Integrand size = 21, antiderivative size = 65 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\log (\cos (c+d x))}{a^2 d}+\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}+\frac {\sec ^4(c+d x)}{4 a^2 d} \] Output:
ln(cos(d*x+c))/a^2/d+2*sec(d*x+c)/a^2/d-2/3*sec(d*x+c)^3/a^2/d+1/4*sec(d*x +c)^4/a^2/d
Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {(20 \cos (c+d x)+3 (2+4 \cos (3 (c+d x))+3 \log (\cos (c+d x))+4 \cos (2 (c+d x)) \log (\cos (c+d x))+\cos (4 (c+d x)) \log (\cos (c+d x)))) \sec ^4(c+d x)}{24 a^2 d} \] Input:
Integrate[Tan[c + d*x]^7/(a + a*Sec[c + d*x])^2,x]
Output:
((20*Cos[c + d*x] + 3*(2 + 4*Cos[3*(c + d*x)] + 3*Log[Cos[c + d*x]] + 4*Co s[2*(c + d*x)]*Log[Cos[c + d*x]] + Cos[4*(c + d*x)]*Log[Cos[c + d*x]]))*Se c[c + d*x]^4)/(24*a^2*d)
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77, 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.
\(\displaystyle \int \frac {\tan ^7(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^7}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^7}{\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {\int a^4 (1-\cos (c+d x))^3 (\cos (c+d x)+1) \sec ^5(c+d x)d\cos (c+d x)}{a^6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int (1-\cos (c+d x))^3 (\cos (c+d x)+1) \sec ^5(c+d x)d\cos (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle -\frac {\int \left (\sec ^5(c+d x)-2 \sec ^4(c+d x)+2 \sec ^2(c+d x)-\sec (c+d x)\right )d\cos (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{4} \sec ^4(c+d x)+\frac {2}{3} \sec ^3(c+d x)-2 \sec (c+d x)-\log (\cos (c+d x))}{a^2 d}\) |
Input:
Int[Tan[c + d*x]^7/(a + a*Sec[c + d*x])^2,x]
Output:
-((-Log[Cos[c + d*x]] - 2*Sec[c + d*x] + (2*Sec[c + d*x]^3)/3 - Sec[c + d* x]^4/4)/(a^2*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 = 0.53 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \cos \left (d x +c \right )^{4}}+\ln \left (\cos \left (d x +c \right )\right )-\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}}{d \,a^{2}}\) | \(46\) |
default | \(\frac {\frac {1}{4 \cos \left (d x +c \right )^{4}}+\ln \left (\cos \left (d x +c \right )\right )-\frac {2}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}}{d \,a^{2}}\) | \(46\) |
risch | \(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}+\frac {4 \,{\mathrm e}^{7 i \left (d x +c \right )}+\frac {20 \,{\mathrm e}^{5 i \left (d x +c \right )}}{3}+4 \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {20 \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+4 \,{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(115\) |
Input:
int(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(1/4/cos(d*x+c)^4+ln(cos(d*x+c))-2/3/cos(d*x+c)^3+2/cos(d*x+c))
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {12 \, \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 24 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right ) + 3}{12 \, a^{2} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/12*(12*cos(d*x + c)^4*log(-cos(d*x + c)) + 24*cos(d*x + c)^3 - 8*cos(d*x + c) + 3)/(a^2*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(tan(d*x+c)**7/(a+a*sec(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**7/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {12 \, \log \left (\cos \left (d x + c\right )\right )}{a^{2}} + \frac {24 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right ) + 3}{a^{2} \cos \left (d x + c\right )^{4}}}{12 \, d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
1/12*(12*log(cos(d*x + c))/a^2 + (24*cos(d*x + c)^3 - 8*cos(d*x + c) + 3)/ (a^2*cos(d*x + c)^4))/d
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {24 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right ) + 3}{\cos \left (d x + c\right )^{4}} + 12 \, \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{12 \, a^{2} d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/12*((24*cos(d*x + c)^3 - 8*cos(d*x + c) + 3)/cos(d*x + c)^4 + 12*log(abs (cos(d*x + c))))/(a^2*d)
Time = 14.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.08 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {8}{3}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^2\,d} \] Input:
int(tan(c + d*x)^7/(a + a/cos(c + d*x))^2,x)
Output:
(8*tan(c/2 + (d*x)/2)^4 - (26*tan(c/2 + (d*x)/2)^2)/3 + 2*tan(c/2 + (d*x)/ 2)^6 + 8/3)/(d*(6*a^2*tan(c/2 + (d*x)/2)^4 - 4*a^2*tan(c/2 + (d*x)/2)^2 - 4*a^2*tan(c/2 + (d*x)/2)^6 + a^2*tan(c/2 + (d*x)/2)^8 + a^2)) - (2*atanh(t an(c/2 + (d*x)/2)^2))/(a^2*d)
Time = 0.17 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.94 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+16 \cos \left (d x +c \right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-19 \sin \left (d x +c \right )^{4}+38 \sin \left (d x +c \right )^{2}-16}{12 a^{2} d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(tan(d*x+c)^7/(a+a*sec(d*x+c))^2,x)
Output:
( - 24*cos(c + d*x)*sin(c + d*x)**2 + 16*cos(c + d*x) - 12*log(tan((c + d* x)/2)**2 + 1)*sin(c + d*x)**4 + 24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d* x)**2 - 12*log(tan((c + d*x)/2)**2 + 1) + 12*log(tan((c + d*x)/2) - 1)*sin (c + d*x)**4 - 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 12*log(tan(( c + d*x)/2) - 1) + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 24*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2 + 12*log(tan((c + d*x)/2) + 1) - 19*s in(c + d*x)**4 + 38*sin(c + d*x)**2 - 16)/(12*a**2*d*(sin(c + d*x)**4 - 2* sin(c + d*x)**2 + 1))