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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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])

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 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 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 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
]])

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*}

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Mathematica [A]  time = 0.10, 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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fricas [B]  time = 1.59, size = 633, normalized size = 6.09 \[ \left [-\frac {2 \, \sqrt {-b} \arctan \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) - \sqrt {-b} \log \left (-\frac {b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - 2 \, b}{\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4}}\right ) + 8 \, \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, d}, \frac {2 \, \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) + \sqrt {b} \log \left (2 \, b \cosh \left (d x + c\right )^{4} + 8 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 12 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 8 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b \sinh \left (d x + c\right )^{4} + 2 \, {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + {\left (6 \, \cosh \left (d x + c\right )^{2} - 1\right )} \sinh \left (d x + c\right )^{2} - \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cosh \left (d x + c\right )^{3} - \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - b\right ) - 8 \, \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, d}\right ] \]

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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giac [B]  time = 0.29, size = 269, normalized size = 2.59 \[ -\frac {2 \, \sqrt {b} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right ) \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) + \sqrt {b} \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right ) \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) - \frac {8 \, b \mathrm {sgn}\left (e^{\left (6 \, d x + 6 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}{\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(b)*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))*sgn(e^(6*d*x + 6*c) -
3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*d*x + 4*c) - 1) + sqrt(b)*log(abs(-sqrt(b)*e^(2*d*x + 2*c
) + sqrt(b*e^(4*d*x + 4*c) - b)))*sgn(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*d*
x + 4*c) - 1) - 8*b*sgn(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1)*sgn(e^(4*d*x + 4*c) - 1)/
(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) - sqrt(b)))/d

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maple [A]  time = 0.17, size = 89, normalized size = 0.86 \[ -\frac {\sqrt {b \left (\coth ^{3}\left (d x +c \right )\right )}\, \left (2 \sqrt {b \coth \left (d x +c \right )}-\sqrt {b}\, \arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )-\sqrt {b}\, \arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )\right )}{d \coth \left (d x +c \right ) \sqrt {b \coth \left (d x +c \right )}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \coth \left (d x + c\right )^{3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x)^3)^(1/2),x)

[Out]

int((b*coth(c + d*x)^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \coth ^{3}{\left (c + d x \right )}}\, dx \]

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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