3.230 \(\int \frac{\tan ^6(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)
Optimal. Leaf size=130 \[ -\frac{a^{3/2} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 b^{5/2} f (a-b)^2}+\frac{(3 a-2 b) \tan (e+f x)}{2 b^2 f (a-b)}-\frac{a \tan ^3(e+f x)}{2 b f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
[Out]
-(x/(a - b)^2) - (a^(3/2)*(3*a - 5*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*(a - b)^2*b^(5/2)*f) + ((3*a
- 2*b)*Tan[e + f*x])/(2*(a - b)*b^2*f) - (a*Tan[e + f*x]^3)/(2*(a - b)*b*f*(a + b*Tan[e + f*x]^2))
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Rubi [A] time = 0.196597, antiderivative size = 130, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used =
{3670, 470, 582, 522, 203, 205} \[ -\frac{a^{3/2} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 b^{5/2} f (a-b)^2}+\frac{(3 a-2 b) \tan (e+f x)}{2 b^2 f (a-b)}-\frac{a \tan ^3(e+f x)}{2 b f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
-(x/(a - b)^2) - (a^(3/2)*(3*a - 5*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*(a - b)^2*b^(5/2)*f) + ((3*a
- 2*b)*Tan[e + f*x])/(2*(a - b)*b^2*f) - (a*Tan[e + f*x]^3)/(2*(a - b)*b*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 470
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 582
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan ^3(e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(3 a-2 b) x^2\right )}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 (a-b) b f}\\ &=\frac{(3 a-2 b) \tan (e+f x)}{2 (a-b) b^2 f}-\frac{a \tan ^3(e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{a (3 a-2 b)+\left (3 a^2-2 a b-2 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 (a-b) b^2 f}\\ &=\frac{(3 a-2 b) \tan (e+f x)}{2 (a-b) b^2 f}-\frac{a \tan ^3(e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}-\frac{\left (a^2 (3 a-5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^2 b^2 f}\\ &=-\frac{x}{(a-b)^2}-\frac{a^{3/2} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 (a-b)^2 b^{5/2} f}+\frac{(3 a-2 b) \tan (e+f x)}{2 (a-b) b^2 f}-\frac{a \tan ^3(e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.22763, size = 118, normalized size = 0.91 \[ \frac{-\frac{a^{3/2} (3 a-5 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{b^{5/2} (a-b)^2}+\frac{a^2 \sin (2 (e+f x))}{b^2 (a-b) ((a-b) \cos (2 (e+f x))+a+b)}-\frac{2 (e+f x)}{(a-b)^2}+\frac{2 \tan (e+f x)}{b^2}}{2 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
((-2*(e + f*x))/(a - b)^2 - (a^(3/2)*(3*a - 5*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)^2*b^(5/2)) +
(a^2*Sin[2*(e + f*x)])/((a - b)*b^2*(a + b + (a - b)*Cos[2*(e + f*x)])) + (2*Tan[e + f*x])/b^2)/(2*f)
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Maple [A] time = 0.023, size = 184, normalized size = 1.4 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{f{b}^{2}}}+{\frac{{a}^{3}\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2}{b}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2}b \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,{a}^{3}}{2\,f \left ( a-b \right ) ^{2}{b}^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{a}^{2}}{2\,f \left ( a-b \right ) ^{2}b}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f \left ( a-b \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x)
[Out]
1/f/b^2*tan(f*x+e)+1/2/f*a^3/(a-b)^2/b^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)-1/2/f*a^2/(a-b)^2/b*tan(f*x+e)/(a+b*tan
(f*x+e)^2)-3/2/f*a^3/(a-b)^2/b^2/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+5/2/f*a^2/(a-b)^2/b/(a*b)^(1/2)*
arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/f/(a-b)^2*arctan(tan(f*x+e))
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.28956, size = 1065, normalized size = 8.19 \begin{align*} \left [-\frac{8 \, b^{3} f x \tan \left (f x + e\right )^{2} + 8 \, a b^{2} f x - 8 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3} +{\left (3 \, a^{3} - 5 \, a^{2} b +{\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} + 4 \,{\left (b^{2} \tan \left (f x + e\right )^{3} - a b \tan \left (f x + e\right )\right )} \sqrt{-\frac{a}{b}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) - 4 \,{\left (3 \, a^{3} - 5 \, a^{2} b + 2 \, a b^{2}\right )} \tan \left (f x + e\right )}{8 \,{\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}, -\frac{4 \, b^{3} f x \tan \left (f x + e\right )^{2} + 4 \, a b^{2} f x - 4 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3} +{\left (3 \, a^{3} - 5 \, a^{2} b +{\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt{\frac{a}{b}}}{2 \, a \tan \left (f x + e\right )}\right ) - 2 \,{\left (3 \, a^{3} - 5 \, a^{2} b + 2 \, a b^{2}\right )} \tan \left (f x + e\right )}{4 \,{\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
[Out]
[-1/8*(8*b^3*f*x*tan(f*x + e)^2 + 8*a*b^2*f*x - 8*(a^2*b - 2*a*b^2 + b^3)*tan(f*x + e)^3 + (3*a^3 - 5*a^2*b +
(3*a^2*b - 5*a*b^2)*tan(f*x + e)^2)*sqrt(-a/b)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 + 4*(b^2*t
an(f*x + e)^3 - a*b*tan(f*x + e))*sqrt(-a/b))/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)) - 4*(3*a^3 -
5*a^2*b + 2*a*b^2)*tan(f*x + e))/((a^2*b^3 - 2*a*b^4 + b^5)*f*tan(f*x + e)^2 + (a^3*b^2 - 2*a^2*b^3 + a*b^4)*f
), -1/4*(4*b^3*f*x*tan(f*x + e)^2 + 4*a*b^2*f*x - 4*(a^2*b - 2*a*b^2 + b^3)*tan(f*x + e)^3 + (3*a^3 - 5*a^2*b
+ (3*a^2*b - 5*a*b^2)*tan(f*x + e)^2)*sqrt(a/b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a/b)/(a*tan(f*x + e)))
- 2*(3*a^3 - 5*a^2*b + 2*a*b^2)*tan(f*x + e))/((a^2*b^3 - 2*a*b^4 + b^5)*f*tan(f*x + e)^2 + (a^3*b^2 - 2*a^2*b
^3 + a*b^4)*f)]
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Sympy [A] time = 147.662, size = 2914, normalized size = 22.42 \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)**6/(a+b*tan(f*x+e)**2)**2,x)
[Out]
Piecewise((zoo*x*tan(e)**2, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((-x + tan(e + f*x)**5/(5*f) - tan(e + f*x)**3/(3
*f) + tan(e + f*x)/f)/a**2, Eq(b, 0)), ((-x + tan(e + f*x)/f)/b**2, Eq(a, 0)), (x*tan(e)**6/(a + b*tan(e)**2)*
*2, Eq(f, 0)), (6*I*a**(7/2)*b*sqrt(1/b)*tan(e + f*x)/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqr
t(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a
**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) + 4*I*a**(5/2)*b**2*sqrt(1/b)*tan(e +
f*x)**3/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*
sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f
*sqrt(1/b)*tan(e + f*x)**2) - 10*I*a**(5/2)*b**2*sqrt(1/b)*tan(e + f*x)/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a
**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e
+ f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 4*I*a**(3/2)*b**3
*f*x*sqrt(1/b)/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b
**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*
b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 8*I*a**(3/2)*b**3*sqrt(1/b)*tan(e + f*x)**3/(4*I*a**(7/2)*b**3*f*sqrt(1/b)
+ 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/
b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) + 4*I*a**(3
/2)*b**3*sqrt(1/b)*tan(e + f*x)/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2
- 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/
b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 4*I*sqrt(a)*b**4*f*x*sqrt(1/b)*tan(e + f*x)**2/(4*I*a**(7
/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**
(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e +
f*x)**2) + 4*I*sqrt(a)*b**4*sqrt(1/b)*tan(e + f*x)**3/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqr
t(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a
**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a**4*log(-I*sqrt(a)*sqrt(1/b) + t
an(e + f*x))/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**
4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b*
*6*f*sqrt(1/b)*tan(e + f*x)**2) + 3*a**4*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b**3*f*sqrt(1/b
) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1
/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a**3*b
*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f
*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4
*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) + 5*a**3*b*log(-I*sqrt(a)*sqrt(1/
b) + tan(e + f*x))/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/
2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt
(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) + 3*a**3*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*
a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*
I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan
(e + f*x)**2) - 5*a**3*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)
*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)
**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) + 5*a**2*b**2*log(-I*sqrt(
a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*ta
n(e + f*x)**2 - 8*I*a**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b
**5*f*sqrt(1/b) + 4*I*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2) - 5*a**2*b**2*log(I*sqrt(a)*sqrt(1/b) + tan(e
+ f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**3*f*sqrt(1/b) + 4*I*a**(5/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a
**(5/2)*b**4*f*sqrt(1/b) - 8*I*a**(3/2)*b**5*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**5*f*sqrt(1/b) + 4*I
*sqrt(a)*b**6*f*sqrt(1/b)*tan(e + f*x)**2), True))
________________________________________________________________________________________
Giac [A] time = 4.35374, size = 201, normalized size = 1.55 \begin{align*} \frac{\frac{a^{2} \tan \left (f x + e\right )}{{\left (a b^{2} - b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}} - \frac{{\left (3 \, a^{3} - 5 \, a^{2} b\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )}}{{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} \sqrt{a b}} - \frac{2 \,{\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac{2 \, \tan \left (f x + e\right )}{b^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
[Out]
1/2*(a^2*tan(f*x + e)/((a*b^2 - b^3)*(b*tan(f*x + e)^2 + a)) - (3*a^3 - 5*a^2*b)*(pi*floor((f*x + e)/pi + 1/2)
*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/((a^2*b^2 - 2*a*b^3 + b^4)*sqrt(a*b)) - 2*(f*x + e)/(a^2 - 2*a*b +
b^2) + 2*tan(f*x + e)/b^2)/f