Integrand size = 23, antiderivative size = 50 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\log (\cos (e+f x))}{(a-b) f}+\frac {a \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b) b f} \] Output:
ln(cos(f*x+e))/(a-b)/f+1/2*a*ln(a+b*tan(f*x+e)^2)/(a-b)/b/f
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {2 b \log (\cos (e+f x))+a \log \left (a+b \tan ^2(e+f x)\right )}{2 a b f-2 b^2 f} \] Input:
Integrate[Tan[e + f*x]^3/(a + b*Tan[e + f*x]^2),x]
Output:
(2*b*Log[Cos[e + f*x]] + a*Log[a + b*Tan[e + f*x]^2])/(2*a*b*f - 2*b^2*f)
Time = 0.47 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4153, 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.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{a+b \tan (e+f x)^2}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {a}{(a-b) \left (b \tan ^2(e+f x)+a\right )}-\frac {1}{(a-b) \left (\tan ^2(e+f x)+1\right )}\right )d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a \log \left (a+b \tan ^2(e+f x)\right )}{b (a-b)}-\frac {\log \left (\tan ^2(e+f x)+1\right )}{a-b}}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/(a + b*Tan[e + f*x]^2),x]
Output:
(-(Log[1 + Tan[e + f*x]^2]/(a - b)) + (a*Log[a + b*Tan[e + f*x]^2])/((a - b)*b))/(2*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[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.67 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right ) b -a \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 b f \left (a -b \right )}\) | \(46\) |
derivativedivides | \(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )}+\frac {a \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right ) b}}{f}\) | \(52\) |
default | \(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )}+\frac {a \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right ) b}}{f}\) | \(52\) |
norman | \(-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a -b \right )}+\frac {a \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right ) b f}\) | \(54\) |
risch | \(\frac {i x}{a -b}+\frac {2 i x}{b}+\frac {2 i e}{b f}-\frac {2 i a x}{b \left (a -b \right )}-\frac {2 i a e}{b f \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}+\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 b f \left (a -b \right )}\) | \(132\) |
Input:
int(tan(f*x+e)^3/(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
-1/2*(ln(1+tan(f*x+e)^2)*b-a*ln(a+b*tan(f*x+e)^2))/b/f/(a-b)
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {a \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a - b\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b - b^{2}\right )} f} \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="fricas")
Output:
1/2*(a*log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)) - (a - b)*log(1/(t an(f*x + e)^2 + 1)))/((a*b - b^2)*f)
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (36) = 72\).
Time = 1.54 (sec) , antiderivative size = 230, normalized size of antiderivative = 4.60 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\begin {cases} \tilde {\infty } x \tan {\left (e \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {- \frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\tan ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {1}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{3}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text {for}\: f = 0 \\\frac {a \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} + \frac {a \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} - \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b f - 2 b^{2} f} & \text {otherwise} \end {cases} \] Input:
integrate(tan(f*x+e)**3/(a+b*tan(f*x+e)**2),x)
Output:
Piecewise((zoo*x*tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((-log(tan(e + f *x)**2 + 1)/(2*f) + tan(e + f*x)**2/(2*f))/a, Eq(b, 0)), (log(tan(e + f*x) **2 + 1)*tan(e + f*x)**2/(2*b*f*tan(e + f*x)**2 + 2*b*f) + log(tan(e + f*x )**2 + 1)/(2*b*f*tan(e + f*x)**2 + 2*b*f) + 1/(2*b*f*tan(e + f*x)**2 + 2*b *f), Eq(a, b)), (x*tan(e)**3/(a + b*tan(e)**2), Eq(f, 0)), (a*log(-sqrt(-a /b) + tan(e + f*x))/(2*a*b*f - 2*b**2*f) + a*log(sqrt(-a/b) + tan(e + f*x) )/(2*a*b*f - 2*b**2*f) - b*log(tan(e + f*x)**2 + 1)/(2*a*b*f - 2*b**2*f), True))
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\frac {a \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b - b^{2}} - \frac {\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b}}{2 \, f} \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="maxima")
Output:
1/2*(a*log(-(a - b)*sin(f*x + e)^2 + a)/(a*b - b^2) - log(sin(f*x + e)^2 - 1)/b)/f
Time = 0.88 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {a \log \left ({\left | b \tan \left (f x + e\right )^{2} + a \right |}\right )}{2 \, {\left (a b f - b^{2} f\right )}} - \frac {\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, {\left (a f - b f\right )}} \] Input:
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2),x, algorithm="giac")
Output:
1/2*a*log(abs(b*tan(f*x + e)^2 + a))/(a*b*f - b^2*f) - 1/2*log(tan(f*x + e )^2 + 1)/(a*f - b*f)
Time = 8.00 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{2\,f\,\left (a\,b-b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,\left (a-b\right )} \] Input:
int(tan(e + f*x)^3/(a + b*tan(e + f*x)^2),x)
Output:
(a*log(a + b*tan(e + f*x)^2))/(2*f*(a*b - b^2)) - log(tan(e + f*x)^2 + 1)/ (2*f*(a - b))
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) b +\mathrm {log}\left (\tan \left (f x +e \right )^{2} b +a \right ) a}{2 b f \left (a -b \right )} \] Input:
int(tan(f*x+e)^3/(a+b*tan(f*x+e)^2),x)
Output:
( - log(tan(e + f*x)**2 + 1)*b + log(tan(e + f*x)**2*b + a)*a)/(2*b*f*(a - b))