3.87 \(\int \sec ^4(x) \tan ^2(x) \, dx\)

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

[Out]

1/3*tan(x)^3+1/5*tan(x)^5

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, 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 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]]

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])

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.43, size = 20, normalized size = 1.18 \[ -\frac {{\left (2 \, \cos \relax (x)^{4} + \cos \relax (x)^{2} - 3\right )} \sin \relax (x)}{15 \, \cos \relax (x)^{5}} \]

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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giac [A]  time = 0.77, size = 13, normalized size = 0.76 \[ \frac {1}{5} \, \tan \relax (x)^{5} + \frac {1}{3} \, \tan \relax (x)^{3} \]

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

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maple [A]  time = 0.03, size = 22, normalized size = 1.29 \[ \frac {2 \left (\sin ^{3}\relax (x )\right )}{15 \cos \relax (x )^{3}}+\frac {\sin ^{3}\relax (x )}{5 \cos \relax (x )^{5}} \]

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 = 0.44, size = 13, normalized size = 0.76 \[ \frac {1}{5} \, \tan \relax (x)^{5} + \frac {1}{3} \, \tan \relax (x)^{3} \]

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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mupad [B]  time = 0.17, size = 13, normalized size = 0.76 \[ \frac {{\mathrm {tan}\relax (x)}^5}{5}+\frac {{\mathrm {tan}\relax (x)}^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^2/cos(x)^4,x)

[Out]

tan(x)^3/3 + tan(x)^5/5

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sympy [B]  time = 0.07, size = 29, normalized size = 1.71 \[ - \frac {2 \sin {\relax (x )}}{15 \cos {\relax (x )}} - \frac {\sin {\relax (x )}}{15 \cos ^{3}{\relax (x )}} + \frac {\sin {\relax (x )}}{5 \cos ^{5}{\relax (x )}} \]

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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