3.12 \(\int \sqrt{c \cot (a+b x)} \, dx\)

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

[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)

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Rubi [A]  time = 0.11515, antiderivative size = 192, normalized size of antiderivative = 1., 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 = {3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\sqrt{c} \log \left (\sqrt{c} \cot (a+b x)-\sqrt{2} \sqrt{c \cot (a+b x)}+\sqrt{c}\right )}{2 \sqrt{2} b}+\frac{\sqrt{c} \log \left (\sqrt{c} \cot (a+b x)+\sqrt{2} \sqrt{c \cot (a+b x)}+\sqrt{c}\right )}{2 \sqrt{2} b}+\frac{\sqrt{c} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{c \cot (a+b x)}}{\sqrt{c}}\right )}{\sqrt{2} b}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c \cot (a+b x)}}{\sqrt{c}}+1\right )}{\sqrt{2} b} \]

Antiderivative was successfully verified.

[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 3476

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]

Rule 329

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 297

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int \sqrt{c \cot (a+b x)} \, dx &=-\frac{c \operatorname{Subst}\left (\int \frac{\sqrt{x}}{c^2+x^2} \, dx,x,c \cot (a+b x)\right )}{b}\\ &=-\frac{(2 c) \operatorname{Subst}\left (\int \frac{x^2}{c^2+x^4} \, dx,x,\sqrt{c \cot (a+b x)}\right )}{b}\\ &=\frac{c \operatorname{Subst}\left (\int \frac{c-x^2}{c^2+x^4} \, dx,x,\sqrt{c \cot (a+b x)}\right )}{b}-\frac{c \operatorname{Subst}\left (\int \frac{c+x^2}{c^2+x^4} \, dx,x,\sqrt{c \cot (a+b x)}\right )}{b}\\ &=-\frac{\sqrt{c} \operatorname{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} \operatorname{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 \operatorname{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 \operatorname{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} \operatorname{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} \operatorname{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} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{c \cot (a+b x)}}{\sqrt{c}}\right )}{\sqrt{2} b}-\frac{\sqrt{c} \tan ^{-1}\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 [C]  time = 0.0391497, size = 40, normalized size = 0.21 \[ -\frac{2 (c \cot (a+b x))^{3/2} \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\cot ^2(a+b x)\right )}{3 b c} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*Cot[a + b*x]],x]

[Out]

(-2*(c*Cot[a + b*x])^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[a + b*x]^2])/(3*b*c)

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Maple [A]  time = 0.053, size = 160, normalized size = 0.8 \begin{align*} -{\frac{c\sqrt{2}}{4\,b}\ln \left ({ \left ( c\cot \left ( bx+a \right ) -\sqrt [4]{{c}^{2}}\sqrt{c\cot \left ( bx+a \right ) }\sqrt{2}+\sqrt{{c}^{2}} \right ) \left ( c\cot \left ( bx+a \right ) +\sqrt [4]{{c}^{2}}\sqrt{c\cot \left ( bx+a \right ) }\sqrt{2}+\sqrt{{c}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{c}^{2}}}}}-{\frac{c\sqrt{2}}{2\,b}\arctan \left ({\sqrt{2}\sqrt{c\cot \left ( bx+a \right ) }{\frac{1}{\sqrt [4]{{c}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{c}^{2}}}}}+{\frac{c\sqrt{2}}{2\,b}\arctan \left ( -{\sqrt{2}\sqrt{c\cot \left ( bx+a \right ) }{\frac{1}{\sqrt [4]{{c}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{c}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c*cot(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c \cot \left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*cot(b*x + a)), x)