3.1.17 \(\int \frac {\tan (x)}{\sqrt {a+a \cot ^2(x)}} \, dx\) [17]

Optimal. Leaf size=36 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a \csc ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {1}{\sqrt {a \csc ^2(x)}} \]

[Out]

arctanh((a*csc(x)^2)^(1/2)/a^(1/2))/a^(1/2)-1/(a*csc(x)^2)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3738, 4209, 53, 65, 213} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a \csc ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {1}{\sqrt {a \csc ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + a*Cot[x]^2],x]

[Out]

ArcTanh[Sqrt[a*Csc[x]^2]/Sqrt[a]]/Sqrt[a] - 1/Sqrt[a*Csc[x]^2]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

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

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Mathematica [A]
time = 0.04, size = 49, normalized size = 1.36 \begin {gather*} -\frac {\csc (x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\sin (x)\right )}{\sqrt {a \csc ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Csc[x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]] + Sin[x]))/Sqrt[a*Csc[x]^2])

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Maple [A]
time = 0.57, size = 56, normalized size = 1.56

method	result	size
default	\(-\frac {\left (\sin \left (x \right )-\ln \left (-\frac {\cos \left (x \right )-1-\sin \left (x \right )}{\sin \left (x \right )}\right )+\ln \left (-\frac {\cos \left (x \right )-1+\sin \left (x \right )}{\sin \left (x \right )}\right )\right ) \sqrt {4}}{2 \sin \left (x \right ) \sqrt {-\frac {a}{\cos ^{2}\left (x \right )-1}}}\)	\(56\)
risch	\(-\frac {{\mathrm e}^{2 i x}}{2 \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}-1\right )}+\frac {1}{2 \left ({\mathrm e}^{2 i x}-1\right ) \sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}}-\frac {i {\mathrm e}^{i x} \ln \left ({\mathrm e}^{i x}-i\right )}{\sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}-1\right )}+\frac {i {\mathrm e}^{i x} \ln \left ({\mathrm e}^{i x}+i\right )}{\sqrt {-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}-1\right )}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sin(x)-ln(-(cos(x)-1-sin(x))/sin(x))+ln(-(cos(x)-1+sin(x))/sin(x)))/sin(x)/(-1/(cos(x)^2-1)*a)^(1/2)*4^(
1/2)

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Maxima [A]
time = 0.50, size = 52, normalized size = 1.44 \begin {gather*} -\frac {1}{2} \, a {\left (\frac {\log \left (-\frac {\sqrt {a} - \sqrt {\frac {a}{\sin \left (x\right )^{2}}}}{\sqrt {a} + \sqrt {\frac {a}{\sin \left (x\right )^{2}}}}\right )}{a^{\frac {3}{2}}} + \frac {2}{a \sqrt {\frac {a}{\sin \left (x\right )^{2}}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(log(-(sqrt(a) - sqrt(a/sin(x)^2))/(sqrt(a) + sqrt(a/sin(x)^2)))/a^(3/2) + 2/(a*sqrt(a/sin(x)^2)))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (28) = 56\).
time = 2.95, size = 78, normalized size = 2.17 \begin {gather*} \frac {{\left (\tan \left (x\right )^{2} + 1\right )} \sqrt {a} \log \left (2 \, a \tan \left (x\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {a \tan \left (x\right )^{2} + a}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2} + a\right ) - 2 \, \sqrt {\frac {a \tan \left (x\right )^{2} + a}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2}}{2 \, {\left (a \tan \left (x\right )^{2} + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((tan(x)^2 + 1)*sqrt(a)*log(2*a*tan(x)^2 + 2*sqrt(a)*sqrt((a*tan(x)^2 + a)/tan(x)^2)*tan(x)^2 + a) - 2*sqr
t((a*tan(x)^2 + a)/tan(x)^2)*tan(x)^2)/(a*tan(x)^2 + a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (x \right )}}{\sqrt {a \left (\cot ^{2}{\left (x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)/sqrt(a*(cot(x)**2 + 1)), x)

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Giac [A]
time = 0.46, size = 12, normalized size = 0.33 \begin {gather*} -\frac {\sin \left (x\right )}{\sqrt {a} \mathrm {sgn}\left (\sin \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(x)/(sqrt(a)*sgn(sin(x)))

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Mupad [B]
time = 0.42, size = 20, normalized size = 0.56 \begin {gather*} \frac {\mathrm {atanh}\left (\sqrt {\frac {1}{{\sin \left (x\right )}^2}}\right )-\sqrt {{\sin \left (x\right )}^2}}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + a*cot(x)^2)^(1/2),x)

[Out]

(atanh((1/sin(x)^2)^(1/2)) - (sin(x)^2)^(1/2))/a^(1/2)

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