Integrand size = 23, antiderivative size = 81 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b (a+b)}{4 a^3 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {a+2 b}{2 a^3 f \left (b+a \cos ^2(e+f x)\right )}+\frac {\log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 f} \] Output:
-1/4*b*(a+b)/a^3/f/(b+a*cos(f*x+e)^2)^2+1/2*(a+2*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 = 0.71 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.62 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {2 \left (a^2+3 a b+3 b^2\right )+(a+2 b)^2 \log (a+2 b+a \cos (2 (e+f x)))+a^2 \cos ^2(2 (e+f x)) \log (a+2 b+a \cos (2 (e+f x)))+2 a (a+2 b) \cos (2 (e+f x)) (1+\log (a+2 b+a \cos (2 (e+f x))))}{2 a^3 f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
(2*(a^2 + 3*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*(a + 2*b)* Cos[2*(e + f*x)]*(1 + Log[a + 2*b + a*Cos[2*(e + f*x)]]))/(2*a^3*f*(a + 2* b + a*Cos[2*(e + f*x)])^2)
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 86, 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.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cos ^3(e+f x) \left (1-\cos ^2(e+f x)\right )}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\cos ^2(e+f x) \left (1-\cos ^2(e+f x)\right )}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {\int \left (-\frac {b (a+b)}{a^2 \left (a \cos ^2(e+f x)+b\right )^3}-\frac {1}{a^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {a+2 b}{a^2 \left (a \cos ^2(e+f x)+b\right )^2}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {b (a+b)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )^2}-\frac {a+2 b}{a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\log \left (a \cos ^2(e+f x)+b\right )}{a^3}}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/2*((b*(a + b))/(2*a^3*(b + a*Cos[e + f*x]^2)^2) - (a + 2*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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
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 = 6.82 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {\left (a +b \right ) b}{4 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}-\frac {-a -2 b}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 a^{3}}}{f}\) | \(73\) |
default | \(\frac {-\frac {\left (a +b \right ) b}{4 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}-\frac {-a -2 b}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 a^{3}}}{f}\) | \(73\) |
risch | \(-\frac {i x}{a^{3}}-\frac {2 i e}{a^{3} f}+\frac {2 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+4 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+12 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+12 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{2 i \left (f x +e \right )}}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{3} f}\) | \(199\) |
Input:
int(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/4*(a+b)*b/a^3/(b+a*cos(f*x+e)^2)^2-1/2*(-a-2*b)/a^3/(b+a*cos(f*x+e )^2)+1/2/a^3*ln(b+a*cos(f*x+e)^2))
Time = 0.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {2 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 3 \, b^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right )}{4 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
1/4*(2*(a^2 + 2*a*b)*cos(f*x + e)^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. \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(tan(f*x+e)**3/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.40 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {2 \, {\left (a^{2} + 2 \, a b\right )} \sin \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b - 3 \, b^{2}}{a^{5} \sin \left (f x + e\right )^{4} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} - 2 \, {\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} - \frac {2 \, \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3}}}{4 \, f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
-1/4*((2*(a^2 + 2*a*b)*sin(f*x + e)^2 - 2*a^2 - 5*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*l og(a*sin(f*x + e)^2 - a - b)/a^3)/f
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, a^{3} f} + \frac {2 \, {\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} + \frac {a b + 3 \, b^{2}}{a}}{4 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{2} f} \] Input:
integrate(tan(f*x+e)^3/(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*(2*(a + 2*b)*cos(f*x + e) ^2 + (a*b + 3*b^2)/a)/((a*cos(f*x + e)^2 + b)^2*a^2*f)
Time = 16.76 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {a^2+3\,a\,b+2\,b^2}{4\,a^2\,b}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\mathrm {atanh}\left (\frac {4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,b^2+\frac {8\,b^3}{a}+4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a}}\right )}{a^3\,f} \] Input:
int(tan(e + f*x)^3/(a + b/cos(e + f*x)^2)^3,x)
Output:
- ((3*a*b + a^2 + 2*b^2)/(4*a^2*b) + (b*tan(e + f*x)^2)/(2*a^2))/(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)) - ata nh((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)
Time = 0.15 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.79 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) \sec \left (f x +e \right )^{4} b^{2}-4 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) \sec \left (f x +e \right )^{2} a b -2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) a^{2}+2 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) \sec \left (f x +e \right )^{4} b^{2}+4 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) \sec \left (f x +e \right )^{2} a b +2 \,\mathrm {log}\left (\sec \left (f x +e \right )^{2} b +a \right ) a^{2}+\sec \left (f x +e \right )^{4} b^{2}+\sec \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2} a b +\sec \left (f x +e \right )^{2} a b +2 \tan \left (f x +e \right )^{2} a^{2}}{4 a^{3} f \left (\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}\right )} \] Input:
int(tan(f*x+e)^3/(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 + sec(e + f*x)**4*b**2 + sec(e + f*x)**2*tan(e + f*x)**2*a*b + sec(e + f*x)**2*a*b + 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))