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

Optimal. Leaf size=95 \[ \frac{a^2 \tan ^5(e+f x)}{5 f}-\frac{a^2 \tan ^3(e+f x)}{3 f}+\frac{a^2 \tan (e+f x)}{f}-a^2 x+\frac{b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac{b^2 \tan ^9(e+f x)}{9 f} \]

[Out]

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

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Rubi [A]  time = 0.107516, antiderivative size = 95, 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} \[ \frac{a^2 \tan ^5(e+f x)}{5 f}-\frac{a^2 \tan ^3(e+f x)}{3 f}+\frac{a^2 \tan (e+f x)}{f}-a^2 x+\frac{b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac{b^2 \tan ^9(e+f x)}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 2.1231, size = 275, normalized size = 2.89 \[ -\frac{4 \sec ^9(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (\left (231 a^2-270 a b+5 b^2\right ) \tan (e) \cos ^7(e+f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \tan (e) \cos ^5(e+f x)-\left (483 a^2-90 a b-10 b^2\right ) \sec (e) \sin (f x) \cos ^8(e+f x)+\left (231 a^2-270 a b+5 b^2\right ) \sec (e) \sin (f x) \cos ^6(e+f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \sec (e) \sin (f x) \cos ^4(e+f x)+315 a^2 f x \cos ^9(e+f x)-5 b (18 a-19 b) \tan (e) \cos ^3(e+f x)-5 b (18 a-19 b) \sec (e) \sin (f x) \cos ^2(e+f x)-35 b^2 \tan (e) \cos (e+f x)-35 b^2 \sec (e) \sin (f x)\right )}{315 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(b + a*Cos[e + f*x]^2)^2*Sec[e + f*x]^9*(315*a^2*f*x*Cos[e + f*x]^9 - 35*b^2*Sec[e]*Sin[f*x] - 5*(18*a - 1
9*b)*b*Cos[e + f*x]^2*Sec[e]*Sin[f*x] - 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e + f*x]^4*Sec[e]*Sin[f*x] + (231*a^2
- 270*a*b + 5*b^2)*Cos[e + f*x]^6*Sec[e]*Sin[f*x] - (483*a^2 - 90*a*b - 10*b^2)*Cos[e + f*x]^8*Sec[e]*Sin[f*x
] - 35*b^2*Cos[e + f*x]*Tan[e] - 5*(18*a - 19*b)*b*Cos[e + f*x]^3*Tan[e] - 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e
+ f*x]^5*Tan[e] + (231*a^2 - 270*a*b + 5*b^2)*Cos[e + f*x]^7*Tan[e]))/(315*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

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Maple [A]  time = 0.07, size = 105, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ({\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{5}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}+\tan \left ( fx+e \right ) -fx-e \right ) +{\frac{2\,ab \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{7\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{9\, \left ( \cos \left ( fx+e \right ) \right ) ^{9}}}+{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{63\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}} \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)^6,x)

[Out]

1/f*(a^2*(1/5*tan(f*x+e)^5-1/3*tan(f*x+e)^3+tan(f*x+e)-f*x-e)+2/7*a*b*sin(f*x+e)^7/cos(f*x+e)^7+b^2*(1/9*sin(f
*x+e)^7/cos(f*x+e)^9+2/63*sin(f*x+e)^7/cos(f*x+e)^7))

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Maxima [A]  time = 1.59323, size = 113, normalized size = 1.19 \begin{align*} \frac{35 \, b^{2} \tan \left (f x + e\right )^{9} + 45 \,{\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \,{\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, 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)^6,x, algorithm="maxima")

[Out]

1/315*(35*b^2*tan(f*x + e)^9 + 45*(2*a*b + b^2)*tan(f*x + e)^7 + 63*a^2*tan(f*x + e)^5 - 105*a^2*tan(f*x + e)^
3 - 315*(f*x + e)*a^2 + 315*a^2*tan(f*x + e))/f

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Fricas [A]  time = 0.548164, size = 342, normalized size = 3.6 \begin{align*} -\frac{315 \, a^{2} f x \cos \left (f x + e\right )^{9} -{\left ({\left (483 \, a^{2} - 90 \, a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{8} -{\left (231 \, a^{2} - 270 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \,{\left (21 \, a^{2} - 90 \, a b + 25 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \,{\left (18 \, a b - 19 \, 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((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^6,x, algorithm="fricas")

[Out]

-1/315*(315*a^2*f*x*cos(f*x + e)^9 - ((483*a^2 - 90*a*b - 10*b^2)*cos(f*x + e)^8 - (231*a^2 - 270*a*b + 5*b^2)
*cos(f*x + e)^6 + 3*(21*a^2 - 90*a*b + 25*b^2)*cos(f*x + e)^4 + 5*(18*a*b - 19*b^2)*cos(f*x + e)^2 + 35*b^2)*s
in(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} \tan ^{6}{\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)**6,x)

[Out]

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

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Giac [A]  time = 4.32276, size = 132, normalized size = 1.39 \begin{align*} \frac{35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \, a b \tan \left (f x + e\right )^{7} + 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \,{\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, 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)^6,x, algorithm="giac")

[Out]

1/315*(35*b^2*tan(f*x + e)^9 + 90*a*b*tan(f*x + e)^7 + 45*b^2*tan(f*x + e)^7 + 63*a^2*tan(f*x + e)^5 - 105*a^2
*tan(f*x + e)^3 - 315*(f*x + e)*a^2 + 315*a^2*tan(f*x + e))/f