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

Optimal result

Integrand size = 15, antiderivative size = 12 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {1}{\sqrt {a \sec ^2(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3738, 4209, 32} \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {1}{\sqrt {a \sec ^2(x)}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {1}{\sqrt {a \sec ^2(x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {1}{\sqrt {a \tan \left (x\right )^{2} + a}} \]

[In]

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

[Out]

-1/sqrt(a*tan(x)^2 + a)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=- \frac {1}{\sqrt {a \tan ^{2}{\left (x \right )} + a}} \]

[In]

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

[Out]

-1/sqrt(a*tan(x)**2 + a)

Maxima [F]

\[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=\int { \frac {\tan \left (x\right )}{\sqrt {a \tan \left (x\right )^{2} + a}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {1}{\sqrt {a \tan \left (x\right )^{2} + a}} \]

[In]

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

[Out]

-1/sqrt(a*tan(x)^2 + a)

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {\tan (x)}{\sqrt {a+a \tan ^2(x)}} \, dx=-\frac {\sqrt {{\cos \left (x\right )}^2}}{\sqrt {a}} \]

[In]

int(tan(x)/(a + a*tan(x)^2)^(1/2),x)

[Out]

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