### 3.372 $$\int \sec ^4(x) \tan ^2(x) \, dx$$

Optimal. Leaf size=17 $\frac{\tan ^5(x)}{5}+\frac{\tan ^3(x)}{3}$

[Out]

Tan[x]^3/3 + Tan[x]^5/5

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Rubi [A]  time = 0.0232369, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {2607, 14} $\frac{\tan ^5(x)}{5}+\frac{\tan ^3(x)}{3}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sec[x]^4*Tan[x]^2,x]

[Out]

Tan[x]^3/3 + Tan[x]^5/5

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0157197, size = 27, normalized size = 1.59 $-\frac{2 \tan (x)}{15}+\frac{1}{5} \tan (x) \sec ^4(x)-\frac{1}{15} \tan (x) \sec ^2(x)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sec[x]^4*Tan[x]^2,x]

[Out]

(-2*Tan[x])/15 - (Sec[x]^2*Tan[x])/15 + (Sec[x]^4*Tan[x])/5

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Maple [A]  time = 0., size = 22, normalized size = 1.3 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{3}}{5\, \left ( \cos \left ( x \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( x \right ) \right ) ^{3}}{15\, \left ( \cos \left ( x \right ) \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4*tan(x)^2,x)

[Out]

1/5*sin(x)^3/cos(x)^5+2/15*sin(x)^3/cos(x)^3

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Maxima [A]  time = 0.923833, size = 18, normalized size = 1.06 \begin{align*} \frac{1}{5} \, \tan \left (x\right )^{5} + \frac{1}{3} \, \tan \left (x\right )^{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4*tan(x)^2,x, algorithm="maxima")

[Out]

1/5*tan(x)^5 + 1/3*tan(x)^3

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Fricas [A]  time = 2.22829, size = 69, normalized size = 4.06 \begin{align*} -\frac{{\left (2 \, \cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 3\right )} \sin \left (x\right )}{15 \, \cos \left (x\right )^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4*tan(x)^2,x, algorithm="fricas")

[Out]

-1/15*(2*cos(x)^4 + cos(x)^2 - 3)*sin(x)/cos(x)^5

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Sympy [B]  time = 0.06336, size = 29, normalized size = 1.71 \begin{align*} - \frac{2 \sin{\left (x \right )}}{15 \cos{\left (x \right )}} - \frac{\sin{\left (x \right )}}{15 \cos ^{3}{\left (x \right )}} + \frac{\sin{\left (x \right )}}{5 \cos ^{5}{\left (x \right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**4*tan(x)**2,x)

[Out]

-2*sin(x)/(15*cos(x)) - sin(x)/(15*cos(x)**3) + sin(x)/(5*cos(x)**5)

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Giac [A]  time = 1.06102, size = 18, normalized size = 1.06 \begin{align*} \frac{1}{5} \, \tan \left (x\right )^{5} + \frac{1}{3} \, \tan \left (x\right )^{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4*tan(x)^2,x, algorithm="giac")

[Out]

1/5*tan(x)^5 + 1/3*tan(x)^3