3.41 \(\int \sqrt{b \coth ^4(c+d x)} \, dx\)

Optimal. Leaf size=50 \[ x \tanh ^2(c+d x) \sqrt{b \coth ^4(c+d x)}-\frac{\tanh (c+d x) \sqrt{b \coth ^4(c+d x)}}{d} \]

[Out]

-((Sqrt[b*Coth[c + d*x]^4]*Tanh[c + d*x])/d) + x*Sqrt[b*Coth[c + d*x]^4]*Tanh[c + d*x]^2

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Rubi [A]  time = 0.0228098, antiderivative size = 50, 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, 8} \[ x \tanh ^2(c+d x) \sqrt{b \coth ^4(c+d x)}-\frac{\tanh (c+d x) \sqrt{b \coth ^4(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Coth[c + d*x]^4],x]

[Out]

-((Sqrt[b*Coth[c + d*x]^4]*Tanh[c + d*x])/d) + x*Sqrt[b*Coth[c + d*x]^4]*Tanh[c + d*x]^2

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.0306745, size = 41, normalized size = 0.82 \[ -\frac{\tanh (c+d x) \sqrt{b \coth ^4(c+d x)} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\tanh ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Coth[c + d*x]^4],x]

[Out]

-((Sqrt[b*Coth[c + d*x]^4]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[c + d*x]^2]*Tanh[c + d*x])/d)

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Maple [A]  time = 0.041, size = 55, normalized size = 1.1 \begin{align*} -{\frac{2\,{\rm coth} \left (dx+c\right )+\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 ) ^{2}}\sqrt{b \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c)^4)^(1/2),x)

[Out]

-1/2/d*(b*coth(d*x+c)^4)^(1/2)*(2*coth(d*x+c)+ln(coth(d*x+c)-1)-ln(coth(d*x+c)+1))/coth(d*x+c)^2

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Maxima [A]  time = 1.81973, size = 46, normalized size = 0.92 \begin{align*} \frac{{\left (d x + c\right )} \sqrt{b}}{d} + \frac{2 \, \sqrt{b}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

(d*x + c)*sqrt(b)/d + 2*sqrt(b)/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 2.15774, size = 1046, normalized size = 20.92 \begin{align*} \frac{{\left (d x \cosh \left (d x + c\right )^{2} +{\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x\right )} \sinh \left (d x + c\right )^{2} - d x +{\left (d x \cosh \left (d x + c\right )^{2} - d x - 2\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \,{\left (d x \cosh \left (d x + c\right )^{2} - d x - 2\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d x \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2\right )} \sqrt{\frac{b e^{\left (8 \, d x + 8 \, c\right )} + 4 \, b e^{\left (6 \, d x + 6 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (8 \, d x + 8 \, c\right )} - 4 \, e^{\left (6 \, d x + 6 \, c\right )} + 6 \, e^{\left (4 \, d x + 4 \, c\right )} - 4 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}}}{d \cosh \left (d x + c\right )^{2} +{\left (d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d\right )} \sinh \left (d x + c\right )^{2} +{\left (d \cosh \left (d x + c\right )^{2} - d\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \,{\left (d \cosh \left (d x + c\right )^{2} - d\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (d \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} + 2 \, d \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

(d*x*cosh(d*x + c)^2 + (d*x*e^(4*d*x + 4*c) - 2*d*x*e^(2*d*x + 2*c) + d*x)*sinh(d*x + c)^2 - d*x + (d*x*cosh(d
*x + c)^2 - d*x - 2)*e^(4*d*x + 4*c) - 2*(d*x*cosh(d*x + c)^2 - d*x - 2)*e^(2*d*x + 2*c) + 2*(d*x*cosh(d*x + c
)*e^(4*d*x + 4*c) - 2*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + d*x*cosh(d*x + c))*sinh(d*x + c) - 2)*sqrt((b*e^(8*d
*x + 8*c) + 4*b*e^(6*d*x + 6*c) + 6*b*e^(4*d*x + 4*c) + 4*b*e^(2*d*x + 2*c) + b)/(e^(8*d*x + 8*c) - 4*e^(6*d*x
+ 6*c) + 6*e^(4*d*x + 4*c) - 4*e^(2*d*x + 2*c) + 1))/(d*cosh(d*x + c)^2 + (d*e^(4*d*x + 4*c) + 2*d*e^(2*d*x +
2*c) + d)*sinh(d*x + c)^2 + (d*cosh(d*x + c)^2 - d)*e^(4*d*x + 4*c) + 2*(d*cosh(d*x + c)^2 - d)*e^(2*d*x + 2*
c) + 2*(d*cosh(d*x + c)*e^(4*d*x + 4*c) + 2*d*cosh(d*x + c)*e^(2*d*x + 2*c) + d*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 ^{4}{\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)**4)**(1/2),x)

[Out]

Integral(sqrt(b*coth(c + d*x)**4), x)

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Giac [A]  time = 1.14477, size = 43, normalized size = 0.86 \begin{align*} \sqrt{b}{\left (\frac{d x + c}{d} - \frac{2}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

sqrt(b)*((d*x + c)/d - 2/(d*(e^(2*d*x + 2*c) - 1)))