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

Optimal. Leaf size=85 \[ \frac{(a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} f \sqrt{a+b}}-\frac{x}{a^2}+\frac{\tan (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.15566, antiderivative size = 85, 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 = {4141, 1975, 471, 522, 203, 205} \[ \frac{(a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} f \sqrt{a+b}}-\frac{x}{a^2}+\frac{\tan (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

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 \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=\frac{\tan (e+f x)}{2 a f \left (a+b+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^2 f}+\frac{(a+2 b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=-\frac{x}{a^2}+\frac{(a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 \sqrt{b} \sqrt{a+b} f}+\frac{\tan (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 7.70941, size = 346, normalized size = 4.07 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac{\frac{(a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}-\frac{a \sqrt{b} \sin (2 (e+f x))}{(a+b) (a \cos (2 (e+f x))+a+2 b)}}{b^{3/2} f}-\frac{\frac{\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e)) (a \cos (2 (e+f x))+a+2 b)}+\frac{\left (6 a^2 b-a^3+24 a b^2+16 b^3\right ) (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{b f (a+b)^{3/2} \sqrt{b (\cos (e)-i \sin (e))^4}}+16 x}{a^2}\right )}{64 \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*(-((16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(Sec[
f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin
[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b^2)*
((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + S
in[e])))/a^2) + (((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2*(e +
f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))/(b^(3/2)*f)))/(64*(a + b*Sec[e + f*x]^2)^2)

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Maple [A]  time = 0.096, size = 108, normalized size = 1.3 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{2}}}+{\frac{\tan \left ( fx+e \right ) }{2\,af \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{1}{2\,af}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{b}{f{a}^{2}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

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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*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.597478, size = 1052, normalized size = 12.38 \begin{align*} \left [-\frac{8 \,{\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 8 \,{\left (a b^{2} + b^{3}\right )} f x - 4 \,{\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt{-a b - b^{2}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \,{\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac{4 \,{\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \,{\left (a b^{2} + b^{3}\right )} f x - 2 \,{\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt{a b + b^{2}} \arctan \left (\frac{{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt{a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \,{\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (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*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/8*(8*(a^2*b + a*b^2)*f*x*cos(f*x + e)^2 + 8*(a*b^2 + b^3)*f*x - 4*(a^2*b + a*b^2)*cos(f*x + e)*sin(f*x + e
) + ((a^2 + 2*a*b)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 -
2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e)
+ b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^3*b^2 + a
^2*b^3)*f), -1/4*(4*(a^2*b + a*b^2)*f*x*cos(f*x + e)^2 + 4*(a*b^2 + b^3)*f*x - 2*(a^2*b + a*b^2)*cos(f*x + e)*
sin(f*x + e) + ((a^2 + 2*a*b)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)
^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))))/((a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^3*b^2 + a^2*b^3
)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral(tan(e + f*x)**2/(a + b*sec(e + f*x)**2)**2, x)

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Giac [A]  time = 1.58837, size = 134, normalized size = 1.58 \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 + b^{2}}}\right )\right )}{\left (a + 2 \, b\right )}}{\sqrt{a b + b^{2}} a^{2}} - \frac{2 \,{\left (f x + e\right )}}{a^{2}} + \frac{\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*sec(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 + b^2)))*(a + 2*b)/(sqrt(a*b + b^2)
*a^2) - 2*(f*x + e)/a^2 + tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)*a))/f