3.171 \(\int \sec ^6(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=106 \[ \frac{\left (a^2+6 a b+6 b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{2 b (a+2 b) \tan ^7(e+f x)}{7 f}+\frac{2 (a+b) (a+2 b) \tan ^3(e+f x)}{3 f}+\frac{(a+b)^2 \tan (e+f x)}{f}+\frac{b^2 \tan ^9(e+f x)}{9 f} \]

[Out]

((a + b)^2*Tan[e + f*x])/f + (2*(a + b)*(a + 2*b)*Tan[e + f*x]^3)/(3*f) + ((a^2 + 6*a*b + 6*b^2)*Tan[e + f*x]^
5)/(5*f) + (2*b*(a + 2*b)*Tan[e + f*x]^7)/(7*f) + (b^2*Tan[e + f*x]^9)/(9*f)

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Rubi [A]  time = 0.089263, antiderivative size = 106, 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, 373} \[ \frac{\left (a^2+6 a b+6 b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{2 b (a+2 b) \tan ^7(e+f x)}{7 f}+\frac{2 (a+b) (a+2 b) \tan ^3(e+f x)}{3 f}+\frac{(a+b)^2 \tan (e+f x)}{f}+\frac{b^2 \tan ^9(e+f x)}{9 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + b)^2*Tan[e + f*x])/f + (2*(a + b)*(a + 2*b)*Tan[e + f*x]^3)/(3*f) + ((a^2 + 6*a*b + 6*b^2)*Tan[e + f*x]^
5)/(5*f) + (2*b*(a + 2*b)*Tan[e + f*x]^7)/(7*f) + (b^2*Tan[e + f*x]^9)/(9*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 373

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

Rubi steps

\begin{align*} \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right )^2 \left (a+b+b x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left ((a+b)^2+2 (a+b) (a+2 b) x^2+\left (a^2+6 a b+6 b^2\right ) x^4+2 b (a+2 b) x^6+b^2 x^8\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a+b)^2 \tan (e+f x)}{f}+\frac{2 (a+b) (a+2 b) \tan ^3(e+f x)}{3 f}+\frac{\left (a^2+6 a b+6 b^2\right ) \tan ^5(e+f x)}{5 f}+\frac{2 b (a+2 b) \tan ^7(e+f x)}{7 f}+\frac{b^2 \tan ^9(e+f x)}{9 f}\\ \end{align*}

Mathematica [A]  time = 0.416444, size = 96, normalized size = 0.91 \[ \frac{63 \left (a^2+6 a b+6 b^2\right ) \tan ^5(e+f x)+210 \left (a^2+3 a b+2 b^2\right ) \tan ^3(e+f x)+90 b (a+2 b) \tan ^7(e+f x)+315 (a+b)^2 \tan (e+f x)+35 b^2 \tan ^9(e+f x)}{315 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(315*(a + b)^2*Tan[e + f*x] + 210*(a^2 + 3*a*b + 2*b^2)*Tan[e + f*x]^3 + 63*(a^2 + 6*a*b + 6*b^2)*Tan[e + f*x]
^5 + 90*b*(a + 2*b)*Tan[e + f*x]^7 + 35*b^2*Tan[e + f*x]^9)/(315*f)

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Maple [A]  time = 0.039, size = 134, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -{a}^{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 ) -2\,ab \left ( -{\frac{16}{35}}-1/7\, \left ( \sec \left ( fx+e \right ) \right ) ^{6}-{\frac{6\, \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{35}} \right ) \tan \left ( fx+e \right ) -{b}^{2} \left ( -{\frac{128}{315}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{8}}{9}}-{\frac{8\, \left ( \sec \left ( fx+e \right ) \right ) ^{6}}{63}}-{\frac{16\, \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{105}}-{\frac{64\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{315}} \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)^6*(a+b*sec(f*x+e)^2)^2,x)

[Out]

1/f*(-a^2*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)-2*a*b*(-16/35-1/7*sec(f*x+e)^6-6/35*sec(f*x+e)
^4-8/35*sec(f*x+e)^2)*tan(f*x+e)-b^2*(-128/315-1/9*sec(f*x+e)^8-8/63*sec(f*x+e)^6-16/105*sec(f*x+e)^4-64/315*s
ec(f*x+e)^2)*tan(f*x+e))

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Maxima [A]  time = 1.01043, size = 139, normalized size = 1.31 \begin{align*} \frac{35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \,{\left (a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \,{\left (a^{2} + 6 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 210 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 315 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/315*(35*b^2*tan(f*x + e)^9 + 90*(a*b + 2*b^2)*tan(f*x + e)^7 + 63*(a^2 + 6*a*b + 6*b^2)*tan(f*x + e)^5 + 210
*(a^2 + 3*a*b + 2*b^2)*tan(f*x + e)^3 + 315*(a^2 + 2*a*b + b^2)*tan(f*x + e))/f

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Fricas [A]  time = 0.508377, size = 300, normalized size = 2.83 \begin{align*} \frac{{\left (8 \,{\left (21 \, a^{2} + 36 \, a b + 16 \, b^{2}\right )} \cos \left (f x + e\right )^{8} + 4 \,{\left (21 \, a^{2} + 36 \, a b + 16 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \,{\left (21 \, a^{2} + 36 \, a b + 16 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 10 \,{\left (9 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/315*(8*(21*a^2 + 36*a*b + 16*b^2)*cos(f*x + e)^8 + 4*(21*a^2 + 36*a*b + 16*b^2)*cos(f*x + e)^6 + 3*(21*a^2 +
36*a*b + 16*b^2)*cos(f*x + e)^4 + 10*(9*a*b + 4*b^2)*cos(f*x + e)^2 + 35*b^2)*sin(f*x + e)/(f*cos(f*x + e)^9)

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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 ^{6}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**2*sec(e + f*x)**6, x)

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Giac [A]  time = 1.28304, size = 221, normalized size = 2.08 \begin{align*} \frac{35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \, a b \tan \left (f x + e\right )^{7} + 180 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} + 378 \, a b \tan \left (f x + e\right )^{5} + 378 \, b^{2} \tan \left (f x + e\right )^{5} + 210 \, a^{2} \tan \left (f x + e\right )^{3} + 630 \, a b \tan \left (f x + e\right )^{3} + 420 \, b^{2} \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right ) + 630 \, a b \tan \left (f x + e\right ) + 315 \, b^{2} \tan \left (f x + e\right )}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/315*(35*b^2*tan(f*x + e)^9 + 90*a*b*tan(f*x + e)^7 + 180*b^2*tan(f*x + e)^7 + 63*a^2*tan(f*x + e)^5 + 378*a*
b*tan(f*x + e)^5 + 378*b^2*tan(f*x + e)^5 + 210*a^2*tan(f*x + e)^3 + 630*a*b*tan(f*x + e)^3 + 420*b^2*tan(f*x
+ e)^3 + 315*a^2*tan(f*x + e) + 630*a*b*tan(f*x + e) + 315*b^2*tan(f*x + e))/f