3.331 \(\int (a+b \sec ^2(e+f x))^2 \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=77 \[ \frac{a^2 \tan ^3(e+f x)}{3 f}-\frac{a^2 \tan (e+f x)}{f}+a^2 x+\frac{b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac{b^2 \tan ^7(e+f x)}{7 f} \]

[Out]

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

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Rubi [A]  time = 0.097092, antiderivative size = 77, 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} \[ \frac{a^2 \tan ^3(e+f x)}{3 f}-\frac{a^2 \tan (e+f x)}{f}+a^2 x+\frac{b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac{b^2 \tan ^7(e+f x)}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^4,x]

[Out]

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

Mathematica [B]  time = 1.14626, size = 395, normalized size = 5.13 \[ \frac{\sec (e) \sec ^7(e+f x) \left (4480 a^2 \sin (2 e+f x)-3780 a^2 \sin (2 e+3 f x)+2100 a^2 \sin (4 e+3 f x)-1540 a^2 \sin (4 e+5 f x)+420 a^2 \sin (6 e+5 f x)-280 a^2 \sin (6 e+7 f x)+3675 a^2 f x \cos (2 e+f x)+2205 a^2 f x \cos (2 e+3 f x)+2205 a^2 f x \cos (4 e+3 f x)+735 a^2 f x \cos (4 e+5 f x)+735 a^2 f x \cos (6 e+5 f x)+105 a^2 f x \cos (6 e+7 f x)+105 a^2 f x \cos (8 e+7 f x)-5320 a^2 \sin (f x)+3675 a^2 f x \cos (f x)-1260 a b \sin (2 e+f x)+924 a b \sin (2 e+3 f x)-840 a b \sin (4 e+3 f x)+168 a b \sin (4 e+5 f x)-420 a b \sin (6 e+5 f x)+84 a b \sin (6 e+7 f x)+1680 a b \sin (f x)+420 b^2 \sin (2 e+f x)-168 b^2 \sin (2 e+3 f x)-420 b^2 \sin (4 e+3 f x)+84 b^2 \sin (4 e+5 f x)+12 b^2 \sin (6 e+7 f x)+840 b^2 \sin (f x)\right )}{13440 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^4,x]

[Out]

(Sec[e]*Sec[e + f*x]^7*(3675*a^2*f*x*Cos[f*x] + 3675*a^2*f*x*Cos[2*e + f*x] + 2205*a^2*f*x*Cos[2*e + 3*f*x] +
2205*a^2*f*x*Cos[4*e + 3*f*x] + 735*a^2*f*x*Cos[4*e + 5*f*x] + 735*a^2*f*x*Cos[6*e + 5*f*x] + 105*a^2*f*x*Cos[
6*e + 7*f*x] + 105*a^2*f*x*Cos[8*e + 7*f*x] - 5320*a^2*Sin[f*x] + 1680*a*b*Sin[f*x] + 840*b^2*Sin[f*x] + 4480*
a^2*Sin[2*e + f*x] - 1260*a*b*Sin[2*e + f*x] + 420*b^2*Sin[2*e + f*x] - 3780*a^2*Sin[2*e + 3*f*x] + 924*a*b*Si
n[2*e + 3*f*x] - 168*b^2*Sin[2*e + 3*f*x] + 2100*a^2*Sin[4*e + 3*f*x] - 840*a*b*Sin[4*e + 3*f*x] - 420*b^2*Sin
[4*e + 3*f*x] - 1540*a^2*Sin[4*e + 5*f*x] + 168*a*b*Sin[4*e + 5*f*x] + 84*b^2*Sin[4*e + 5*f*x] + 420*a^2*Sin[6
*e + 5*f*x] - 420*a*b*Sin[6*e + 5*f*x] - 280*a^2*Sin[6*e + 7*f*x] + 84*a*b*Sin[6*e + 7*f*x] + 12*b^2*Sin[6*e +
7*f*x]))/(13440*f)

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Maple [A]  time = 0.054, size = 94, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ({\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 ) ^{5}}{5\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}}+{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{7\, \left ( \cos \left ( fx+e \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{35\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}} \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)^4,x)

[Out]

1/f*(a^2*(1/3*tan(f*x+e)^3-tan(f*x+e)+f*x+e)+2/5*a*b*sin(f*x+e)^5/cos(f*x+e)^5+b^2*(1/7*sin(f*x+e)^5/cos(f*x+e
)^7+2/35*sin(f*x+e)^5/cos(f*x+e)^5))

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Maxima [A]  time = 1.46163, size = 96, normalized size = 1.25 \begin{align*} \frac{15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \,{\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 105 \,{\left (f x + e\right )} a^{2} - 105 \, a^{2} \tan \left (f x + e\right )}{105 \, 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)^4,x, algorithm="maxima")

[Out]

1/105*(15*b^2*tan(f*x + e)^7 + 21*(2*a*b + b^2)*tan(f*x + e)^5 + 35*a^2*tan(f*x + e)^3 + 105*(f*x + e)*a^2 - 1
05*a^2*tan(f*x + e))/f

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Fricas [A]  time = 0.520851, size = 273, normalized size = 3.55 \begin{align*} \frac{105 \, a^{2} f x \cos \left (f x + e\right )^{7} -{\left (2 \,{\left (70 \, a^{2} - 21 \, a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} -{\left (35 \, a^{2} - 84 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 6 \,{\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \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)^4,x, algorithm="fricas")

[Out]

1/105*(105*a^2*f*x*cos(f*x + e)^7 - (2*(70*a^2 - 21*a*b - 3*b^2)*cos(f*x + e)^6 - (35*a^2 - 84*a*b + 3*b^2)*co
s(f*x + e)^4 - 6*(7*a*b - 4*b^2)*cos(f*x + e)^2 - 15*b^2)*sin(f*x + e))/(f*cos(f*x + e)^7)

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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 ^{4}{\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)**4,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**2*tan(e + f*x)**4, x)

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Giac [A]  time = 2.46368, size = 113, normalized size = 1.47 \begin{align*} \frac{15 \, b^{2} \tan \left (f x + e\right )^{7} + 42 \, a b \tan \left (f x + e\right )^{5} + 21 \, b^{2} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 105 \,{\left (f x + e\right )} a^{2} - 105 \, a^{2} \tan \left (f x + e\right )}{105 \, 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)^4,x, algorithm="giac")

[Out]

1/105*(15*b^2*tan(f*x + e)^7 + 42*a*b*tan(f*x + e)^5 + 21*b^2*tan(f*x + e)^5 + 35*a^2*tan(f*x + e)^3 + 105*(f*
x + e)*a^2 - 105*a^2*tan(f*x + e))/f