3.30 \(\int \sqrt{b \coth ^3(c+d x)} \, dx\)

Optimal. Leaf size=104 \[ \frac{\sqrt{b \coth ^3(c+d x)} \tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \coth ^{\frac{3}{2}}(c+d x)}+\frac{\sqrt{b \coth ^3(c+d x)} \tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \coth ^{\frac{3}{2}}(c+d x)}-\frac{2 \tanh (c+d x) \sqrt{b \coth ^3(c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.0474185, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3658, 3473, 3476, 329, 212, 206, 203} \[ \frac{\sqrt{b \coth ^3(c+d x)} \tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \coth ^{\frac{3}{2}}(c+d x)}+\frac{\sqrt{b \coth ^3(c+d x)} \tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )}{d \coth ^{\frac{3}{2}}(c+d x)}-\frac{2 \tanh (c+d x) \sqrt{b \coth ^3(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Coth[c + d*x]^3],x]

[Out]

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

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

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

Mathematica [A]  time = 0.0962737, size = 63, normalized size = 0.61 \[ \frac{\sqrt{b \coth ^3(c+d x)} \left (-2 \sqrt{\coth (c+d x)}+\tan ^{-1}\left (\sqrt{\coth (c+d x)}\right )+\tanh ^{-1}\left (\sqrt{\coth (c+d x)}\right )\right )}{d \coth ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Coth[c + d*x]^3],x]

[Out]

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

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Maple [A]  time = 0.045, size = 86, normalized size = 0.8 \begin{align*}{\frac{1}{d{\rm coth} \left (dx+c\right )}\sqrt{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}} \left ( -2\,\sqrt{b{\rm coth} \left (dx+c\right )}+\sqrt{b}{\it Artanh} \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) +\sqrt{b}\arctan \left ({\sqrt{b{\rm coth} \left (dx+c\right )}{\frac{1}{\sqrt{b}}}} \right ) \right ){\frac{1}{\sqrt{b{\rm coth} \left (dx+c\right )}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c)^3)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*coth(d*x + c)^3), x)

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Fricas [B]  time = 2.18796, size = 1743, normalized size = 16.76 \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)^(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)) + 8*sqrt(
b*cosh(d*x + c)/sinh(d*x + c)))/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*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)*sqrt(b*
cosh(d*x + c)/sinh(d*x + c)) - b) - 8*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/d]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \coth ^{3}{\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)**3)**(1/2),x)

[Out]

Integral(sqrt(b*coth(c + d*x)**3), 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 )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c)^3)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*coth(d*x + c)^3), x)