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

Optimal. Leaf size=59 \[ \frac{a^2 \tan (e+f x)}{f}-a^2 x+\frac{b (2 a+b) \tan ^3(e+f x)}{3 f}+\frac{b^2 \tan ^5(e+f x)}{5 f} \]

[Out]

-(a^2*x) + (a^2*Tan[e + f*x])/f + (b*(2*a + b)*Tan[e + f*x]^3)/(3*f) + (b^2*Tan[e + f*x]^5)/(5*f)

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Rubi [A]  time = 0.0926242, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4141, 1802, 203} \[ \frac{a^2 \tan (e+f x)}{f}-a^2 x+\frac{b (2 a+b) \tan ^3(e+f x)}{3 f}+\frac{b^2 \tan ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) + (a^2*Tan[e + f*x])/f + (b*(2*a + b)*Tan[e + f*x]^3)/(3*f) + (b^2*Tan[e + f*x]^5)/(5*f)

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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b \left (1+x^2\right )\right )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+b (2 a+b) x^2+b^2 x^4-\frac{a^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^2 \tan (e+f x)}{f}+\frac{b (2 a+b) \tan ^3(e+f x)}{3 f}+\frac{b^2 \tan ^5(e+f x)}{5 f}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-a^2 x+\frac{a^2 \tan (e+f x)}{f}+\frac{b (2 a+b) \tan ^3(e+f x)}{3 f}+\frac{b^2 \tan ^5(e+f x)}{5 f}\\ \end{align*}

Mathematica [B]  time = 0.816515, size = 281, normalized size = 4.76 \[ -\frac{\sec (e) \sec ^5(e+f x) \left (120 a^2 \sin (2 e+f x)-120 a^2 \sin (2 e+3 f x)+30 a^2 \sin (4 e+3 f x)-30 a^2 \sin (4 e+5 f x)+150 a^2 f x \cos (2 e+f x)+75 a^2 f x \cos (2 e+3 f x)+75 a^2 f x \cos (4 e+3 f x)+15 a^2 f x \cos (4 e+5 f x)+15 a^2 f x \cos (6 e+5 f x)-180 a^2 \sin (f x)+150 a^2 f x \cos (f x)-120 a b \sin (2 e+f x)+40 a b \sin (2 e+3 f x)-60 a b \sin (4 e+3 f x)+20 a b \sin (4 e+5 f x)+80 a b \sin (f x)-60 b^2 \sin (2 e+f x)+20 b^2 \sin (2 e+3 f x)+4 b^2 \sin (4 e+5 f x)-20 b^2 \sin (f x)\right )}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[e]*Sec[e + f*x]^5*(150*a^2*f*x*Cos[f*x] + 150*a^2*f*x*Cos[2*e + f*x] + 75*a^2*f*x*Cos[2*e + 3*f*x] + 75*
a^2*f*x*Cos[4*e + 3*f*x] + 15*a^2*f*x*Cos[4*e + 5*f*x] + 15*a^2*f*x*Cos[6*e + 5*f*x] - 180*a^2*Sin[f*x] + 80*a
*b*Sin[f*x] - 20*b^2*Sin[f*x] + 120*a^2*Sin[2*e + f*x] - 120*a*b*Sin[2*e + f*x] - 60*b^2*Sin[2*e + f*x] - 120*
a^2*Sin[2*e + 3*f*x] + 40*a*b*Sin[2*e + 3*f*x] + 20*b^2*Sin[2*e + 3*f*x] + 30*a^2*Sin[4*e + 3*f*x] - 60*a*b*Si
n[4*e + 3*f*x] - 30*a^2*Sin[4*e + 5*f*x] + 20*a*b*Sin[4*e + 5*f*x] + 4*b^2*Sin[4*e + 5*f*x]))/(480*f)

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Maple [A]  time = 0.05, size = 85, normalized size = 1.4 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( \tan \left ( fx+e \right ) -fx-e \right ) +{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}ab}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}+{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{5\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{15\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a^2*(tan(f*x+e)-f*x-e)+2/3*a*b*sin(f*x+e)^3/cos(f*x+e)^3+b^2*(1/5*sin(f*x+e)^3/cos(f*x+e)^5+2/15*sin(f*x+
e)^3/cos(f*x+e)^3))

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Maxima [A]  time = 1.49729, size = 78, normalized size = 1.32 \begin{align*} \frac{3 \, b^{2} \tan \left (f x + e\right )^{5} + 5 \,{\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3} - 15 \,{\left (f x + e\right )} a^{2} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*tan(f*x + e)^5 + 5*(2*a*b + b^2)*tan(f*x + e)^3 - 15*(f*x + e)*a^2 + 15*a^2*tan(f*x + e))/f

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Fricas [A]  time = 0.504371, size = 205, normalized size = 3.47 \begin{align*} -\frac{15 \, a^{2} f x \cos \left (f x + e\right )^{5} -{\left ({\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{4} +{\left (10 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*f*x*cos(f*x + e)^5 - ((15*a^2 - 10*a*b - 2*b^2)*cos(f*x + e)^4 + (10*a*b - b^2)*cos(f*x + e)^2 +
3*b^2)*sin(f*x + e))/(f*cos(f*x + e)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.48211, size = 95, normalized size = 1.61 \begin{align*} \frac{3 \, b^{2} \tan \left (f x + e\right )^{5} + 10 \, a b \tan \left (f x + e\right )^{3} + 5 \, b^{2} \tan \left (f x + e\right )^{3} - 15 \,{\left (f x + e\right )} a^{2} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*tan(f*x + e)^5 + 10*a*b*tan(f*x + e)^3 + 5*b^2*tan(f*x + e)^3 - 15*(f*x + e)*a^2 + 15*a^2*tan(f*x
+ e))/f