Integrand size = 23, antiderivative size = 45 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\log (\cos (e+f x))}{b f}+\frac {(a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a b f} \] Output:
-ln(cos(f*x+e))/b/f+1/2*(a+b)*ln(b+a*cos(f*x+e)^2)/a/b/f
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-2 a \log (\cos (e+f x))+(a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a b f} \] Input:
Integrate[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
(-2*a*Log[Cos[e + f*x]] + (a + b)*Log[b + a*Cos[e + f*x]^2])/(2*a*b*f)
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right ) \sec (e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right ) \sec (e+f x)}{a \cos ^2(e+f x)+b}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {\int \left (\frac {-a-b}{b \left (a \cos ^2(e+f x)+b\right )}+\frac {\sec (e+f x)}{b}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\log \left (\cos ^2(e+f x)\right )}{b}-\frac {(a+b) \log \left (a \cos ^2(e+f x)+b\right )}{a b}}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*(Log[Cos[e + f*x]^2]/b - ((a + b)*Log[b + a*Cos[e + f*x]^2])/(a*b))/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 = 0.72 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\cos \left (f x +e \right )\right )}{b}+\frac {\left (a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b a}}{f}\) | \(42\) |
default | \(\frac {-\frac {\ln \left (\cos \left (f x +e \right )\right )}{b}+\frac {\left (a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b a}}{f}\) | \(42\) |
risch | \(-\frac {i x}{a}-\frac {2 i e}{a f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}+\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 b f}+\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f}\) | \(117\) |
Input:
int(tan(f*x+e)^3/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/b*ln(cos(f*x+e))+1/2*(a+b)/b/a*ln(b+a*cos(f*x+e)^2))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\left (a + b\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, a \log \left (-\cos \left (f x + e\right )\right )}{2 \, a b f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/2*((a + b)*log(a*cos(f*x + e)^2 + b) - 2*a*log(-cos(f*x + e)))/(a*b*f)
\[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(tan(f*x+e)**3/(a+b*sec(f*x+e)**2),x)
Output:
Integral(tan(e + f*x)**3/(a + b*sec(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (a + b\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b} - \frac {\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b}}{2 \, f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/2*((a + b)*log(a*sin(f*x + e)^2 - a - b)/(a*b) - log(sin(f*x + e)^2 - 1) /b)/f
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\left (a + b\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, a b f} - \frac {\log \left ({\left | \cos \left (f x + e\right ) \right |}\right )}{b f} \] Input:
integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/2*(a + b)*log(abs(a*cos(f*x + e)^2 + b))/(a*b*f) - log(abs(cos(f*x + e)) )/(b*f)
Time = 16.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.42 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f}+\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,b\,f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f} \] Input:
int(tan(e + f*x)^3/(a + b/cos(e + f*x)^2),x)
Output:
log(a + b + b*tan(e + f*x)^2)/(2*a*f) + log(a + b + b*tan(e + f*x)^2)/(2*b *f) - log(tan(e + f*x)^2 + 1)/(2*a*f)
Time = 0.16 (sec) , antiderivative size = 207, normalized size of antiderivative = 4.60 \[ \int \frac {\tan ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) b -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b}{2 a b f} \] Input:
int(tan(f*x+e)^3/(a+b*sec(f*x+e)^2),x)
Output:
( - 2*log(tan((e + f*x)/2)**2 + 1)*b - 2*log(tan((e + f*x)/2) - 1)*a - 2*l og(tan((e + f*x)/2) + 1)*a + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*b + log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*a + log(sqrt(a + b) *tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*b)/(2*a*b *f)