3.245 \(\int \frac{\tan ^2(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=144 \[ \frac{\left (3 a^2+6 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} \sqrt{b} f (a-b)^3}+\frac{(3 a+b) \tan (e+f x)}{8 a f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{\tan (e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]

[Out]

-(x/(a - b)^3) + ((3*a^2 + 6*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(3/2)*(a - b)^3*Sqrt[b]*f
) + Tan[e + f*x]/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) + ((3*a + b)*Tan[e + f*x])/(8*a*(a - b)^2*f*(a + b*Tan
[e + f*x]^2))

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Rubi [A]  time = 0.15395, antiderivative size = 144, 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, 471, 527, 522, 203, 205} \[ \frac{\left (3 a^2+6 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} \sqrt{b} f (a-b)^3}+\frac{(3 a+b) \tan (e+f x)}{8 a f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{\tan (e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{x}{(a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-(x/(a - b)^3) + ((3*a^2 + 6*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(3/2)*(a - b)^3*Sqrt[b]*f
) + Tan[e + f*x]/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) + ((3*a + b)*Tan[e + f*x])/(8*a*(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 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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -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 ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{1-3 x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=\frac{\tan (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{(3 a+b) \tan (e+f x)}{8 a (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{5 a-b+(-3 a-b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a (a-b)^2 f}\\ &=\frac{\tan (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{(3 a+b) \tan (e+f x)}{8 a (a-b)^2 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)^3 f}+\frac{\left (3 a^2+6 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a (a-b)^3 f}\\ &=-\frac{x}{(a-b)^3}+\frac{\left (3 a^2+6 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{3/2} (a-b)^3 \sqrt{b} f}+\frac{\tan (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{(3 a+b) \tan (e+f x)}{8 a (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.9767, size = 139, normalized size = 0.97 \[ \frac{\frac{\left (3 a^2+6 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} \sqrt{b}}+\frac{(a-b) \sin (2 (e+f x)) \left (\left (5 a^2-4 a b-b^2\right ) \cos (2 (e+f x))+5 a^2+2 a b+b^2\right )}{a ((a-b) \cos (2 (e+f x))+a+b)^2}-8 (e+f x)}{8 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(-8*(e + f*x) + ((3*a^2 + 6*a*b - b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*Sqrt[b]) + ((a - b)*(5
*a^2 + 2*a*b + b^2 + (5*a^2 - 4*a*b - b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(a*(a + b + (a - b)*Cos[2*(e +
f*x)])^2))/(8*(a - b)^3*f)

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Maple [B]  time = 0.023, size = 339, normalized size = 2.4 \begin{align*}{\frac{3\,ab \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}a}}+{\frac{5\,{a}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ) ab}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,a}{8\,f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,b}{4\,f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{{b}^{2}}{8\,f \left ( a-b \right ) ^{3}a}\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 ) ^{3}}} \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)^3,x)

[Out]

3/8/f*a/(a-b)^3/(a+b*tan(f*x+e)^2)^2*b*tan(f*x+e)^3-1/4/f*b^2/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-1/8/f*
b^3/(a-b)^3/(a+b*tan(f*x+e)^2)^2/a*tan(f*x+e)^3+5/8/f*a^2/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)-3/4/f*b/(a-b
)^3/(a+b*tan(f*x+e)^2)^2*a*tan(f*x+e)+1/8/f*b^2/(a-b)^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)+3/8/f*a/(a-b)^3/(a*b)^
(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+3/4/f*b/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/8/f*b^2/
(a-b)^3/a/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/f/(a-b)^3*arctan(tan(f*x+e))

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.62202, size = 1638, normalized size = 11.38 \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)^3,x, algorithm="fricas")

[Out]

[-1/32*(32*a^2*b^3*f*x*tan(f*x + e)^4 + 64*a^3*b^2*f*x*tan(f*x + e)^2 + 32*a^4*b*f*x - 4*(3*a^3*b^2 - 2*a^2*b^
3 - a*b^4)*tan(f*x + e)^3 + ((3*a^2*b^2 + 6*a*b^3 - b^4)*tan(f*x + e)^4 + 3*a^4 + 6*a^3*b - a^2*b^2 + 2*(3*a^3
*b + 6*a^2*b^2 - a*b^3)*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
*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*(5*a^4*b
- 6*a^3*b^2 + a^2*b^3)*tan(f*x + e))/((a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f*tan(f*x + e)^4 + 2*(a^6*b^
2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^2 + (a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f), -1/16*(1
6*a^2*b^3*f*x*tan(f*x + e)^4 + 32*a^3*b^2*f*x*tan(f*x + e)^2 + 16*a^4*b*f*x - 2*(3*a^3*b^2 - 2*a^2*b^3 - a*b^4
)*tan(f*x + e)^3 - ((3*a^2*b^2 + 6*a*b^3 - b^4)*tan(f*x + e)^4 + 3*a^4 + 6*a^3*b - a^2*b^2 + 2*(3*a^3*b + 6*a^
2*b^2 - a*b^3)*tan(f*x + e)^2)*sqrt(a*b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a*b)/(a*b*tan(f*x + e))) - 2*(
5*a^4*b - 6*a^3*b^2 + a^2*b^3)*tan(f*x + e))/((a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f*tan(f*x + e)^4 + 2
*(a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^2 + (a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.75252, size = 270, normalized size = 1.88 \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 (3 \, a^{2} + 6 \, a b - b^{2}\right )}}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt{a b}} - \frac{8 \,{\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{3 \, a b \tan \left (f x + e\right )^{3} + b^{2} \tan \left (f x + e\right )^{3} + 5 \, a^{2} \tan \left (f x + e\right ) - a b \tan \left (f x + e\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}}}{8 \, 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)^3,x, algorithm="giac")

[Out]

1/8*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(3*a^2 + 6*a*b - b^2)/((a^4 - 3*
a^3*b + 3*a^2*b^2 - a*b^3)*sqrt(a*b)) - 8*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b*tan(f*x + e)^3 +
b^2*tan(f*x + e)^3 + 5*a^2*tan(f*x + e) - a*b*tan(f*x + e))/((a^3 - 2*a^2*b + a*b^2)*(b*tan(f*x + e)^2 + a)^2)
)/f