3.238 \(\int \frac{\tan ^3(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)
Optimal. Leaf size=97 \[ -\frac{a}{4 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]
[Out]
Log[a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2]/(2*(a - b)^3*f) - a/(4*(a - b)*b*f*(a + b*Tan[e + f*x]^2)^2) - 1/(2*(
a - b)^2*f*(a + b*Tan[e + f*x]^2))
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Rubi [A] time = 0.11735, antiderivative size = 97, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{3670, 446, 77} \[ -\frac{a}{4 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^3/(a + b*Tan[e + f*x]^2)^3,x]
[Out]
Log[a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2]/(2*(a - b)^3*f) - a/(4*(a - b)*b*f*(a + b*Tan[e + f*x]^2)^2) - 1/(2*(
a - b)^2*f*(a + b*Tan[e + f*x]^2))
Rule 3670
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]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^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] && RationalQ[n]))
Rule 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 77
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))
Rubi steps
\begin{align*} \int \frac{\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1+x) (a+b x)^3} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a-b)^3 (1+x)}+\frac{a}{(a-b) (a+b x)^3}+\frac{b}{(a-b)^2 (a+b x)^2}+\frac{b}{(a-b)^3 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 (a-b)^3 f}-\frac{a}{4 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{1}{2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.749997, size = 87, normalized size = 0.9 \[ \frac{-\frac{a (a-b)^2}{b \left (a+b \tan ^2(e+f x)\right )^2}-\frac{2 (a-b)}{a+b \tan ^2(e+f x)}+2 \log \left (a+b \tan ^2(e+f x)\right )+4 \log (\cos (e+f x))}{4 f (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^3/(a + b*Tan[e + f*x]^2)^3,x]
[Out]
(4*Log[Cos[e + f*x]] + 2*Log[a + b*Tan[e + f*x]^2] - (a*(a - b)^2)/(b*(a + b*Tan[e + f*x]^2)^2) - (2*(a - b))/
(a + b*Tan[e + f*x]^2))/(4*(a - b)^3*f)
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Maple [B] time = 0.023, size = 193, normalized size = 2. \begin{align*}{\frac{\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{3}}}-{\frac{{a}^{3}}{4\,f \left ( a-b \right ) ^{3}b \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{a}^{2}}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{ab}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{a}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{b}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x)
[Out]
1/2/f/(a-b)^3*ln(a+b*tan(f*x+e)^2)-1/4/f/(a-b)^3*a^3/b/(a+b*tan(f*x+e)^2)^2+1/2/f/(a-b)^3*a^2/(a+b*tan(f*x+e)^
2)^2-1/4/f/(a-b)^3*a*b/(a+b*tan(f*x+e)^2)^2-1/2/f/(a-b)^3*a/(a+b*tan(f*x+e)^2)+1/2/f/(a-b)^3/(a+b*tan(f*x+e)^2
)*b-1/2/f/(a-b)^3*ln(1+tan(f*x+e)^2)
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Maxima [B] time = 1.17617, size = 262, normalized size = 2.7 \begin{align*} -\frac{\frac{2 \,{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )^{2} - 2 \, a^{2} - a b}{a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3} +{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sin \left (f x + e\right )^{4} - 2 \,{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (f x + e\right )^{2}} - \frac{2 \, \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")
[Out]
-1/4*((2*(a^2 - b^2)*sin(f*x + e)^2 - 2*a^2 - a*b)/(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^5 - 5*a^4*b + 10*
a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sin(f*x + e)^4 - 2*(a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*sin(f
*x + e)^2) - 2*log(-(a - b)*sin(f*x + e)^2 + a)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/f
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Fricas [B] time = 1.57221, size = 473, normalized size = 4.88 \begin{align*} \frac{{\left (a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{2} + a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2} + a b + 2 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")
[Out]
1/4*((a*b + 2*b^2)*tan(f*x + e)^4 + 2*(a^2 + a*b + b^2)*tan(f*x + e)^2 + 2*a^2 + a*b + 2*(b^2*tan(f*x + e)^4 +
2*a*b*tan(f*x + e)^2 + a^2)*log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)))/((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4
- b^5)*f*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3
*b^2 - a^2*b^3)*f)
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Sympy [A] time = 170.769, size = 2849, normalized size = 29.37 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)**3/(a+b*tan(f*x+e)**2)**3,x)
[Out]
Piecewise((zoo*x/tan(e)**3, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (-3*tan(e + f*x)**2/(12*b**3*f*tan(e + f*x)**6 +
36*b**3*f*tan(e + f*x)**4 + 36*b**3*f*tan(e + f*x)**2 + 12*b**3*f) - 1/(12*b**3*f*tan(e + f*x)**6 + 36*b**3*f*
tan(e + f*x)**4 + 36*b**3*f*tan(e + f*x)**2 + 12*b**3*f), Eq(a, b)), ((-log(tan(e + f*x)**2 + 1)/(2*f) + tan(e
+ f*x)**2/(2*f))/a**3, Eq(b, 0)), (x*tan(e)**3/(a + b*tan(e)**2)**3, Eq(f, 0)), (-a**3/(4*a**5*b*f + 8*a**4*b
**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a
**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan
(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 2*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + ta
n(e + f*x))/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*
a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2
- 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 2*a*
*2*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*
a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4
+ 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 -
4*b**6*f*tan(e + f*x)**4) - 2*a**2*b*log(tan(e + f*x)**2 + 1)/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 1
2*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4
*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f
*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 4*a*b**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)*
*2/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3
*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2
*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 4*a*b**2*log(
I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2
*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e +
f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*
x)**2 - 4*b**6*f*tan(e + f*x)**4) - 4*a*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(4*a**5*b*f + 8*a**4*b**
2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**
3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e
+ f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) - 2*a*b**2*tan(e + f*x)**2/(4*a**5*b*f + 8
*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2
+ 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**
5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + a*b**2/(4*a**5*b*f + 8*a**4*b**
2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**
3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e
+ f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 2*b**3*log(-I*sqrt(a)*sqrt(1/b) + tan(e
+ f*x))*tan(e + f*x)**4/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f
*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan
(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x
)**4) + 2*b**3*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**4/(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x
)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a
**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*
a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) - 2*b**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**4/(4*a**5
*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*tan(e +
f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b**4*f +
12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4) + 2*b**3*tan(e + f*x)**2/
(4*a**5*b*f + 8*a**4*b**2*f*tan(e + f*x)**2 - 12*a**4*b**2*f + 4*a**3*b**3*f*tan(e + f*x)**4 - 24*a**3*b**3*f*
tan(e + f*x)**2 + 12*a**3*b**3*f - 12*a**2*b**4*f*tan(e + f*x)**4 + 24*a**2*b**4*f*tan(e + f*x)**2 - 4*a**2*b*
*4*f + 12*a*b**5*f*tan(e + f*x)**4 - 8*a*b**5*f*tan(e + f*x)**2 - 4*b**6*f*tan(e + f*x)**4), True))
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Giac [B] time = 2.00028, size = 682, normalized size = 7.03 \begin{align*} \frac{\frac{2 \, \log \left (a + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{4 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{4 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{3 \, a^{3} + \frac{20 \, a^{3}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{32 \, a^{2} b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{34 \, a^{3}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{80 \, a^{2} b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{48 \, a b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{16 \, b^{3}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{20 \, a^{3}{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{32 \, a^{2} b{\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, a^{3}{\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )}{\left (a + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{4 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{2}}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")
[Out]
1/4*(2*log(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(
f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 4*log(-(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a^3 + 20*a^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 32*a^2*b
*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 34*a^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 80*a^2*b*(cos(f*x
+ e) - 1)^2/(cos(f*x + e) + 1)^2 + 48*a*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 16*b^3*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2 + 20*a^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 32*a^2*b*(cos(f*x + e) - 1)^3/
(cos(f*x + e) + 1)^3 + 3*a^3*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(
a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2)^2))/f