[c + d*x]])/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 = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{2} a}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}}{d}\) | \(57\) |
default | \(\frac {\frac {\left (-a^{2}+b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{2} a}+\frac {a \ln \left (\cos \left (d x +c \right )\right )}{b^{2}}+\frac {1}{b \cos \left (d x +c \right )}}{d}\) | \(57\) |
risch | \(-\frac {i x}{a}-\frac {2 i c}{d a}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}\) | \(139\) |
Input:
int(tan(d*x+c)^3/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*((-a^2+b^2)/b^2/a*ln(b+a*cos(d*x+c))+a/b^2*ln(cos(d*x+c))+1/b/cos(d*x+ c))
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \log \left (a \cos \left (d x + c\right ) + b\right ) + a b}{a b^{2} d \cos \left (d x + c\right )} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
(a^2*cos(d*x + c)*log(-cos(d*x + c)) - (a^2 - b^2)*cos(d*x + c)*log(a*cos( d*x + c) + b) + a*b)/(a*b^2*d*cos(d*x + c))
\[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+b*sec(d*x+c)),x)
Output:
Integral(tan(c + d*x)**3/(a + b*sec(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {a \log \left (\cos \left (d x + c\right )\right )}{b^{2}} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{2}} + \frac {1}{b \cos \left (d x + c\right )}}{d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
(a*log(cos(d*x + c))/b^2 - (a^2 - b^2)*log(a*cos(d*x + c) + b)/(a*b^2) + 1 /(b*cos(d*x + c)))/d
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {a \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{b^{2} d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a b^{2} d} + \frac {1}{b d \cos \left (d x + c\right )} \] Input:
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
a*log(abs(cos(d*x + c)))/(b^2*d) - (a^2 - b^2)*log(abs(a*cos(d*x + c) + b) )/(a*b^2*d) + 1/(b*d*cos(d*x + c))
Time = 11.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.95 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{b^2\,d}-\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (\frac {a}{b^2}-\frac {1}{a}\right )}{d} \] Input:
int(tan(c + d*x)^3/(a + b/cos(c + d*x)),x)
Output:
(a*log(tan(c/2 + (d*x)/2)^2 - 1))/(b^2*d) - 2/(b*d*(tan(c/2 + (d*x)/2)^2 - 1)) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - (log(a + b - a*tan(c/2 + (d*x )/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(a/b^2 - 1/a))/d
Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.25 \[ \int \frac {\tan ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) a^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) b^{2}-\cos \left (d x +c \right ) a b +a b}{\cos \left (d x +c \right ) a \,b^{2} d} \] Input:
int(tan(d*x+c)^3/(a+b*sec(d*x+c)),x)
Output:
( - cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*b**2 + cos(c + d*x)*log(tan( (c + d*x)/2) - 1)*a**2 + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2 - cos (c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a**2 + cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)* b**2 - cos(c + d*x)*a*b + a*b)/(cos(c + d*x)*a*b**2*d)