Integrand size = 21, antiderivative size = 99 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\log (\cos (c+d x))}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {\sec ^2(c+d x)}{a^3 d}+\frac {2 \sec ^3(c+d x)}{3 a^3 d}-\frac {3 \sec ^4(c+d x)}{4 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d} \]
-ln(cos(d*x+c))/a^3/d-3*sec(d*x+c)/a^3/d+sec(d*x+c)^2/a^3/d+2/3*sec(d*x+c) ^3/a^3/d-3/4*sec(d*x+c)^4/a^3/d+1/5*sec(d*x+c)^5/a^3/d
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {(142+280 \cos (2 (c+d x))+90 \cos (4 (c+d x))+150 \cos (c+d x) \log (\cos (c+d x))+15 \cos (5 (c+d x)) \log (\cos (c+d x))+15 \cos (3 (c+d x)) (-4+5 \log (\cos (c+d x)))) \sec ^5(c+d x)}{240 a^3 d} \]
-1/240*((142 + 280*Cos[2*(c + d*x)] + 90*Cos[4*(c + d*x)] + 150*Cos[c + d* x]*Log[Cos[c + d*x]] + 15*Cos[5*(c + d*x)]*Log[Cos[c + d*x]] + 15*Cos[3*(c + d*x)]*(-4 + 5*Log[Cos[c + d*x]]))*Sec[c + d*x]^5)/(a^3*d)
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4367, 27, 84, 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.
\(\displaystyle \int \frac {\tan ^9(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^9}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^9}{\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^5 (1-\cos (c+d x))^4 (\cos (c+d x)+1) \sec ^6(c+d x)d\cos (c+d x)}{a^8 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int (1-\cos (c+d x))^4 (\cos (c+d x)+1) \sec ^6(c+d x)d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle -\frac {\int \left (\sec ^6(c+d x)-3 \sec ^5(c+d x)+2 \sec ^4(c+d x)+2 \sec ^3(c+d x)-3 \sec ^2(c+d x)+\sec (c+d x)\right )d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \sec ^5(c+d x)+\frac {3}{4} \sec ^4(c+d x)-\frac {2}{3} \sec ^3(c+d x)-\sec ^2(c+d x)+3 \sec (c+d x)+\log (\cos (c+d x))}{a^3 d}\) |
-((Log[Cos[c + d*x]] + 3*Sec[c + d*x] - Sec[c + d*x]^2 - (2*Sec[c + d*x]^3 )/3 + (3*Sec[c + d*x]^4)/4 - Sec[c + d*x]^5/5)/(a^3*d))
3.1.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 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 = 2.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {\frac {1}{5 \cos \left (d x +c \right )^{5}}-\ln \left (\cos \left (d x +c \right )\right )+\frac {2}{3 \cos \left (d x +c \right )^{3}}-\frac {3}{4 \cos \left (d x +c \right )^{4}}-\frac {3}{\cos \left (d x +c \right )}+\frac {1}{\cos \left (d x +c \right )^{2}}}{d \,a^{3}}\) | \(66\) |
default | \(\frac {\frac {1}{5 \cos \left (d x +c \right )^{5}}-\ln \left (\cos \left (d x +c \right )\right )+\frac {2}{3 \cos \left (d x +c \right )^{3}}-\frac {3}{4 \cos \left (d x +c \right )^{4}}-\frac {3}{\cos \left (d x +c \right )}+\frac {1}{\cos \left (d x +c \right )^{2}}}{d \,a^{3}}\) | \(66\) |
risch | \(\frac {i x}{a^{3}}+\frac {2 i c}{a^{3} d}-\frac {2 \left (45 \,{\mathrm e}^{9 i \left (d x +c \right )}-30 \,{\mathrm e}^{8 i \left (d x +c \right )}+140 \,{\mathrm e}^{7 i \left (d x +c \right )}+142 \,{\mathrm e}^{5 i \left (d x +c \right )}+140 \,{\mathrm e}^{3 i \left (d x +c \right )}-30 \,{\mathrm e}^{2 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\) | \(138\) |
1/d/a^3*(1/5/cos(d*x+c)^5-ln(cos(d*x+c))+2/3/cos(d*x+c)^3-3/4/cos(d*x+c)^4 -3/cos(d*x+c)+1/cos(d*x+c)^2)
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {60 \, \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) + 180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{60 \, a^{3} d \cos \left (d x + c\right )^{5}} \]
-1/60*(60*cos(d*x + c)^5*log(-cos(d*x + c)) + 180*cos(d*x + c)^4 - 60*cos( d*x + c)^3 - 40*cos(d*x + c)^2 + 45*cos(d*x + c) - 12)/(a^3*d*cos(d*x + c) ^5)
\[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\tan ^{9}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} + \frac {180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{a^{3} \cos \left (d x + c\right )^{5}}}{60 \, d} \]
-1/60*(60*log(cos(d*x + c))/a^3 + (180*cos(d*x + c)^4 - 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 + 45*cos(d*x + c) - 12)/(a^3*cos(d*x + c)^5))/d
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (93) = 186\).
Time = 6.25 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.04 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a^{3}} - \frac {\frac {475 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {590 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {50 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {805 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 119}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{5}}}{60 \, d} \]
1/60*(60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 - 60*log (abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/a^3 - (475*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 590*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 50*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 805*(cos(d*x + c) - 1)^4/( cos(d*x + c) + 1)^4 - 137*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 119) /(a^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^5))/d
Time = 18.70 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {98\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {64}{15}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \]
(2*atanh(tan(c/2 + (d*x)/2)^2))/(a^3*d) - ((58*tan(c/2 + (d*x)/2)^2)/3 - ( 98*tan(c/2 + (d*x)/2)^4)/3 + 22*tan(c/2 + (d*x)/2)^6 + 2*tan(c/2 + (d*x)/2 )^8 - 64/15)/(d*(5*a^3*tan(c/2 + (d*x)/2)^2 - 10*a^3*tan(c/2 + (d*x)/2)^4 + 10*a^3*tan(c/2 + (d*x)/2)^6 - 5*a^3*tan(c/2 + (d*x)/2)^8 + a^3*tan(c/2 + (d*x)/2)^10 - a^3))