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

Optimal. Leaf size=53 \[ \frac{2 b (a+b) \tan ^3(e+f x)}{3 f}+\frac{(a+b)^2 \tan (e+f x)}{f}+\frac{b^2 \tan ^5(e+f x)}{5 f} \]

[Out]

((a + b)^2*Tan[e + f*x])/f + (2*b*(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 = 0.06289, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {4146, 194} \[ \frac{2 b (a+b) \tan ^3(e+f x)}{3 f}+\frac{(a+b)^2 \tan (e+f x)}{f}+\frac{b^2 \tan ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4146

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.265406, size = 48, normalized size = 0.91 \[ \frac{10 b (a+b) \tan ^3(e+f x)+15 (a+b)^2 \tan (e+f x)+3 b^2 \tan ^5(e+f x)}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(a + b)^2*Tan[e + f*x] + 10*b*(a + b)*Tan[e + f*x]^3 + 3*b^2*Tan[e + f*x]^5)/(15*f)

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Maple [A]  time = 0.032, size = 71, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ({a}^{2}\tan \left ( fx+e \right ) -2\,ab \left ( -2/3-1/3\, \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) \tan \left ( fx+e \right ) -{b}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15}} \right ) \tan \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a^2*tan(f*x+e)-2*a*b*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)-b^2*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*ta
n(f*x+e))

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Maxima [A]  time = 0.986876, size = 96, normalized size = 1.81 \begin{align*} \frac{10 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a b +{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} b^{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(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.477477, size = 167, normalized size = 3.15 \begin{align*} \frac{{\left ({\left (15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (5 \, a b + 2 \, 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(sec(f*x+e)^2*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/15*((15*a^2 + 20*a*b + 8*b^2)*cos(f*x + e)^4 + 2*(5*a*b + 2*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} \sec ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.35357, size = 111, normalized size = 2.09 \begin{align*} \frac{3 \, b^{2} \tan \left (f x + e\right )^{5} + 10 \, a b \tan \left (f x + e\right )^{3} + 10 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 30 \, a b \tan \left (f x + e\right ) + 15 \, b^{2} \tan \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*tan(f*x + e)^5 + 10*a*b*tan(f*x + e)^3 + 10*b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 30*a*b*tan(
f*x + e) + 15*b^2*tan(f*x + e))/f