Integrand size = 14, antiderivative size = 264 \[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^2(c+d x)}}{2 d \coth ^{\frac {2}{3}}(c+d x)}-\frac {\sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right ) \sqrt [3]{b \coth ^2(c+d x)}}{2 d \coth ^{\frac {2}{3}}(c+d x)}+\frac {\text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right ) \sqrt [3]{b \coth ^2(c+d x)}}{d \coth ^{\frac {2}{3}}(c+d x)}-\frac {\sqrt [3]{b \coth ^2(c+d x)} \log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {2}{3}}(c+d x)}+\frac {\sqrt [3]{b \coth ^2(c+d x)} \log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )}{4 d \coth ^{\frac {2}{3}}(c+d x)} \]
arctanh(coth(d*x+c)^(1/3))*(b*coth(d*x+c)^2)^(1/3)/d/coth(d*x+c)^(2/3)-1/4 *(b*coth(d*x+c)^2)^(1/3)*ln(1-coth(d*x+c)^(1/3)+coth(d*x+c)^(2/3))/d/coth( d*x+c)^(2/3)+1/4*(b*coth(d*x+c)^2)^(1/3)*ln(1+coth(d*x+c)^(1/3)+coth(d*x+c )^(2/3))/d/coth(d*x+c)^(2/3)+1/2*arctan(1/3*(1-2*coth(d*x+c)^(1/3))*3^(1/2 ))*(b*coth(d*x+c)^2)^(1/3)*3^(1/2)/d/coth(d*x+c)^(2/3)-1/2*arctan(1/3*(1+2 *coth(d*x+c)^(1/3))*3^(1/2))*(b*coth(d*x+c)^2)^(1/3)*3^(1/2)/d/coth(d*x+c) ^(2/3)
Time = 0.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.57 \[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\frac {\sqrt [3]{b \coth ^2(c+d x)} \left (2 \sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{\coth (c+d x)}}{\sqrt {3}}\right )+4 \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )-\log \left (1-\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )+\log \left (1+\sqrt [3]{\coth (c+d x)}+\coth ^{\frac {2}{3}}(c+d x)\right )\right )}{4 d \coth ^{\frac {2}{3}}(c+d x)} \]
((b*Coth[c + d*x]^2)^(1/3)*(2*Sqrt[3]*ArcTan[(1 - 2*Coth[c + d*x]^(1/3))/S qrt[3]] - 2*Sqrt[3]*ArcTan[(1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]] + 4*ArcTan h[Coth[c + d*x]^(1/3)] - Log[1 - Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3) ] + Log[1 + Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)]))/(4*d*Coth[c + d*x ]^(2/3))
Time = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.63, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4141, 3042, 3957, 25, 266, 825, 27, 219, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sqrt [3]{b \coth ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt [3]{-b \tan \left (i c+i d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4141 |
| \(\displaystyle \frac {\sqrt [3]{b \coth ^2(c+d x)} \int \coth ^{\frac {2}{3}}(c+d x)dx}{\coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{b \coth ^2(c+d x)} \int \left (-i \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{\coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle -\frac {\sqrt [3]{b \coth ^2(c+d x)} \int -\frac {\coth ^{\frac {2}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt [3]{b \coth ^2(c+d x)} \int \frac {\coth ^{\frac {2}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \int \frac {\coth ^{\frac {4}{3}}(c+d x)}{1-\coth ^2(c+d x)}d\sqrt [3]{\coth (c+d x)}}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 825 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{3} \int \frac {1}{1-\coth ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \int -\frac {\sqrt [3]{\coth (c+d x)}+1}{2 \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt [3]{\coth (c+d x)}}{2 \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )}d\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{3} \int \frac {1}{1-\coth ^{\frac {2}{3}}(c+d x)}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {\sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (-\frac {1}{6} \int \frac {\sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {1}{2} \int -\frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\frac {3}{2} \int \frac {1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{6} \left (3 \int \frac {1}{-\coth ^{\frac {2}{3}}(c+d x)-3}d\left (2 \sqrt [3]{\coth (c+d x)}-1\right )+\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{6} \left (3 \int \frac {1}{-\coth ^{\frac {2}{3}}(c+d x)-3}d\left (2 \sqrt [3]{\coth (c+d x)}+1\right )+\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1-2 \sqrt [3]{\coth (c+d x)}}{\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}-1}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {2 \sqrt [3]{\coth (c+d x)}+1}{\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1}d\sqrt [3]{\coth (c+d x)}-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {3 \sqrt [3]{b \coth ^2(c+d x)} \left (\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (\coth ^{\frac {2}{3}}(c+d x)-\sqrt [3]{\coth (c+d x)}+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \log \left (\coth ^{\frac {2}{3}}(c+d x)+\sqrt [3]{\coth (c+d x)}+1\right )-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{\coth (c+d x)}+1}{\sqrt {3}}\right )\right )+\frac {1}{3} \text {arctanh}\left (\sqrt [3]{\coth (c+d x)}\right )\right )}{d \coth ^{\frac {2}{3}}(c+d x)}\) |
(3*(b*Coth[c + d*x]^2)^(1/3)*(ArcTanh[Coth[c + d*x]^(1/3)]/3 + (-(Sqrt[3]* ArcTan[(-1 + 2*Coth[c + d*x]^(1/3))/Sqrt[3]]) - Log[1 - Coth[c + d*x]^(1/3 ) + Coth[c + d*x]^(2/3)]/2)/6 + (-(Sqrt[3]*ArcTan[(1 + 2*Coth[c + d*x]^(1/ 3))/Sqrt[3]]) + Log[1 + Coth[c + d*x]^(1/3) + Coth[c + d*x]^(2/3)]/2)/6))/ (d*Coth[c + d*x]^(2/3))
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator [Rt[-a/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r*Cos[2*k *m*(Pi/n)] - s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[2*k*m*(Pi/n)] + s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x]; 2*(r^(m + 2)/(a*n*s^m)) Int[1/ (r^2 - s^2*x^2), x] + 2*(r^(m + 1)/(a*n*s^m)) Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1 ] && NegQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[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]])
\[\int \left (\coth \left (d x +c \right )^{2} b \right )^{\frac {1}{3}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1618 vs. \(2 (216) = 432\).
Time = 0.28 (sec) , antiderivative size = 1618, normalized size of antiderivative = 6.13 \[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\text {Too large to display} \]
-1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(1/3*(sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt( 3)*b*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*b*sinh(d*x + c)^2 + 2*(sqrt(3)* cosh(d*x + c)^2 + 2*sqrt(3)*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*sinh(d*x + c)^2 - sqrt(3))*(-b)^(2/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b) /(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3) + sqrt(3)*b)/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) - 2*sqrt (3)*b^(1/3)*arctan(-1/3*(sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt(3)*b*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*b*sinh(d*x + c)^2 - 2*(sqrt(3)*cosh(d*x + c)^ 2 + 2*sqrt(3)*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*sinh(d*x + c)^2 - sqrt (3))*b^(2/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3) + sqrt(3)*b)/(b*cosh(d*x + c)^2 + 2*b*cosh( d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) + (-b)^(1/3)*log(((cosh(d *x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2*(3*cosh( d*x + c)^2 + 1)*sinh(d*x + c)^2 + 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)*(-b)^(2/3) - (cosh(d*x + c)^4 + 4*cosh( d*x + c)^3*sinh(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 - 1)*(-b)^(1/3)*((b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + b)/(cosh(d*x + c)^2 + sinh(d*x + c)^2 - 1))^(1/3) + (cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + 2* (3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d...
\[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\int \sqrt [3]{b \coth ^{2}{\left (c + d x \right )}}\, dx \]
\[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\int { \left (b \coth \left (d x + c\right )^{2}\right )^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\int { \left (b \coth \left (d x + c\right )^{2}\right )^{\frac {1}{3}} \,d x } \]
Timed out. \[ \int \sqrt [3]{b \coth ^2(c+d x)} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^2\right )}^{1/3} \,d x \]