3.830 \(\int 4 \sin ^2(x) \tan ^2(x) \, dx\)

Optimal. Leaf size=16 \[ -6 x+6 \tan (x)-2 \sin ^2(x) \tan (x) \]

[Out]

-6*x+6*tan(x)-2*sin(x)^2*tan(x)

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {12, 2591, 288, 321, 203} \[ -6 x+6 \tan (x)-2 \sin ^2(x) \tan (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-6*x + 6*Tan[x] - 2*Sin[x]^2*Tan[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int 4 \sin ^2(x) \tan ^2(x) \, dx &=4 \int \sin ^2(x) \tan ^2(x) \, dx\\ &=4 \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=-2 \sin ^2(x) \tan (x)+6 \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (x)\right )\\ &=6 \tan (x)-2 \sin ^2(x) \tan (x)-6 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-6 x+6 \tan (x)-2 \sin ^2(x) \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.12 \[ 4 \left (-\frac {3 x}{2}+\frac {1}{4} \sin (2 x)+\tan (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*((-3*x)/2 + Sin[2*x]/4 + Tan[x])

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fricas [A]  time = 0.59, size = 22, normalized size = 1.38 \[ -\frac {2 \, {\left (3 \, x \cos \relax (x) - {\left (\cos \relax (x)^{2} + 2\right )} \sin \relax (x)\right )}}{\cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*x*cos(x) - (cos(x)^2 + 2)*sin(x))/cos(x)

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giac [A]  time = 0.14, size = 20, normalized size = 1.25 \[ -6 \, x + \frac {2 \, \tan \relax (x)}{\tan \relax (x)^{2} + 1} + 4 \, \tan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x + 2*tan(x)/(tan(x)^2 + 1) + 4*tan(x)

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maple [A]  time = 0.04, size = 28, normalized size = 1.75 \[ \frac {4 \left (\sin ^{5}\relax (x )\right )}{\cos \relax (x )}+4 \left (\sin ^{3}\relax (x )+\frac {3 \sin \relax (x )}{2}\right ) \cos \relax (x )-6 x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sin(x)^5/cos(x)+4*(sin(x)^3+3/2*sin(x))*cos(x)-6*x

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maxima [A]  time = 0.43, size = 20, normalized size = 1.25 \[ -6 \, x + \frac {2 \, \tan \relax (x)}{\tan \relax (x)^{2} + 1} + 4 \, \tan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x + 2*tan(x)/(tan(x)^2 + 1) + 4*tan(x)

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mupad [B]  time = 2.96, size = 18, normalized size = 1.12 \[ 2\,\cos \relax (x)\,\sin \relax (x)-6\,x+\frac {4\,\sin \relax (x)}{\cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*cos(x)*sin(x) - 6*x + (4*sin(x))/cos(x)

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sympy [A]  time = 0.05, size = 20, normalized size = 1.25 \[ - 6 x + \frac {4 \sin ^{3}{\relax (x )}}{\cos {\relax (x )}} + 6 \sin {\relax (x )} \cos {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x + 4*sin(x)**3/cos(x) + 6*sin(x)*cos(x)

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