3.207 \(\int (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=46 \[ \frac{b (2 a-b) \tan (e+f x)}{f}+x (a-b)^2+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0310278, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3661, 390, 203} \[ \frac{b (2 a-b) \tan (e+f x)}{f}+x (a-b)^2+\frac{b^2 \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x]^2)^2,x]

[Out]

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

Rule 3661

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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])

Rubi steps

\begin{align*} \int \left (a+b \tan ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left ((2 a-b) b+b^2 x^2+\frac{(a-b)^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(2 a-b) b \tan (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f}+\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=(a-b)^2 x+\frac{(2 a-b) b \tan (e+f x)}{f}+\frac{b^2 \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.575648, size = 73, normalized size = 1.59 \[ \frac{\tan (e+f x) \left (b \left (6 a-b \left (3-\tan ^2(e+f x)\right )\right )+\frac{3 (a-b)^2 \tanh ^{-1}\left (\sqrt{-\tan ^2(e+f x)}\right )}{\sqrt{-\tan ^2(e+f x)}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.003, size = 87, normalized size = 1.9 \begin{align*}{\frac{{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3\,f}}+2\,{\frac{\tan \left ( fx+e \right ) ab}{f}}-{\frac{{b}^{2}\tan \left ( fx+e \right ) }{f}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}}{f}}-2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) ab}{f}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.6435, size = 78, normalized size = 1.7 \begin{align*} a^{2} x - \frac{2 \,{\left (f x + e - \tan \left (f x + e\right )\right )} a b}{f} + \frac{{\left (\tan \left (f x + e\right )^{3} + 3 \, f x + 3 \, e - 3 \, \tan \left (f x + e\right )\right )} b^{2}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.06408, size = 117, normalized size = 2.54 \begin{align*} \frac{b^{2} \tan \left (f x + e\right )^{3} + 3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} f x + 3 \,{\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.381832, size = 68, normalized size = 1.48 \begin{align*} \begin{cases} a^{2} x - 2 a b x + \frac{2 a b \tan{\left (e + f x \right )}}{f} + b^{2} x + \frac{b^{2} \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac{b^{2} \tan{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x - 2*a*b*x + 2*a*b*tan(e + f*x)/f + b**2*x + b**2*tan(e + f*x)**3/(3*f) - b**2*tan(e + f*x)/f
, Ne(f, 0)), (x*(a + b*tan(e)**2)**2, True))

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Giac [B]  time = 1.46975, size = 516, normalized size = 11.22 \begin{align*} \frac{3 \, a^{2} f x \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 6 \, a b f x \tan \left (f x\right )^{3} \tan \left (e\right )^{3} + 3 \, b^{2} f x \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 9 \, a^{2} f x \tan \left (f x\right )^{2} \tan \left (e\right )^{2} + 18 \, a b f x \tan \left (f x\right )^{2} \tan \left (e\right )^{2} - 9 \, b^{2} f x \tan \left (f x\right )^{2} \tan \left (e\right )^{2} - 6 \, a b \tan \left (f x\right )^{3} \tan \left (e\right )^{2} + 3 \, b^{2} \tan \left (f x\right )^{3} \tan \left (e\right )^{2} - 6 \, a b \tan \left (f x\right )^{2} \tan \left (e\right )^{3} + 3 \, b^{2} \tan \left (f x\right )^{2} \tan \left (e\right )^{3} + 9 \, a^{2} f x \tan \left (f x\right ) \tan \left (e\right ) - 18 \, a b f x \tan \left (f x\right ) \tan \left (e\right ) + 9 \, b^{2} f x \tan \left (f x\right ) \tan \left (e\right ) - b^{2} \tan \left (f x\right )^{3} + 12 \, a b \tan \left (f x\right )^{2} \tan \left (e\right ) - 9 \, b^{2} \tan \left (f x\right )^{2} \tan \left (e\right ) + 12 \, a b \tan \left (f x\right ) \tan \left (e\right )^{2} - 9 \, b^{2} \tan \left (f x\right ) \tan \left (e\right )^{2} - b^{2} \tan \left (e\right )^{3} - 3 \, a^{2} f x + 6 \, a b f x - 3 \, b^{2} f x - 6 \, a b \tan \left (f x\right ) + 3 \, b^{2} \tan \left (f x\right ) - 6 \, a b \tan \left (e\right ) + 3 \, b^{2} \tan \left (e\right )}{3 \,{\left (f \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 3 \, f \tan \left (f x\right )^{2} \tan \left (e\right )^{2} + 3 \, f \tan \left (f x\right ) \tan \left (e\right ) - f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/3*(3*a^2*f*x*tan(f*x)^3*tan(e)^3 - 6*a*b*f*x*tan(f*x)^3*tan(e)^3 + 3*b^2*f*x*tan(f*x)^3*tan(e)^3 - 9*a^2*f*x
*tan(f*x)^2*tan(e)^2 + 18*a*b*f*x*tan(f*x)^2*tan(e)^2 - 9*b^2*f*x*tan(f*x)^2*tan(e)^2 - 6*a*b*tan(f*x)^3*tan(e
)^2 + 3*b^2*tan(f*x)^3*tan(e)^2 - 6*a*b*tan(f*x)^2*tan(e)^3 + 3*b^2*tan(f*x)^2*tan(e)^3 + 9*a^2*f*x*tan(f*x)*t
an(e) - 18*a*b*f*x*tan(f*x)*tan(e) + 9*b^2*f*x*tan(f*x)*tan(e) - b^2*tan(f*x)^3 + 12*a*b*tan(f*x)^2*tan(e) - 9
*b^2*tan(f*x)^2*tan(e) + 12*a*b*tan(f*x)*tan(e)^2 - 9*b^2*tan(f*x)*tan(e)^2 - b^2*tan(e)^3 - 3*a^2*f*x + 6*a*b
*f*x - 3*b^2*f*x - 6*a*b*tan(f*x) + 3*b^2*tan(f*x) - 6*a*b*tan(e) + 3*b^2*tan(e))/(f*tan(f*x)^3*tan(e)^3 - 3*f
*tan(f*x)^2*tan(e)^2 + 3*f*tan(f*x)*tan(e) - f)