3.851 \(\int \frac {\cot (c+d x)}{\sqrt {a \cos ^2(c+d x)}} \, dx\)

Optimal. Leaf size=31 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

-arctanh((a*cos(d*x+c)^2)^(1/2)/a^(1/2))/d/a^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3205, 63, 206} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/Sqrt[a*Cos[c + d*x]^2],x]

[Out]

-(ArcTanh[Sqrt[a*Cos[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d))

Rule 63

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 206

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

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rubi steps

\begin {align*} \int \frac {\cot (c+d x)}{\sqrt {a \cos ^2(c+d x)}} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a x}} \, dx,x,\cos ^2(c+d x)\right )}{2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a \cos ^2(c+d x)}\right )}{a d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 1.58 \[ \frac {\cos (c+d x) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d \sqrt {a \cos ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/Sqrt[a*Cos[c + d*x]^2],x]

[Out]

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

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fricas [A]  time = 1.21, size = 84, normalized size = 2.71 \[ \left [-\frac {\sqrt {a \cos \left (d x + c\right )^{2}} \log \left (-\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1}\right )}{2 \, a d \cos \left (d x + c\right )}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right )^{2}} \sqrt {-a}}{a}\right )}{a d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(a*cos(d*x + c)^2)*log(-(cos(d*x + c) + 1)/(cos(d*x + c) - 1))/(a*d*cos(d*x + c)), sqrt(-a)*arctan(s
qrt(a*cos(d*x + c)^2)*sqrt(-a)/a)/(a*d)]

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giac [A]  time = 0.25, size = 31, normalized size = 1.00 \[ \frac {\arctan \left (\frac {\sqrt {-a \sin \left (d x + c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(-a*sin(d*x + c)^2 + a)/sqrt(-a))/(sqrt(-a)*d)

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maple [A]  time = 0.16, size = 31, normalized size = 1.00 \[ -\frac {\cos \left (d x +c \right ) \arctanh \left (\cos \left (d x +c \right )\right )}{\sqrt {a \left (\cos ^{2}\left (d x +c \right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a*cos(d*x+c)^2)^(1/2),x)

[Out]

-1/(a*cos(d*x+c)^2)^(1/2)*cos(d*x+c)*arctanh(cos(d*x+c))/d

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maxima [B]  time = 0.32, size = 51, normalized size = 1.65 \[ -\frac {\log \left (\frac {2 \, \sqrt {-a \sin \left (d x + c\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (d x + c\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (d x + c\right ) \right |}}\right )}{\sqrt {a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(2*sqrt(-a*sin(d*x + c)^2 + a)*sqrt(a)/abs(sin(d*x + c)) + 2*a/abs(sin(d*x + c)))/(sqrt(a)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {cot}\left (c+d\,x\right )}{\sqrt {a\,{\cos \left (c+d\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)/(a*cos(c + d*x)^2)^(1/2),x)

[Out]

int(cot(c + d*x)/(a*cos(c + d*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (c + d x \right )}}{\sqrt {a \cos ^{2}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a*cos(d*x+c)**2)**(1/2),x)

[Out]

Integral(cot(c + d*x)/sqrt(a*cos(c + d*x)**2), x)

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