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 = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ -\frac {1}{3 \left (a \sec ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.40, size = 35, normalized size = 2.50 \[ -\frac {\sqrt {a \tan \relax (x)^{2} + a}}{3 \, {\left (a^{2} \tan \relax (x)^{4} + 2 \, a^{2} \tan \relax (x)^{2} + a^{2}\right )}} \]

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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giac [A]  time = 0.26, size = 12, normalized size = 0.86 \[ -\frac {1}{3 \, {\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}}} \]

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)

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maple [A]  time = 0.12, size = 13, normalized size = 0.93 \[ -\frac {1}{3 \left (a +a \left (\tan ^{2}\relax (x )\right )\right )^{\frac {3}{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \relax (x)}{{\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [B]  time = 0.15, size = 23, normalized size = 1.64 \[ -\frac {\sqrt {a\,{\mathrm {tan}\relax (x)}^2+a}}{3\,a^2\,{\left ({\mathrm {tan}\relax (x)}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + a*tan(x)^2)^(3/2),x)

[Out]

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

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sympy [A]  time = 3.32, size = 15, normalized size = 1.07 \[ - \frac {1}{3 \left (a \tan ^{2}{\relax (x )} + a\right )^{\frac {3}{2}}} \]

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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