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

Optimal. Leaf size=61 \[ \frac{b \tanh (c+d x) \sqrt{b \coth ^2(c+d x)} \log (\sinh (c+d x))}{d}-\frac{b \coth (c+d x) \sqrt{b \coth ^2(c+d x)}}{2 d} \]

[Out]

-(b*Coth[c + d*x]*Sqrt[b*Coth[c + d*x]^2])/(2*d) + (b*Sqrt[b*Coth[c + d*x]^2]*Log[Sinh[c + d*x]]*Tanh[c + d*x]
)/d

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Rubi [A]  time = 0.0368725, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ \frac{b \tanh (c+d x) \sqrt{b \coth ^2(c+d x)} \log (\sinh (c+d x))}{d}-\frac{b \coth (c+d x) \sqrt{b \coth ^2(c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x]^2)^(3/2),x]

[Out]

-(b*Coth[c + d*x]*Sqrt[b*Coth[c + d*x]^2])/(2*d) + (b*Sqrt[b*Coth[c + d*x]^2]*Log[Sinh[c + d*x]]*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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.118465, size = 56, normalized size = 0.92 \[ -\frac{\tanh ^3(c+d x) \left (b \coth ^2(c+d x)\right )^{3/2} \left (\coth ^2(c+d x)-2 \log (\tanh (c+d x))-2 \log (\cosh (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x]^2)^(3/2),x]

[Out]

-((b*Coth[c + d*x]^2)^(3/2)*(Coth[c + d*x]^2 - 2*Log[Cosh[c + d*x]] - 2*Log[Tanh[c + d*x]])*Tanh[c + d*x]^3)/(
2*d)

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Maple [A]  time = 0.026, size = 53, normalized size = 0.9 \begin{align*} -{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}+\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) +\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) }{2\,d \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}} \left ( b \left ({\rm coth} \left (dx+c\right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.60448, size = 131, normalized size = 2.15 \begin{align*} -\frac{{\left (d x + c\right )} b^{\frac{3}{2}}}{d} - \frac{b^{\frac{3}{2}} \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{b^{\frac{3}{2}} \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac{2 \, b^{\frac{3}{2}} e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x + c)*b^(3/2)/d - b^(3/2)*log(e^(-d*x - c) + 1)/d - b^(3/2)*log(e^(-d*x - c) - 1)/d - 2*b^(3/2)*e^(-2*d*x
- 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) - 1))

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Fricas [B]  time = 1.98088, size = 2068, normalized size = 33.9 \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)^2)^(3/2),x, algorithm="fricas")

[Out]

(b*d*x*cosh(d*x + c)^4 - (b*d*x*e^(2*d*x + 2*c) - b*d*x)*sinh(d*x + c)^4 - 4*(b*d*x*cosh(d*x + c)*e^(2*d*x + 2
*c) - b*d*x*cosh(d*x + c))*sinh(d*x + c)^3 + b*d*x - 2*(b*d*x - b)*cosh(d*x + c)^2 + 2*(3*b*d*x*cosh(d*x + c)^
2 - b*d*x - (3*b*d*x*cosh(d*x + c)^2 - b*d*x + b)*e^(2*d*x + 2*c) + b)*sinh(d*x + c)^2 - (b*d*x*cosh(d*x + c)^
4 + b*d*x - 2*(b*d*x - b)*cosh(d*x + c)^2)*e^(2*d*x + 2*c) - (b*cosh(d*x + c)^4 - (b*e^(2*d*x + 2*c) - b)*sinh
(d*x + c)^4 - 4*(b*cosh(d*x + c)*e^(2*d*x + 2*c) - b*cosh(d*x + c))*sinh(d*x + c)^3 - 2*b*cosh(d*x + c)^2 + 2*
(3*b*cosh(d*x + c)^2 - (3*b*cosh(d*x + c)^2 - b)*e^(2*d*x + 2*c) - b)*sinh(d*x + c)^2 - (b*cosh(d*x + c)^4 - 2
*b*cosh(d*x + c)^2 + b)*e^(2*d*x + 2*c) + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c) - (b*cosh(d*x + c)^3 - b*cosh
(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) + b)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(b*d*x
*cosh(d*x + c)^3 - (b*d*x - b)*cosh(d*x + c) - (b*d*x*cosh(d*x + c)^3 - (b*d*x - b)*cosh(d*x + c))*e^(2*d*x +
2*c))*sinh(d*x + c))*sqrt((b*e^(4*d*x + 4*c) + 2*b*e^(2*d*x + 2*c) + b)/(e^(4*d*x + 4*c) - 2*e^(2*d*x + 2*c) +
1))/(d*cosh(d*x + c)^4 + (d*e^(2*d*x + 2*c) + d)*sinh(d*x + c)^4 + 4*(d*cosh(d*x + c)*e^(2*d*x + 2*c) + d*cos
h(d*x + c))*sinh(d*x + c)^3 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 + (3*d*cosh(d*x + c)^2 - d)*e^(2*d*
x + 2*c) - d)*sinh(d*x + c)^2 + (d*cosh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + d)*e^(2*d*x + 2*c) + 4*(d*cosh(d*x
+ c)^3 - d*cosh(d*x + c) + (d*cosh(d*x + c)^3 - d*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) + d)

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

[Out]

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

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Giac [A]  time = 1.20253, size = 122, normalized size = 2. \begin{align*} -\frac{{\left ({\left (d x + c\right )} \mathrm{sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) - \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) \mathrm{sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) + \frac{2 \, e^{\left (2 \, d x + 2 \, c\right )} \mathrm{sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}\right )} b^{\frac{3}{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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