3.3.75 \(\int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx\) [275]

Optimal. Leaf size=12 \[ -\frac {1}{\sqrt {a \sec ^2(x)}} \]

[Out]

-1/(a*sec(x)^2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 12, 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 = {3738, 4209, 32} \begin {gather*} -\frac {1}{\sqrt {a \sec ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

-(1/Sqrt[a*Sec[x]^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 3738

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 4209

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

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{\sqrt {a \sec ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/Sqrt[a + a*Tan[x]^2],x]

[Out]

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

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Maple [A]
time = 0.04, size = 13, normalized size = 1.08

method	result	size
derivativedivides	\(-\frac {1}{\sqrt {a +a \left (\tan ^{2}\left (x \right )\right )}}\)	\(13\)
default	\(-\frac {1}{\sqrt {a +a \left (\tan ^{2}\left (x \right )\right )}}\)	\(13\)
risch	\(-\frac {{\mathrm e}^{2 i x}}{2 \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}+1\right )}-\frac {1}{2 \left ({\mathrm e}^{2 i x}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a+a*tan(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.62, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{\sqrt {a \tan \left (x\right )^{2} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/sqrt(a*tan(x)^2 + a)

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Sympy [A]
time = 0.33, size = 14, normalized size = 1.17 \begin {gather*} - \frac {1}{\sqrt {a \tan ^{2}{\left (x \right )} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)**2)**(1/2),x)

[Out]

-1/sqrt(a*tan(x)**2 + a)

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Giac [A]
time = 0.40, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{\sqrt {a \tan \left (x\right )^{2} + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(a+a*tan(x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/sqrt(a*tan(x)^2 + a)

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Mupad [B]
time = 11.80, size = 11, normalized size = 0.92 \begin {gather*} -\frac {\sqrt {{\cos \left (x\right )}^2}}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + a*tan(x)^2)^(1/2),x)

[Out]

-(cos(x)^2)^(1/2)/a^(1/2)

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