Integrand size = 21, antiderivative size = 99 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\log (\cos (c+d x))}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {\sec ^2(c+d x)}{a^3 d}+\frac {2 \sec ^3(c+d x)}{3 a^3 d}-\frac {3 \sec ^4(c+d x)}{4 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d} \] Output:
-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 = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {(142+280 \cos (2 (c+d x))+90 \cos (4 (c+d x))+150 \cos (c+d x) \log (\cos (c+d x))+15 \cos (5 (c+d x)) \log (\cos (c+d x))+15 \cos (3 (c+d x)) (-4+5 \log (\cos (c+d x)))) \sec ^5(c+d x)}{240 a^3 d} \] Input:
Integrate[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]
Output:
-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 = 0.27 (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 |
| \(\displaystyle -\frac {-\frac {1}{5} \sec ^5(c+d x)+\frac {3}{4} \sec ^4(c+d x)-\frac {2}{3} \sec ^3(c+d x)-\sec ^2(c+d x)+3 \sec (c+d x)+\log (\cos (c+d x))}{a^3 d}\) |
Input:
Int[Tan[c + d*x]^9/(a + a*Sec[c + d*x])^3,x]
Output:
-((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))
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 = 1.99 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {-\frac {3}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{\cos \left (d x +c \right )^{2}}-\frac {3}{\cos \left (d x +c \right )}-\ln \left (\cos \left (d x +c \right )\right )+\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{5 \cos \left (d x +c \right )^{5}}}{d \,a^{3}}\) | \(66\) |
| default | \(\frac {-\frac {3}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{\cos \left (d x +c \right )^{2}}-\frac {3}{\cos \left (d x +c \right )}-\ln \left (\cos \left (d x +c \right )\right )+\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{5 \cos \left (d x +c \right )^{5}}}{d \,a^{3}}\) | \(66\) |
| risch | \(\frac {i x}{a^{3}}+\frac {2 i c}{d \,a^{3}}-\frac {2 \left (45 \,{\mathrm e}^{9 i \left (d x +c \right )}-30 \,{\mathrm e}^{8 i \left (d x +c \right )}+140 \,{\mathrm e}^{7 i \left (d x +c \right )}+142 \,{\mathrm e}^{5 i \left (d x +c \right )}+140 \,{\mathrm e}^{3 i \left (d x +c \right )}-30 \,{\mathrm e}^{2 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(138\) |
Input:
int(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-3/4/cos(d*x+c)^4+1/cos(d*x+c)^2-3/cos(d*x+c)-ln(cos(d*x+c))+2/3/ cos(d*x+c)^3+1/5/cos(d*x+c)^5)
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {60 \, \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) + 180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{60 \, a^{3} d \cos \left (d x + c\right )^{5}} \] Input:
integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
-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)
\[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\tan ^{9}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(tan(d*x+c)**9/(a+a*sec(d*x+c))**3,x)
Output:
Integral(tan(c + d*x)**9/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a^{3}} + \frac {180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{a^{3} \cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-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
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {180 \, \cos \left (d x + c\right )^{4} - 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 12}{\cos \left (d x + c\right )^{5}} + 60 \, \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{60 \, a^{3} d} \] Input:
integrate(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
-1/60*((180*cos(d*x + c)^4 - 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 + 45*co s(d*x + c) - 12)/cos(d*x + c)^5 + 60*log(abs(cos(d*x + c))))/(a^3*d)
Time = 15.96 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {98\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {58\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {64}{15}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )} \] Input:
int(tan(c + d*x)^9/(a + a/cos(c + d*x))^3,x)
Output:
(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))
Time = 0.15 (sec) , antiderivative size = 334, normalized size of antiderivative = 3.37 \[ \int \frac {\tan ^9(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+113 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-286 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+128 \cos \left (d x +c \right )-180 \sin \left (d x +c \right )^{4}+320 \sin \left (d x +c \right )^{2}-128}{60 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(tan(d*x+c)^9/(a+a*sec(d*x+c))^3,x)
Output:
(60*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 - 120*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 + 60*cos(c + d*x)*log( tan((c + d*x)/2)**2 + 1) - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1) - 60*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**4 + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s in(c + d*x)**2 - 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1) + 113*cos(c + d *x)*sin(c + d*x)**4 - 286*cos(c + d*x)*sin(c + d*x)**2 + 128*cos(c + d*x) - 180*sin(c + d*x)**4 + 320*sin(c + d*x)**2 - 128)/(60*cos(c + d*x)*a**3*d *(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))