3.1.13 \(\int \frac {1}{\sqrt {c \cot (a+b x)}} \, dx\) [13]

Optimal. Leaf size=192 \[ \frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}-\frac {\text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}-\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}} \]

[Out]

1/2*arctan(1-2^(1/2)*(c*cot(b*x+a))^(1/2)/c^(1/2))/b*2^(1/2)/c^(1/2)-1/2*arctan(1+2^(1/2)*(c*cot(b*x+a))^(1/2)
/c^(1/2))/b*2^(1/2)/c^(1/2)+1/4*ln(c^(1/2)+cot(b*x+a)*c^(1/2)-2^(1/2)*(c*cot(b*x+a))^(1/2))/b*2^(1/2)/c^(1/2)-
1/4*ln(c^(1/2)+cot(b*x+a)*c^(1/2)+2^(1/2)*(c*cot(b*x+a))^(1/2))/b*2^(1/2)/c^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}-\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b \sqrt {c}}+\frac {\log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b \sqrt {c}}-\frac {\log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c*Cot[a + b*x]],x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[c*Cot[a + b*x]])/Sqrt[c]]/(Sqrt[2]*b*Sqrt[c]) - ArcTan[1 + (Sqrt[2]*Sqrt[c*Cot[a + b*
x]])/Sqrt[c]]/(Sqrt[2]*b*Sqrt[c]) + Log[Sqrt[c] + Sqrt[c]*Cot[a + b*x] - Sqrt[2]*Sqrt[c*Cot[a + b*x]]]/(2*Sqrt
[2]*b*Sqrt[c]) - Log[Sqrt[c] + Sqrt[c]*Cot[a + b*x] + Sqrt[2]*Sqrt[c*Cot[a + b*x]]]/(2*Sqrt[2]*b*Sqrt[c])

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {c \cot (a+b x)}} \, dx &=-\frac {c \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (c^2+x^2\right )} \, dx,x,c \cot (a+b x)\right )}{b}\\ &=-\frac {(2 c) \text {Subst}\left (\int \frac {1}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {c-x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b}-\frac {\text {Subst}\left (\int \frac {c+x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {1}{c-\sqrt {2} \sqrt {c} x+x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 b}-\frac {\text {Subst}\left (\int \frac {1}{c+\sqrt {2} \sqrt {c} x+x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {c}+2 x}{-c-\sqrt {2} \sqrt {c} x-x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {c}-2 x}{-c+\sqrt {2} \sqrt {c} x-x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}\\ &=\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}-\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}\\ &=\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}-\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}-\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 131, normalized size = 0.68 \begin {gather*} \frac {\sqrt {\cot (a+b x)} \left (2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (a+b x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (a+b x)}\right )+\log \left (1-\sqrt {2} \sqrt {\cot (a+b x)}+\cot (a+b x)\right )-\log \left (1+\sqrt {2} \sqrt {\cot (a+b x)}+\cot (a+b x)\right )\right )}{2 \sqrt {2} b \sqrt {c \cot (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c*Cot[a + b*x]],x]

[Out]

(Sqrt[Cot[a + b*x]]*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[a + b*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Cot[a + b*x]]] + Log
[1 - Sqrt[2]*Sqrt[Cot[a + b*x]] + Cot[a + b*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[a + b*x]] + Cot[a + b*x]]))/(2*Sqrt
[2]*b*Sqrt[c*Cot[a + b*x]])

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Maple [A]
time = 0.29, size = 138, normalized size = 0.72

method	result	size
derivativedivides	\(-\frac {\left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 b c}\)	\(138\)
default	\(-\frac {\left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 b c}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b/c*(c^2)^(1/4)*2^(1/2)*(ln((c*cot(b*x+a)+(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)*2^(1/2)+(c^2)^(1/2))/(c*cot(b*
x+a)-(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)*2^(1/2)+(c^2)^(1/2)))+2*arctan(2^(1/2)/(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)+
1)-2*arctan(-2^(1/2)/(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)+1))

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Maxima [A]
time = 0.51, size = 165, normalized size = 0.86 \begin {gather*} -\frac {c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} + 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} - 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{c^{\frac {3}{2}}}\right )}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(c) + 2*sqrt(c/tan(b*x + a)))/sqrt(c))/c^(3/2) + 2*sqrt(2)*a
rctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(c) - 2*sqrt(c/tan(b*x + a)))/sqrt(c))/c^(3/2) + sqrt(2)*log(sqrt(2)*sqrt(c)*s
qrt(c/tan(b*x + a)) + c + c/tan(b*x + a))/c^(3/2) - sqrt(2)*log(-sqrt(2)*sqrt(c)*sqrt(c/tan(b*x + a)) + c + c/
tan(b*x + a))/c^(3/2))/b

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c*cot(a + b*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(c*cot(b*x + a)), x)

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Mupad [B]
time = 0.34, size = 57, normalized size = 0.30 \begin {gather*} \frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{b\,\sqrt {c}}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{b\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cot(a + b*x))^(1/2),x)

[Out]

((-1)^(1/4)*atan(((-1)^(1/4)*(c*cot(a + b*x))^(1/2))/c^(1/2))*1i)/(b*c^(1/2)) + ((-1)^(1/4)*atanh(((-1)^(1/4)*
(c*cot(a + b*x))^(1/2))/c^(1/2))*1i)/(b*c^(1/2))

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