3.9 \(\int (b \coth (c+d x))^{4/3} \, dx\)

Optimal. Leaf size=236 \[ -\frac{b^{4/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}+\frac{b^{4/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}-\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}+1}{\sqrt{3}}\right )}{2 d}+\frac{b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{d}-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d} \]

[Out]

-(Sqrt[3]*b^(4/3)*ArcTan[(1 - (2*(b*Coth[c + d*x])^(1/3))/b^(1/3))/Sqrt[3]])/(2*d) + (Sqrt[3]*b^(4/3)*ArcTan[(
1 + (2*(b*Coth[c + d*x])^(1/3))/b^(1/3))/Sqrt[3]])/(2*d) + (b^(4/3)*ArcTanh[(b*Coth[c + d*x])^(1/3)/b^(1/3)])/
d - (3*b*(b*Coth[c + d*x])^(1/3))/d - (b^(4/3)*Log[b^(2/3) - b^(1/3)*(b*Coth[c + d*x])^(1/3) + (b*Coth[c + d*x
])^(2/3)])/(4*d) + (b^(4/3)*Log[b^(2/3) + b^(1/3)*(b*Coth[c + d*x])^(1/3) + (b*Coth[c + d*x])^(2/3)])/(4*d)

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Rubi [A]  time = 0.281901, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3473, 3476, 329, 210, 634, 618, 204, 628, 206} \[ -\frac{b^{4/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}+\frac{b^{4/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}-\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}+1}{\sqrt{3}}\right )}{2 d}+\frac{b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{d}-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(4/3),x]

[Out]

-(Sqrt[3]*b^(4/3)*ArcTan[(1 - (2*(b*Coth[c + d*x])^(1/3))/b^(1/3))/Sqrt[3]])/(2*d) + (Sqrt[3]*b^(4/3)*ArcTan[(
1 + (2*(b*Coth[c + d*x])^(1/3))/b^(1/3))/Sqrt[3]])/(2*d) + (b^(4/3)*ArcTanh[(b*Coth[c + d*x])^(1/3)/b^(1/3)])/
d - (3*b*(b*Coth[c + d*x])^(1/3))/d - (b^(4/3)*Log[b^(2/3) - b^(1/3)*(b*Coth[c + d*x])^(1/3) + (b*Coth[c + d*x
])^(2/3)])/(4*d) + (b^(4/3)*Log[b^(2/3) + b^(1/3)*(b*Coth[c + d*x])^(1/3) + (b*Coth[c + d*x])^(2/3)])/(4*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 210

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt[-(a/b
), n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r +
s*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 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])

Rubi steps

\begin{align*} \int (b \coth (c+d x))^{4/3} \, dx &=-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}+b^2 \int \frac{1}{(b \coth (c+d x))^{2/3}} \, dx\\ &=-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (-b^2+x^2\right )} \, dx,x,b \coth (c+d x)\right )}{d}\\ &=-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-b^2+x^6} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{d}\\ &=-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}+\frac{b^{4/3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{b}-\frac{x}{2}}{b^{2/3}-\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{d}+\frac{b^{4/3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{b}+\frac{x}{2}}{b^{2/3}+\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{d}+\frac{b^{5/3} \operatorname{Subst}\left (\int \frac{1}{b^{2/3}-x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{d}\\ &=\frac{b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{d}-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}-\frac{b^{4/3} \operatorname{Subst}\left (\int \frac{-\sqrt [3]{b}+2 x}{b^{2/3}-\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{4 d}+\frac{b^{4/3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{b}+2 x}{b^{2/3}+\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{4 d}+\frac{\left (3 b^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{b^{2/3}-\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{4 d}+\frac{\left (3 b^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{b^{2/3}+\sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \coth (c+d x)}\right )}{4 d}\\ &=\frac{b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{d}-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}-\frac{b^{4/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}+\frac{b^{4/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}+\frac{\left (3 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{2 d}-\frac{\left (3 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{2 d}\\ &=-\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} b^{4/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{2 d}+\frac{b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{d}-\frac{3 b \sqrt [3]{b \coth (c+d x)}}{d}-\frac{b^{4/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}+\frac{b^{4/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}+(b \coth (c+d x))^{2/3}\right )}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.0259547, size = 36, normalized size = 0.15 \[ \frac{3 b \sqrt [3]{b \coth (c+d x)} \left (\, _2F_1\left (\frac{1}{6},1;\frac{7}{6};\coth ^2(c+d x)\right )-1\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(4/3),x]

[Out]

(3*b*(b*Coth[c + d*x])^(1/3)*(-1 + Hypergeometric2F1[1/6, 1, 7/6, Coth[c + d*x]^2]))/d

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Maple [A]  time = 0.02, size = 209, normalized size = 0.9 \begin{align*} -3\,{\frac{b\sqrt [3]{b{\rm coth} \left (dx+c\right )}}{d}}+{\frac{1}{2\,d}{b}^{{\frac{4}{3}}}\ln \left ( \sqrt [3]{b{\rm coth} \left (dx+c\right )}+\sqrt [3]{b} \right ) }-{\frac{1}{4\,d}{b}^{{\frac{4}{3}}}\ln \left ({b}^{{\frac{2}{3}}}-\sqrt [3]{b}\sqrt [3]{b{\rm coth} \left (dx+c\right )}+ \left ( b{\rm coth} \left (dx+c\right ) \right ) ^{{\frac{2}{3}}} \right ) }+{\frac{\sqrt{3}}{2\,d}{b}^{{\frac{4}{3}}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt [3]{b{\rm coth} \left (dx+c\right )}}{\sqrt [3]{b}}}-1 \right ) } \right ) }-{\frac{1}{2\,d}{b}^{{\frac{4}{3}}}\ln \left ( \sqrt [3]{b{\rm coth} \left (dx+c\right )}-\sqrt [3]{b} \right ) }+{\frac{1}{4\,d}{b}^{{\frac{4}{3}}}\ln \left ({b}^{{\frac{2}{3}}}+\sqrt [3]{b}\sqrt [3]{b{\rm coth} \left (dx+c\right )}+ \left ( b{\rm coth} \left (dx+c\right ) \right ) ^{{\frac{2}{3}}} \right ) }+{\frac{\sqrt{3}}{2\,d}{b}^{{\frac{4}{3}}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 1+2\,{\frac{\sqrt [3]{b{\rm coth} \left (dx+c\right )}}{\sqrt [3]{b}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c))^(4/3),x)

[Out]

-3*b*(b*coth(d*x+c))^(1/3)/d+1/2*b^(4/3)/d*ln((b*coth(d*x+c))^(1/3)+b^(1/3))-1/4*b^(4/3)*ln(b^(2/3)-b^(1/3)*(b
*coth(d*x+c))^(1/3)+(b*coth(d*x+c))^(2/3))/d+1/2*b^(4/3)/d*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b*coth(d*x+c))^(1/3)
/b^(1/3)-1))-1/2*b^(4/3)/d*ln((b*coth(d*x+c))^(1/3)-b^(1/3))+1/4*b^(4/3)*ln(b^(2/3)+b^(1/3)*(b*coth(d*x+c))^(1
/3)+(b*coth(d*x+c))^(2/3))/d+1/2*b^(4/3)*arctan(1/3*(1+2*(b*coth(d*x+c))^(1/3)/b^(1/3))*3^(1/2))*3^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \coth \left (d x + c\right )\right )^{\frac{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*coth(d*x + c))^(4/3), x)

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Fricas [A]  time = 1.96127, size = 865, normalized size = 3.67 \begin{align*} -\frac{2 \, \sqrt{3} \left (-b\right )^{\frac{1}{3}} b \arctan \left (\frac{\sqrt{3} b + 2 \, \sqrt{3} \left (-b\right )^{\frac{2}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{3 \, b}\right ) - 2 \, \sqrt{3} b^{\frac{4}{3}} \arctan \left (-\frac{\sqrt{3} b - 2 \, \sqrt{3} b^{\frac{2}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{3 \, b}\right ) + \left (-b\right )^{\frac{1}{3}} b \log \left (\left (-b\right )^{\frac{2}{3}} - \left (-b\right )^{\frac{1}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{2}{3}}\right ) + b^{\frac{4}{3}} \log \left (b^{\frac{2}{3}} - b^{\frac{1}{3}} \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{2}{3}}\right ) - 2 \, \left (-b\right )^{\frac{1}{3}} b \log \left (\left (-b\right )^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}\right ) - 2 \, b^{\frac{4}{3}} \log \left (b^{\frac{1}{3}} + \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}\right ) + 12 \, b \left (\frac{b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac{1}{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

-1/4*(2*sqrt(3)*(-b)^(1/3)*b*arctan(1/3*(sqrt(3)*b + 2*sqrt(3)*(-b)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3
))/b) - 2*sqrt(3)*b^(4/3)*arctan(-1/3*(sqrt(3)*b - 2*sqrt(3)*b^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/b)
+ (-b)^(1/3)*b*log((-b)^(2/3) - (-b)^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x
+ c))^(2/3)) + b^(4/3)*log(b^(2/3) - b^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3) + (b*cosh(d*x + c)/sinh(d*x
+ c))^(2/3)) - 2*(-b)^(1/3)*b*log((-b)^(1/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) - 2*b^(4/3)*log(b^(1/3)
+ (b*cosh(d*x + c)/sinh(d*x + c))^(1/3)) + 12*b*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))**(4/3),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(4/3),x, algorithm="giac")

[Out]

Exception raised: TypeError