3.4 \(\int \sqrt{b \coth (c+d x)} \, dx\)

Optimal. Leaf size=58 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d} \]

[Out]

-((Sqrt[b]*ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d) + (Sqrt[b]*ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d

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Rubi [A]  time = 0.0356433, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3476, 329, 298, 203, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Coth[c + d*x]],x]

[Out]

-((Sqrt[b]*ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d) + (Sqrt[b]*ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d

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 298

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

Rule 203

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

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])

Rubi steps

\begin{align*} \int \sqrt{b \coth (c+d x)} \, dx &=-\frac{b \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-b^2+x^2} \, dx,x,b \coth (c+d x)\right )}{d}\\ &=-\frac{(2 b) \operatorname{Subst}\left (\int \frac{x^2}{-b^2+x^4} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}\\ &=-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0419602, size = 51, normalized size = 0.88 \[ \frac{\sqrt{b \coth (c+d x)} \left (\tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )-\tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )\right )}{d \sqrt{\coth (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Coth[c + d*x]],x]

[Out]

((-ArcTan[Sqrt[Coth[c + d*x]]] + ArcTanh[Sqrt[Coth[c + d*x]]])*Sqrt[b*Coth[c + d*x]])/(d*Sqrt[Coth[c + d*x]])

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Maple [A]  time = 0.023, size = 47, normalized size = 0.8 \begin{align*} -{\frac{1}{d}\arctan \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) \sqrt{b}}+{\frac{1}{d}{\it Artanh} \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c))^(1/2),x)

[Out]

-arctan((b*coth(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d+arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*coth(d*x + c)), x)

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Fricas [B]  time = 2.21435, size = 1636, normalized size = 28.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(-b)*sqrt(b*c
osh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) - s
qrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*
b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sin
h(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sin
h(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)))/d, -1/4*
(2*sqrt(b)*arctan(sqrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x
+ c) + b*sinh(d*x + c)^2 + b)) - sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*co
sh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 +
4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2
+ 2*(2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - b))/d]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(b*coth(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*coth(d*x + c)), x)