\(\int \sqrt {c \cot (a+b x)} \, dx\) [12]

Optimal result

Integrand size = 12, antiderivative size = 192 \[ \int \sqrt {c \cot (a+b x)} \, dx=\frac {\sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {\sqrt {c} \arctan \left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b}+\frac {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \sqrt {c \cot (a+b x)} \, dx=\frac {\sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b}-\frac {\sqrt {c} \log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}+\frac {\sqrt {c} \log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b} \]

[In]

Int[Sqrt[c*Cot[a + b*x]],x]

[Out]

(Sqrt[c]*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)

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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*} \text {integral}& = -\frac {c \text {Subst}\left (\int \frac {\sqrt {x}}{c^2+x^2} \, dx,x,c \cot (a+b x)\right )}{b} \\ & = -\frac {(2 c) \text {Subst}\left (\int \frac {x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b} \\ & = \frac {c \text {Subst}\left (\int \frac {c-x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b}-\frac {c \text {Subst}\left (\int \frac {c+x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b} \\ & = -\frac {\sqrt {c} \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}-\frac {\sqrt {c} \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}-\frac {c \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 {c \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 {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b}+\frac {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b}-\frac {\sqrt {c} \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}+\frac {\sqrt {c} \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} \\ & = \frac {\sqrt {c} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {\sqrt {c} \arctan \left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b}+\frac {\sqrt {c} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.37 \[ \int \sqrt {c \cot (a+b x)} \, dx=\frac {\left (-\arctan \left (\sqrt [4]{-\cot ^2(a+b x)}\right )+\text {arctanh}\left (\sqrt [4]{-\cot ^2(a+b x)}\right )\right ) \sqrt [4]{-\cot (a+b x)} \sqrt {c \cot (a+b x)}}{b \cot ^{\frac {3}{4}}(a+b x)} \]

[In]

Integrate[Sqrt[c*Cot[a + b*x]],x]

[Out]

((-ArcTan[(-Cot[a + b*x]^2)^(1/4)] + ArcTanh[(-Cot[a + b*x]^2)^(1/4)])*(-Cot[a + b*x])^(1/4)*Sqrt[c*Cot[a + b*
x]])/(b*Cot[a + b*x]^(3/4))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {c \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 b \left (c^{2}\right )^{\frac {1}{4}}}\)	\(136\)
default	\(-\frac {c \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 b \left (c^{2}\right )^{\frac {1}{4}}}\)	\(136\)

[In]

int((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))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.21 \[ \int \sqrt {c \cot (a+b x)} \, dx=-\frac {1}{2} \, \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {3}{4}} + c \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}}\right ) + \frac {1}{2} i \, \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {1}{4}} \log \left (i \, b^{3} \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {3}{4}} + c \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}}\right ) - \frac {1}{2} i \, \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-i \, b^{3} \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {3}{4}} + c \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}}\right ) + \frac {1}{2} \, \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {c^{2}}{b^{4}}\right )^{\frac {3}{4}} + c \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}}\right ) \]

[In]

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

[Out]

-1/2*(-c^2/b^4)^(1/4)*log(b^3*(-c^2/b^4)^(3/4) + c*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))) + 1/2*I*(-
c^2/b^4)^(1/4)*log(I*b^3*(-c^2/b^4)^(3/4) + c*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))) - 1/2*I*(-c^2/b
^4)^(1/4)*log(-I*b^3*(-c^2/b^4)^(3/4) + c*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))) + 1/2*(-c^2/b^4)^(1
/4)*log(-b^3*(-c^2/b^4)^(3/4) + c*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a)))

Sympy [F]

\[ \int \sqrt {c \cot (a+b x)} \, dx=\int \sqrt {c \cot {\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(c*cot(a + b*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.86 \[ \int \sqrt {c \cot (a+b x)} \, dx=-\frac {c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} + 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} - 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{\sqrt {c}}\right )}}{4 \, b} \]

[In]

integrate((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))/sqrt(c) + 2*sqrt(2)*a
rctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(c) - 2*sqrt(c/tan(b*x + a)))/sqrt(c))/sqrt(c) - sqrt(2)*log(sqrt(2)*sqrt(c)*s
qrt(c/tan(b*x + a)) + c + c/tan(b*x + a))/sqrt(c) + sqrt(2)*log(-sqrt(2)*sqrt(c)*sqrt(c/tan(b*x + a)) + c + c/
tan(b*x + a))/sqrt(c))/b

Giac [F]

\[ \int \sqrt {c \cot (a+b x)} \, dx=\int { \sqrt {c \cot \left (b x + a\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(c*cot(b*x + a)), x)

Mupad [B] (verification not implemented)

Time = 12.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.26 \[ \int \sqrt {c \cot (a+b x)} \, dx=-\frac {{\left (-1\right )}^{1/4}\,\sqrt {c}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )\right )}{b} \]

[In]

int((c*cot(a + b*x))^(1/2),x)

[Out]

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