Integrand size = 23, antiderivative size = 99 \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)}+\frac {2 \tan ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3+n p)}+\frac {\tan ^5(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (5+n p)} \] Output:
tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+1)+2*tan(f*x+e)^3*(b*(c*tan(f*x+e ))^n)^p/f/(n*p+3)+tan(f*x+e)^5*(b*(c*tan(f*x+e))^n)^p/f/(n*p+5)
Time = 1.44 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.23 \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\cot (e+f x) \left (b (c \tan (e+f x))^n\right )^p \left (\left (8+6 n p+n^2 p^2+2 (3+n p) \cos (2 (e+f x))+\cos (4 (e+f x))\right ) \sec ^4(e+f x) \tan ^2(e+f x)+8 \left (-\tan ^2(e+f x)\right )^{\frac {1}{2} (1-n p)}\right )}{f (1+n p) (3+n p) (5+n p)} \] Input:
Integrate[Sec[e + f*x]^6*(b*(c*Tan[e + f*x])^n)^p,x]
Output:
(Cot[e + f*x]*(b*(c*Tan[e + f*x])^n)^p*((8 + 6*n*p + n^2*p^2 + 2*(3 + n*p) *Cos[2*(e + f*x)] + Cos[4*(e + f*x)])*Sec[e + f*x]^4*Tan[e + f*x]^2 + 8*(- Tan[e + f*x]^2)^((1 - n*p)/2)))/(f*(1 + n*p)*(3 + n*p)*(5 + n*p))
Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4142, 3042, 3087, 244, 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.
\(\displaystyle \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (e+f x)^6 \left (b (c \tan (e+f x))^n\right )^pdx\) |
\(\Big \downarrow \) 4142 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sec ^6(e+f x) (c \tan (e+f x))^{n p}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sec (e+f x)^6 (c \tan (e+f x))^{n p}dx\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {(c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int (c \tan (e+f x))^{n p} \left (\tan ^2(e+f x)+1\right )^2d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {(c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \left ((c \tan (e+f x))^{n p}+\frac {2 (c \tan (e+f x))^{n p+2}}{c^2}+\frac {(c \tan (e+f x))^{n p+4}}{c^4}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(c \tan (e+f x))^{-n p} \left (\frac {(c \tan (e+f x))^{n p+5}}{c^5 (n p+5)}+\frac {2 (c \tan (e+f x))^{n p+3}}{c^3 (n p+3)}+\frac {(c \tan (e+f x))^{n p+1}}{c (n p+1)}\right ) \left (b (c \tan (e+f x))^n\right )^p}{f}\) |
Input:
Int[Sec[e + f*x]^6*(b*(c*Tan[e + f*x])^n)^p,x]
Output:
((b*(c*Tan[e + f*x])^n)^p*((c*Tan[e + f*x])^(1 + n*p)/(c*(1 + n*p)) + (2*( c*Tan[e + f*x])^(3 + n*p))/(c^3*(3 + n*p)) + (c*Tan[e + f*x])^(5 + n*p)/(c ^5*(5 + n*p))))/(f*(c*Tan[e + f*x])^(n*p))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr acPart[p])) Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || Ma tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.10 (sec) , antiderivative size = 60672, normalized size of antiderivative = 612.85
\[\text {output too large to display}\]
Input:
int(sec(f*x+e)^6*(b*(c*tan(f*x+e))^n)^p,x)
Output:
result too large to display
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {{\left (n^{2} p^{2} + 8 \, \cos \left (f x + e\right )^{4} + 4 \, {\left (n p + 1\right )} \cos \left (f x + e\right )^{2} + 4 \, n p + 3\right )} e^{\left (n p \log \left (\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + p \log \left (b\right )\right )} \sin \left (f x + e\right )}{{\left (f n^{3} p^{3} + 9 \, f n^{2} p^{2} + 23 \, f n p + 15 \, f\right )} \cos \left (f x + e\right )^{5}} \] Input:
integrate(sec(f*x+e)^6*(b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
Output:
(n^2*p^2 + 8*cos(f*x + e)^4 + 4*(n*p + 1)*cos(f*x + e)^2 + 4*n*p + 3)*e^(n *p*log(c*sin(f*x + e)/cos(f*x + e)) + p*log(b))*sin(f*x + e)/((f*n^3*p^3 + 9*f*n^2*p^2 + 23*f*n*p + 15*f)*cos(f*x + e)^5)
\[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \sec ^{6}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**6*(b*(c*tan(f*x+e))**n)**p,x)
Output:
Integral((b*(c*tan(e + f*x))**n)**p*sec(e + f*x)**6, x)
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.07 \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\frac {b^{p} c^{n p} {\left (\tan \left (f x + e\right )^{n}\right )}^{p} \tan \left (f x + e\right )^{5}}{n p + 5} + \frac {2 \, b^{p} c^{n p} {\left (\tan \left (f x + e\right )^{n}\right )}^{p} \tan \left (f x + e\right )^{3}}{n p + 3} + \frac {b^{p} c^{n p} {\left (\tan \left (f x + e\right )^{n}\right )}^{p} \tan \left (f x + e\right )}{n p + 1}}{f} \] Input:
integrate(sec(f*x+e)^6*(b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
Output:
(b^p*c^(n*p)*(tan(f*x + e)^n)^p*tan(f*x + e)^5/(n*p + 5) + 2*b^p*c^(n*p)*( tan(f*x + e)^n)^p*tan(f*x + e)^3/(n*p + 3) + b^p*c^(n*p)*(tan(f*x + e)^n)^ p*tan(f*x + e)/(n*p + 1))/f
Time = 9.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.31 \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\frac {c^{5} e^{\left (n p \log \left (c \tan \left (f x + e\right )\right ) + p \log \left (b\right )\right )} \tan \left (f x + e\right )^{5}}{c^{4} n p + 5 \, c^{4}} + \frac {2 \, c^{3} e^{\left (n p \log \left (c \tan \left (f x + e\right )\right ) + p \log \left (b\right )\right )} \tan \left (f x + e\right )^{3}}{c^{2} n p + 3 \, c^{2}} + \frac {c e^{\left (n p \log \left (c \tan \left (f x + e\right )\right ) + p \log \left (b\right )\right )} \tan \left (f x + e\right )}{n p + 1}}{c f} \] Input:
integrate(sec(f*x+e)^6*(b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
Output:
(c^5*e^(n*p*log(c*tan(f*x + e)) + p*log(b))*tan(f*x + e)^5/(c^4*n*p + 5*c^ 4) + 2*c^3*e^(n*p*log(c*tan(f*x + e)) + p*log(b))*tan(f*x + e)^3/(c^2*n*p + 3*c^2) + c*e^(n*p*log(c*tan(f*x + e)) + p*log(b))*tan(f*x + e)/(n*p + 1) )/(c*f)
Timed out. \[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \] Input:
int((b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^6,x)
Output:
int((b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^6, x)
\[ \int \sec ^6(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=c^{n p} b^{p} \left (\int \tan \left (f x +e \right )^{n p} \sec \left (f x +e \right )^{6}d x \right ) \] Input:
int(sec(f*x+e)^6*(b*(c*tan(f*x+e))^n)^p,x)
Output:
c**(n*p)*b**p*int(tan(e + f*x)**(n*p)*sec(e + f*x)**6,x)