Integrand size = 21, antiderivative size = 97 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\log (\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^2(c+d x)}{a d}-\frac {2 \sec ^3(c+d x)}{3 a d}-\frac {\sec ^4(c+d x)}{4 a d}+\frac {\sec ^5(c+d x)}{5 a d} \] Output:
ln(cos(d*x+c))/a/d+sec(d*x+c)/a/d+sec(d*x+c)^2/a/d-2/3*sec(d*x+c)^3/a/d-1/ 4*sec(d*x+c)^4/a/d+1/5*sec(d*x+c)^5/a/d
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {(58+40 \cos (2 (c+d x))+60 \cos (3 (c+d x))+30 \cos (4 (c+d x))+75 \cos (3 (c+d x)) \log (\cos (c+d x))+15 \cos (5 (c+d x)) \log (\cos (c+d x))+30 \cos (c+d x) (4+5 \log (\cos (c+d x)))) \sec ^5(c+d x)}{240 a d} \] Input:
Integrate[Tan[c + d*x]^7/(a + a*Sec[c + d*x]),x]
Output:
((58 + 40*Cos[2*(c + d*x)] + 60*Cos[3*(c + d*x)] + 30*Cos[4*(c + d*x)] + 7 5*Cos[3*(c + d*x)]*Log[Cos[c + d*x]] + 15*Cos[5*(c + d*x)]*Log[Cos[c + d*x ]] + 30*Cos[c + d*x]*(4 + 5*Log[Cos[c + d*x]]))*Sec[c + d*x]^5)/(240*a*d)
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74, 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, 99, 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.
\(\displaystyle \int \frac {\tan ^7(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^7}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^7}{\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^5 (1-\cos (c+d x))^3 (\cos (c+d x)+1)^2 \sec ^6(c+d x)d\cos (c+d x)}{a^6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int (1-\cos (c+d x))^3 (\cos (c+d x)+1)^2 \sec ^6(c+d x)d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\sec ^6(c+d x)-\sec ^5(c+d x)-2 \sec ^4(c+d x)+2 \sec ^3(c+d x)+\sec ^2(c+d x)-\sec (c+d x)\right )d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \sec ^5(c+d x)+\frac {1}{4} \sec ^4(c+d x)+\frac {2}{3} \sec ^3(c+d x)-\sec ^2(c+d x)-\sec (c+d x)-\log (\cos (c+d x))}{a d}\) |
Input:
Int[Tan[c + d*x]^7/(a + a*Sec[c + d*x]),x]
Output:
-((-Log[Cos[c + d*x]] - Sec[c + d*x] - Sec[c + d*x]^2 + (2*Sec[c + d*x]^3) /3 + Sec[c + d*x]^4/4 - Sec[c + d*x]^5/5)/(a*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 \cos \left (d x +c \right )^{4}}-\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{5 \cos \left (d x +c \right )^{5}}+\frac {1}{\cos \left (d x +c \right )^{2}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\cos \left (d x +c \right )\right )}{d a}\) | \(62\) |
default | \(\frac {-\frac {1}{4 \cos \left (d x +c \right )^{4}}-\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{5 \cos \left (d x +c \right )^{5}}+\frac {1}{\cos \left (d x +c \right )^{2}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\cos \left (d x +c \right )\right )}{d a}\) | \(62\) |
risch | \(-\frac {i x}{a}-\frac {2 i c}{d a}+\frac {2 \,{\mathrm e}^{9 i \left (d x +c \right )}+4 \,{\mathrm e}^{8 i \left (d x +c \right )}+\frac {8 \,{\mathrm e}^{7 i \left (d x +c \right )}}{3}+8 \,{\mathrm e}^{6 i \left (d x +c \right )}+\frac {116 \,{\mathrm e}^{5 i \left (d x +c \right )}}{15}+8 \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {8 \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+4 \,{\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d a}\) | \(159\) |
Input:
int(tan(d*x+c)^7/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/4/cos(d*x+c)^4-2/3/cos(d*x+c)^3+1/5/cos(d*x+c)^5+1/cos(d*x+c)^2+ 1/cos(d*x+c)+ln(cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {60 \, \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) + 60 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} - 15 \, \cos \left (d x + c\right ) + 12}{60 \, a d \cos \left (d x + c\right )^{5}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/60*(60*cos(d*x + c)^5*log(-cos(d*x + c)) + 60*cos(d*x + c)^4 + 60*cos(d* x + c)^3 - 40*cos(d*x + c)^2 - 15*cos(d*x + c) + 12)/(a*d*cos(d*x + c)^5)
\[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)**7/(a+a*sec(d*x+c)),x)
Output:
Integral(tan(c + d*x)**7/(sec(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {60 \, \log \left (\cos \left (d x + c\right )\right )}{a} + \frac {60 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} - 15 \, \cos \left (d x + c\right ) + 12}{a \cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
1/60*(60*log(cos(d*x + c))/a + (60*cos(d*x + c)^4 + 60*cos(d*x + c)^3 - 40 *cos(d*x + c)^2 - 15*cos(d*x + c) + 12)/(a*cos(d*x + c)^5))/d
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {60 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{3} - 40 \, \cos \left (d x + c\right )^{2} - 15 \, \cos \left (d x + c\right ) + 12}{\cos \left (d x + c\right )^{5}} + 60 \, \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{60 \, a d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/60*((60*cos(d*x + c)^4 + 60*cos(d*x + c)^3 - 40*cos(d*x + c)^2 - 15*cos( d*x + c) + 12)/cos(d*x + c)^5 + 60*log(abs(cos(d*x + c))))/(a*d)
Time = 16.79 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {16}{15}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a\,d} \] Input:
int(tan(c + d*x)^7/(a + a/cos(c + d*x)),x)
Output:
((2*tan(c/2 + (d*x)/2)^4)/3 - (10*tan(c/2 + (d*x)/2)^2)/3 + 10*tan(c/2 + ( d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^8 + 16/15)/(d*(a - 5*a*tan(c/2 + (d*x)/2) ^2 + 10*a*tan(c/2 + (d*x)/2)^4 - 10*a*tan(c/2 + (d*x)/2)^6 + 5*a*tan(c/2 + (d*x)/2)^8 - a*tan(c/2 + (d*x)/2)^10)) - (2*atanh(tan(c/2 + (d*x)/2)^2))/ (a*d)
Time = 0.19 (sec) , antiderivative size = 334, normalized size of antiderivative = 3.44 \[ \int \frac {\tan ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}+120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-77 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+94 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-32 \cos \left (d x +c \right )+60 \sin \left (d x +c \right )^{4}-80 \sin \left (d x +c \right )^{2}+32}{60 \cos \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(tan(d*x+c)^7/(a+a*sec(d*x+c)),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)*l og(tan((c + d*x)/2)**2 + 1) + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*si n(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 )*sin(c + d*x)**2 + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1) - 77*cos(c + d*x)*sin(c + d*x)**4 + 94*cos(c + d*x)*sin(c + d*x)**2 - 32*cos(c + d*x) + 60*sin(c + d*x)**4 - 80*sin(c + d*x)**2 + 32)/(60*cos(c + d*x)*a*d*(sin( c + d*x)**4 - 2*sin(c + d*x)**2 + 1))