3.3 \(\int (b \coth (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=75 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b \sqrt{b \coth (c+d x)}}{d} \]

[Out]

(b^(3/2)*ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d + (b^(3/2)*ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d - (2*b*
Sqrt[b*Coth[c + d*x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.0500011, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3473, 3476, 329, 212, 206, 203} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b \sqrt{b \coth (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(3/2),x]

[Out]

(b^(3/2)*ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d + (b^(3/2)*ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d - (2*b*
Sqrt[b*Coth[c + d*x]])/d

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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 212

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

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])

Rubi steps

\begin{align*} \int (b \coth (c+d x))^{3/2} \, dx &=-\frac{2 b \sqrt{b \coth (c+d x)}}{d}+b^2 \int \frac{1}{\sqrt{b \coth (c+d x)}} \, dx\\ &=-\frac{2 b \sqrt{b \coth (c+d x)}}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-b^2+x^2\right )} \, dx,x,b \coth (c+d x)\right )}{d}\\ &=-\frac{2 b \sqrt{b \coth (c+d x)}}{d}-\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-b^2+x^4} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}\\ &=-\frac{2 b \sqrt{b \coth (c+d x)}}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \coth (c+d x)}\right )}{d}\\ &=\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b \coth (c+d x)}}{\sqrt{b}}\right )}{d}-\frac{2 b \sqrt{b \coth (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.0947051, size = 61, normalized size = 0.81 \[ \frac{(b \coth (c+d x))^{3/2} \left (-2 \sqrt{\coth (c+d x)}+\tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )+\tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )\right )}{d \coth ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(3/2),x]

[Out]

((ArcTan[Sqrt[Coth[c + d*x]]] + ArcTanh[Sqrt[Coth[c + d*x]]] - 2*Sqrt[Coth[c + d*x]])*(b*Coth[c + d*x])^(3/2))
/(d*Coth[c + d*x]^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 62, normalized size = 0.8 \begin{align*}{\frac{1}{d}{b}^{{\frac{3}{2}}}\arctan \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) }+{\frac{1}{d}{b}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) }-2\,{\frac{b\sqrt{b{\rm coth} \left (dx+c\right )}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c))^(3/2),x)

[Out]

b^(3/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/d+b^(3/2)*arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))/d-2*b*(b*coth(d
*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*coth(d*x + c))^(3/2), x)

________________________________________________________________________________________

Fricas [B]  time = 2.27195, size = 1754, normalized size = 23.39 \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))^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(2*sqrt(-b)*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
*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)*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) +
sinh(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
*sinh(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)) + 8*b
*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/d, 1/4*(2*b^(3/2)*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)) + b^(3/2)*log(2*b*cosh(d*x + c)^4
+ 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(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)*s
qrt(b*cosh(d*x + c)/sinh(d*x + c)) - b) - 8*b*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/d]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth{\left (c + d x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))**(3/2),x)

[Out]

Integral((b*coth(c + d*x))**(3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*coth(d*x + c))^(3/2), x)