Integrand size = 23, antiderivative size = 78 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+b)^2}{4 a^3 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {a+b}{a^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 f} \] Output:
1/4*(a+b)^2/a^3/f/(b+a*cos(f*x+e)^2)^2-(a+b)/a^3/f/(b+a*cos(f*x+e)^2)-1/2* ln(b+a*cos(f*x+e)^2)/a^3/f
Time = 1.94 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.74 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {2 \left (a^2+4 a b+3 b^2\right )+(a+2 b)^2 \log (a+2 b+a \cos (2 (e+f x)))+a^2 \cos ^2(2 (e+f x)) \log (a+2 b+a \cos (2 (e+f x)))+2 a \cos (2 (e+f x)) (2 (a+b)+(a+2 b) \log (a+2 b+a \cos (2 (e+f x))))}{2 a^3 f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/2*(2*(a^2 + 4*a*b + 3*b^2) + (a + 2*b)^2*Log[a + 2*b + a*Cos[2*(e + f*x )]] + a^2*Cos[2*(e + f*x)]^2*Log[a + 2*b + a*Cos[2*(e + f*x)]] + 2*a*Cos[2 *(e + f*x)]*(2*(a + b) + (a + 2*b)*Log[a + 2*b + a*Cos[2*(e + f*x)]]))/(a^ 3*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 353, 49, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cos (e+f x) \left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (\frac {(a+b)^2}{a^2 \left (a \cos ^2(e+f x)+b\right )^3}-\frac {2 (a+b)}{a^2 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {1}{a^2 \left (a \cos ^2(e+f x)+b\right )}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {(a+b)^2}{2 a^3 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {2 (a+b)}{a^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {\log \left (a \cos ^2(e+f x)+b\right )}{a^3}}{2 f}\) |
Input:
Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/2*(-1/2*(a + b)^2/(a^3*(b + a*Cos[e + f*x]^2)^2) + (2*(a + b))/(a^3*(b + a*Cos[e + f*x]^2)) + Log[b + a*Cos[e + f*x]^2]/a^3)/f
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 11.79 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 a^{3}}+\frac {a^{2}+2 a b +b^{2}}{4 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {-2 a -2 b}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}}{f}\) | \(80\) |
default | \(\frac {-\frac {\ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 a^{3}}+\frac {a^{2}+2 a b +b^{2}}{4 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {-2 a -2 b}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}}{f}\) | \(80\) |
risch | \(\frac {i x}{a^{3}}+\frac {2 i e}{a^{3} f}-\frac {4 \left (a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+a b \,{\mathrm e}^{6 i \left (f x +e \right )}+a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+3 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+a b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{3} f}\) | \(196\) |
Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/2/a^3*ln(b+a*cos(f*x+e)^2)+1/4*(a^2+2*a*b+b^2)/a^3/(b+a*cos(f*x+e) ^2)^2+1/2*(-2*a-2*b)/a^3/(b+a*cos(f*x+e)^2))
Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - a^{2} + 2 \, a b + 3 \, b^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right )}{4 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
-1/4*(4*(a^2 + a*b)*cos(f*x + e)^2 - a^2 + 2*a*b + 3*b^2 + 2*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*log(a*cos(f*x + e)^2 + b))/(a^5*f*cos (f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)
Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {4 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - 3 \, a^{2} - 6 \, a b - 3 \, b^{2}}{a^{5} \sin \left (f x + e\right )^{4} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} - 2 \, {\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} - \frac {2 \, \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3}}}{4 \, f} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
1/4*((4*(a^2 + a*b)*sin(f*x + e)^2 - 3*a^2 - 6*a*b - 3*b^2)/(a^5*sin(f*x + e)^4 + a^5 + 2*a^4*b + a^3*b^2 - 2*(a^5 + a^4*b)*sin(f*x + e)^2) - 2*log( a*sin(f*x + e)^2 - a - b)/a^3)/f
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, a^{3} f} - \frac {4 \, {\left (a + b\right )} \cos \left (f x + e\right )^{2} - \frac {a^{2} - 2 \, a b - 3 \, b^{2}}{a}}{4 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{2} f} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
Output:
-1/2*log(abs(a*cos(f*x + e)^2 + b))/(a^3*f) - 1/4*(4*(a + b)*cos(f*x + e)^ 2 - (a^2 - 2*a*b - 3*b^2)/a)/((a*cos(f*x + e)^2 + b)^2*a^2*f)
Time = 16.50 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\mathrm {atanh}\left (\frac {4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,b^2+\frac {8\,b^3}{a}+4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a}}\right )}{a^3\,f}+\frac {\frac {-a^3+3\,a\,b^2+2\,b^3}{4\,a^2\,b^2}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a^2-b^2\right )}{2\,a^2\,b}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \] Input:
int(tan(e + f*x)^5/(a + b/cos(e + f*x)^2)^3,x)
Output:
atanh((4*b^2*tan(e + f*x)^2)/(8*b^2 + (8*b^3)/a + 4*b^2*tan(e + f*x)^2 + ( 8*b^3*tan(e + f*x)^2)/a))/(a^3*f) + ((3*a*b^2 - a^3 + 2*b^3)/(4*a^2*b^2) - (tan(e + f*x)^2*(a^2 - b^2))/(2*a^2*b))/(f*(2*a*b + a^2 + b^2 + tan(e + f *x)^2*(2*a*b + 2*b^2) + b^2*tan(e + f*x)^4))
Time = 0.16 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.77 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) \sec \left (f x +e \right )^{4} b^{2}+4 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) \sec \left (f x +e \right )^{2} a b +2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) a^{2}-2 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) \sec \left (f x +e \right )^{4} b^{2}-4 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) \sec \left (f x +e \right )^{2} a b -2 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) a^{2}-2 \sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2} a b +\tan \left (f x +e \right )^{4} a^{2}-2 \tan \left (f x +e \right )^{2} a^{2}}{4 a^{3} f \left (\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}\right )} \] Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x)
Output:
(2*log(tan(e + f*x)**2 + 1)*sec(e + f*x)**4*b**2 + 4*log(tan(e + f*x)**2 + 1)*sec(e + f*x)**2*a*b + 2*log(tan(e + f*x)**2 + 1)*a**2 - 2*log(sec(e + f*x)**2*b + a)*sec(e + f*x)**4*b**2 - 4*log(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*a*b - 2*log(sec(e + f*x)**2*b + a)*a**2 - 2*sec(e + f*x)**2*tan(e + f*x)**2*a*b + tan(e + f*x)**4*a**2 - 2*tan(e + f*x)**2*a**2)/(4*a**3*f* (sec(e + f*x)**4*b**2 + 2*sec(e + f*x)**2*a*b + a**2))