### 3.114 $$\int \sin ^2(x) \tan (x) \, dx$$

Optimal. Leaf size=14 $\frac{\cos ^2(x)}{2}-\log (\cos (x))$

[Out]

Cos[x]^2/2 - Log[Cos[x]]

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Rubi [A]  time = 0.0150215, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {2590, 14} $\frac{\cos ^2(x)}{2}-\log (\cos (x))$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Cos[x]^2/2 - Log[Cos[x]]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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

Mathematica [A]  time = 0.0052021, size = 14, normalized size = 1. $\frac{\cos ^2(x)}{2}-\log (\cos (x))$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Cos[x]^2/2 - Log[Cos[x]]

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Maple [A]  time = 0.011, size = 13, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \sin \left ( x \right ) \right ) ^{2}}{2}}-\ln \left ( \cos \left ( x \right ) \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sin(x)^2-ln(cos(x))

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Maxima [A]  time = 0.92659, size = 22, normalized size = 1.57 \begin{align*} -\frac{1}{2} \, \sin \left (x\right )^{2} - \frac{1}{2} \, \log \left (\sin \left (x\right )^{2} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sin(x)^2 - 1/2*log(sin(x)^2 - 1)

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Fricas [A]  time = 2.16732, size = 39, normalized size = 2.79 \begin{align*} \frac{1}{2} \, \cos \left (x\right )^{2} - \log \left (-\cos \left (x\right )\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cos(x)^2 - log(-cos(x))

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Sympy [A]  time = 0.076593, size = 10, normalized size = 0.71 \begin{align*} - \log{\left (\cos{\left (x \right )} \right )} + \frac{\cos ^{2}{\left (x \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(cos(x)) + cos(x)**2/2

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Giac [A]  time = 1.08055, size = 24, normalized size = 1.71 \begin{align*} -\frac{1}{2} \, \sin \left (x\right )^{2} - \frac{1}{2} \, \log \left (-\sin \left (x\right )^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sin(x)^2 - 1/2*log(-sin(x)^2 + 1)