3.232 \(\int \frac{\tan ^2(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)
Optimal. Leaf size=90 \[ \frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} f (a-b)^2}+\frac{\tan (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
[Out]
-(x/(a - b)^2) + ((a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*Sqrt[a]*(a - b)^2*Sqrt[b]*f) + Tan[e + f*
x]/(2*(a - b)*f*(a + b*Tan[e + f*x]^2))
________________________________________________________________________________________
Rubi [A] time = 0.103679, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{3670, 471, 522, 203, 205} \[ \frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} f (a-b)^2}+\frac{\tan (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{x}{(a-b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
-(x/(a - b)^2) + ((a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*Sqrt[a]*(a - b)^2*Sqrt[b]*f) + Tan[e + f*
x]/(2*(a - 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 471
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
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 ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 (a-b) f}\\ &=\frac{\tan (e+f x)}{2 (a-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{(a+b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^2 f}\\ &=-\frac{x}{(a-b)^2}+\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 \sqrt{a} (a-b)^2 \sqrt{b} f}+\frac{\tan (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.507534, size = 87, normalized size = 0.97 \[ \frac{\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b}}+\frac{(a-b) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}-2 (e+f x)}{2 f (a-b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
(-2*(e + f*x) + ((a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + ((a - b)*Sin[2*(e + f*x)]
)/(a + b + (a - b)*Cos[2*(e + f*x)]))/(2*(a - b)^2*f)
________________________________________________________________________________________
Maple [A] time = 0.022, size = 151, normalized size = 1.7 \begin{align*}{\frac{a\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{b\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{a}{2\,f \left ( a-b \right ) ^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{b}{2\,f \left ( a-b \right ) ^{2}}\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)^2/(a+b*tan(f*x+e)^2)^2,x)
[Out]
1/2/f*a/(a-b)^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)-1/2*b*tan(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)+1/2/f*a/(a-b)^2/(a
*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+1/2/f/(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)^2/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.55971, size = 892, normalized size = 9.91 \begin{align*} \left [-\frac{8 \, a b^{2} f x \tan \left (f x + e\right )^{2} + 8 \, a^{2} b f x +{\left ({\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + a b\right )} \sqrt{-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 \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt{-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 (a^{2} b - a b^{2}\right )} \tan \left (f x + e\right )}{8 \,{\left ({\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac{4 \, a b^{2} f x \tan \left (f x + e\right )^{2} + 4 \, a^{2} b f x -{\left ({\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + a b\right )} \sqrt{a b} \arctan \left (\frac{{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt{a b}}{2 \, a b \tan \left (f x + e\right )}\right ) - 2 \,{\left (a^{2} b - a b^{2}\right )} \tan \left (f x + e\right )}{4 \,{\left ({\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
[Out]
[-1/8*(8*a*b^2*f*x*tan(f*x + e)^2 + 8*a^2*b*f*x + ((a*b + b^2)*tan(f*x + e)^2 + a^2 + a*b)*sqrt(-a*b)*log((b^2
*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 - 4*(b*tan(f*x + e)^3 - a*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*(a^2*b - a*b^2)*tan(f*x + e))/((a^3*b^2 - 2*a^2*b^3 + a*b^4)*f*tan(f
*x + e)^2 + (a^4*b - 2*a^3*b^2 + a^2*b^3)*f), -1/4*(4*a*b^2*f*x*tan(f*x + e)^2 + 4*a^2*b*f*x - ((a*b + b^2)*ta
n(f*x + e)^2 + a^2 + a*b)*sqrt(a*b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a*b)/(a*b*tan(f*x + e))) - 2*(a^2*b
- a*b^2)*tan(f*x + e))/((a^3*b^2 - 2*a^2*b^3 + a*b^4)*f*tan(f*x + e)^2 + (a^4*b - 2*a^3*b^2 + a^2*b^3)*f)]
________________________________________________________________________________________
Sympy [A] time = 50.9058, size = 2145, normalized size = 23.83 \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)**2/(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)/f)/a**2, Eq(b, 0)), ((-x - 1/
(f*tan(e + f*x)))/b**2, Eq(a, 0)), (x*tan(e)**2/(a + b*tan(e)**2)**2, Eq(f, 0)), (-4*I*a**(3/2)*b*f*x*sqrt(1/b
)/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b)
- 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)
*tan(e + f*x)**2) + 2*I*a**(3/2)*b*sqrt(1/b)*tan(e + f*x)/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sq
rt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*
a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) - 4*I*sqrt(a)*b**2*f*x*sqrt(1/b)*tan
(e + f*x)**2/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f
*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*
f*sqrt(1/b)*tan(e + f*x)**2) - 2*I*sqrt(a)*b**2*sqrt(1/b)*tan(e + f*x)/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5
/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f
*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) + a**2*log(-I*sqrt(a)*s
qrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**
(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*s
qrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) - a**2*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b*f*sqrt
(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sq
rt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) + a*b*
log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqr
t(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a
**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) + a*b*log(-I*sqrt(a)*sqrt(1/b) + tan(
e + f*x))/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sq
rt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*s
qrt(1/b)*tan(e + f*x)**2) - a*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b*f*sqrt
(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sq
rt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2) - a*b*
log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*
x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sq
rt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**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*f*sqrt(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sq
rt(1/b) - 8*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*s
qrt(1/b)*tan(e + f*x)**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*f*sqr
t(1/b) + 4*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**2*f*sqrt(1/b) - 8*I*a**(3/2)*b**3*f*s
qrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**3*f*sqrt(1/b) + 4*I*sqrt(a)*b**4*f*sqrt(1/b)*tan(e + f*x)**2), True
))
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Giac [A] time = 1.56828, size = 151, normalized size = 1.68 \begin{align*} \frac{\frac{{\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 + b\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt{a b}} - \frac{2 \,{\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac{\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}{\left (a - b\right )}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
[Out]
1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(a + b)/((a^2 - 2*a*b + b^2)*sqr
t(a*b)) - 2*(f*x + e)/(a^2 - 2*a*b + b^2) + tan(f*x + e)/((b*tan(f*x + e)^2 + a)*(a - b)))/f