Integrand size = 21, antiderivative size = 108 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\log (\cos (c+d x))}{a d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a b^4 d}+\frac {\left (a^2-2 b^2\right ) \sec (c+d x)}{b^3 d}-\frac {a \sec ^2(c+d x)}{2 b^2 d}+\frac {\sec ^3(c+d x)}{3 b d} \] Output:
-ln(cos(d*x+c))/a/d-(a^2-b^2)^2*ln(a+b*sec(d*x+c))/a/b^4/d+(a^2-2*b^2)*sec (d*x+c)/b^3/d-1/2*a*sec(d*x+c)^2/b^2/d+1/3*sec(d*x+c)^3/b/d
Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-\frac {b^4 \log (\cos (c+d x))}{a}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a}+b \left (a^2-2 b^2\right ) \sec (c+d x)-\frac {1}{2} a b^2 \sec ^2(c+d x)+\frac {1}{3} b^3 \sec ^3(c+d x)}{b^4 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + b*Sec[c + d*x]),x]
Output:
(-((b^4*Log[Cos[c + d*x]])/a) - ((a^2 - b^2)^2*Log[a + b*Sec[c + d*x]])/a + b*(a^2 - 2*b^2)*Sec[c + d*x] - (a*b^2*Sec[c + d*x]^2)/2 + (b^3*Sec[c + d *x]^3)/3)/(b^4*d)
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^5}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}{a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle \frac {\int \frac {\cos (c+d x) \left (b^2-b^2 \sec ^2(c+d x)\right )^2}{b (a+b \sec (c+d x))}d(b \sec (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\frac {\cos (c+d x) b^3}{a}+\sec ^2(c+d x) b^2-a \sec (c+d x) b+a^2 \left (1-\frac {2 b^2}{a^2}\right )-\frac {\left (a^2-b^2\right )^2}{a (a+b \sec (c+d x))}\right )d(b \sec (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (a^2-2 b^2\right ) \sec (c+d x)-\frac {\left (a^2-b^2\right )^2 \log (a+b \sec (c+d x))}{a}+\frac {b^4 \log (b \sec (c+d x))}{a}-\frac {1}{2} a b^2 \sec ^2(c+d x)+\frac {1}{3} b^3 \sec ^3(c+d x)}{b^4 d}\) |
Input:
Int[Tan[c + d*x]^5/(a + b*Sec[c + d*x]),x]
Output:
((b^4*Log[b*Sec[c + d*x]])/a - ((a^2 - b^2)^2*Log[a + b*Sec[c + d*x]])/a + b*(a^2 - 2*b^2)*Sec[c + d*x] - (a*b^2*Sec[c + d*x]^2)/2 + (b^3*Sec[c + d* x]^3)/3)/(b^4*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 = 115, normalized size of antiderivative = 1.06
method | result | size |
derivativedivides | \(\frac {-\frac {a}{2 b^{2} \cos \left (d x +c \right )^{2}}-\frac {-a^{2}+2 b^{2}}{b^{3} \cos \left (d x +c \right )}+\frac {\left (a^{2}-2 b^{2}\right ) a \ln \left (\cos \left (d x +c \right )\right )}{b^{4}}+\frac {1}{3 b \cos \left (d x +c \right )^{3}}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{4} a}}{d}\) | \(115\) |
default | \(\frac {-\frac {a}{2 b^{2} \cos \left (d x +c \right )^{2}}-\frac {-a^{2}+2 b^{2}}{b^{3} \cos \left (d x +c \right )}+\frac {\left (a^{2}-2 b^{2}\right ) a \ln \left (\cos \left (d x +c \right )\right )}{b^{4}}+\frac {1}{3 b \cos \left (d x +c \right )^{3}}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{4} a}}{d}\) | \(115\) |
risch | \(\frac {i x}{a}+\frac {2 i c}{d a}+\frac {2 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-\frac {16 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{3}-2 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{b^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{4} d}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{4} d}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{2} d}\) | \(303\) |
Input:
int(tan(d*x+c)^5/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*a/b^2/cos(d*x+c)^2-(-a^2+2*b^2)/b^3/cos(d*x+c)+(a^2-2*b^2)/b^4*a *ln(cos(d*x+c))+1/3/b/cos(d*x+c)^3+(-a^4+2*a^2*b^2-b^4)/b^4/a*ln(b+a*cos(d *x+c)))
Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {3 \, a^{2} b^{2} \cos \left (d x + c\right ) + 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (a \cos \left (d x + c\right ) + b\right ) - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 2 \, a b^{3} - 6 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}}{6 \, a b^{4} d \cos \left (d x + c\right )^{3}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
-1/6*(3*a^2*b^2*cos(d*x + c) + 6*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^3*lo g(a*cos(d*x + c) + b) - 6*(a^4 - 2*a^2*b^2)*cos(d*x + c)^3*log(-cos(d*x + c)) - 2*a*b^3 - 6*(a^3*b - 2*a*b^3)*cos(d*x + c)^2)/(a*b^4*d*cos(d*x + c)^ 3)
\[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+b*sec(d*x+c)),x)
Output:
Integral(tan(c + d*x)**5/(a + b*sec(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {6 \, {\left (a^{3} - 2 \, a b^{2}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{4}} - \frac {6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a b^{4}} - \frac {3 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}}{b^{3} \cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
1/6*(6*(a^3 - 2*a*b^2)*log(cos(d*x + c))/b^4 - 6*(a^4 - 2*a^2*b^2 + b^4)*l og(a*cos(d*x + c) + b)/(a*b^4) - (3*a*b*cos(d*x + c) - 6*(a^2 - 2*b^2)*cos (d*x + c)^2 - 2*b^2)/(b^3*cos(d*x + c)^3))/d
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {{\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{b^{4} d} - \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a b^{4} d} - \frac {3 \, a b^{2} \cos \left (d x + c\right ) - 2 \, b^{3} - 6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{6 \, b^{4} d \cos \left (d x + c\right )^{3}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
(a^3 - 2*a*b^2)*log(abs(cos(d*x + c)))/(b^4*d) - (a^4 - 2*a^2*b^2 + b^4)*l og(abs(a*cos(d*x + c) + b))/(a*b^4*d) - 1/6*(3*a*b^2*cos(d*x + c) - 2*b^3 - 6*(a^2*b - 2*b^3)*cos(d*x + c)^2)/(b^4*d*cos(d*x + c)^3)
Time = 11.52 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.10 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {\frac {2\,\left (3\,a^2-5\,b^2\right )}{3\,b^3}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+a\,b-4\,b^2\right )}{b^3}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+a\,b-b^2\right )}{b^3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (a^2-2\,b^2\right )}{b^4\,d}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,{\left (a^2-b^2\right )}^2}{a\,b^4\,d} \] Input:
int(tan(c + d*x)^5/(a + b/cos(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - ((2*(3*a^2 - 5*b^2))/(3*b^3) - (2*ta n(c/2 + (d*x)/2)^2*(a*b + 2*a^2 - 4*b^2))/b^3 + (2*tan(c/2 + (d*x)/2)^4*(a *b + a^2 - b^2))/b^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)) + (a*log(tan(c/2 + (d*x)/2)^2 - 1)*(a^2 - 2*b ^2))/(b^4*d) - (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^ 2)*(a^2 - b^2)^2)/(a*b^4*d)
Time = 0.18 (sec) , antiderivative size = 750, normalized size of antiderivative = 6.94 \[ \int \frac {\tan ^5(c+d x)}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^5/(a+b*sec(d*x+c)),x)
Output:
(6*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**4 - 6*cos( c + d*x)*log(tan((c + d*x)/2)**2 + 1)*b**4 + 6*cos(c + d*x)*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2*a**4 - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a* *4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**2 + 6*cos(c + d*x)* log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 - 12*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - 6*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*a**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**2 - 6 *cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*s in(c + d*x)**2*a**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*a**2*b**2 - 6*cos(c + d*x)*log(tan( (c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*b**4 + 6*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)* a**4 - 12*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*a**2*b**2 + 6*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d* x)/2)**2*b - a - b)*b**4 - 6*cos(c + d*x)*sin(c + d*x)**2*a**3*b + 3*cos(c + d*x)*sin(c + d*x)**2*a**2*b**2 + 10*cos(c + d*x)*sin(c + d*x)**2*a*b**3 + 6*cos(c + d*x)*a**3*b - 10*cos(c + d*x)*a*b**3 + 6*sin(c + d*x)**2*a**3 *b - 12*sin(c + d*x)**2*a*b**3 - 6*a**3*b + 10*a*b**3)/(6*cos(c + d*x)*a*b **4*d*(sin(c + d*x)**2 - 1))