3.280 \(\int \frac{\tan (x)}{(a+a \tan ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=14 \[ -\frac{1}{3 \left (a \sec ^2(x)\right )^{3/2}} \]

[Out]

-1/(3*(a*Sec[x]^2)^(3/2))

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Rubi [A]  time = 0.0489939, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3657, 4124, 32} \[ -\frac{1}{3 \left (a \sec ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a + a*Tan[x]^2)^(3/2),x]

[Out]

-1/(3*(a*Sec[x]^2)^(3/2))

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4124

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Dist[b/(2*f), Subst[In
t[(-1 + x)^((m - 1)/2)*(b*x)^(p - 1), x], x, Sec[e + f*x]^2], x] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p] &&
IntegerQ[(m - 1)/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0117573, size = 14, normalized size = 1. \[ -\frac{1}{3 \left (a \sec ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(a + a*Tan[x]^2)^(3/2),x]

[Out]

-1/(3*(a*Sec[x]^2)^(3/2))

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Maple [A]  time = 0.011, size = 13, normalized size = 0.9 \begin{align*} -{\frac{1}{3} \left ( a+a \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a+a*tan(x)^2)^(3/2),x)

[Out]

-1/3/(a+a*tan(x)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )}{{\left (a \tan \left (x\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(x)/(a*tan(x)^2 + a)^(3/2), x)

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Fricas [B]  time = 1.35137, size = 88, normalized size = 6.29 \begin{align*} -\frac{\sqrt{a \tan \left (x\right )^{2} + a}}{3 \,{\left (a^{2} \tan \left (x\right )^{4} + 2 \, a^{2} \tan \left (x\right )^{2} + a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/3*sqrt(a*tan(x)^2 + a)/(a^2*tan(x)^4 + 2*a^2*tan(x)^2 + a^2)

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Sympy [A]  time = 2.25182, size = 15, normalized size = 1.07 \begin{align*} - \frac{1}{3 \left (a \tan ^{2}{\left (x \right )} + a\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)**2)**(3/2),x)

[Out]

-1/(3*(a*tan(x)**2 + a)**(3/2))

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Giac [A]  time = 1.08089, size = 16, normalized size = 1.14 \begin{align*} -\frac{1}{3 \,{\left (a \tan \left (x\right )^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/3/(a*tan(x)^2 + a)^(3/2)