Integrand size = 8, antiderivative size = 57 \[ \int \sqrt {\tanh ^3(x)} \, dx=-2 \coth (x) \sqrt {\tanh ^3(x)}+\frac {\arctan \left (\sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)}+\frac {\text {arctanh}\left (\sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)} \]
-2*coth(x)*(tanh(x)^3)^(1/2)+arctan(tanh(x)^(1/2))*(tanh(x)^3)^(1/2)/tanh( x)^(3/2)+arctanh(tanh(x)^(1/2))*(tanh(x)^3)^(1/2)/tanh(x)^(3/2)
Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.67 \[ \int \sqrt {\tanh ^3(x)} \, dx=\frac {\left (\arctan \left (\sqrt {\tanh (x)}\right )+\text {arctanh}\left (\sqrt {\tanh (x)}\right )-2 \sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)} \]
((ArcTan[Sqrt[Tanh[x]]] + ArcTanh[Sqrt[Tanh[x]]] - 2*Sqrt[Tanh[x]])*Sqrt[T anh[x]^3])/Tanh[x]^(3/2)
Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {3042, 4141, 3042, 3954, 3042, 3957, 25, 266, 756, 216, 219}
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.
\(\displaystyle \int \sqrt {\tanh ^3(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {i \tan (i x)^3}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \int \tanh ^{\frac {3}{2}}(x)dx}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \int (-i \tan (i x))^{3/2}dx}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (\int \frac {1}{\sqrt {\tanh (x)}}dx-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (-2 \sqrt {\tanh (x)}+\int \frac {1}{\sqrt {-i \tan (i x)}}dx\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (-\int -\frac {1}{\sqrt {\tanh (x)} \left (1-\tanh ^2(x)\right )}d\tanh (x)-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (\int \frac {1}{\sqrt {\tanh (x)} \left (1-\tanh ^2(x)\right )}d\tanh (x)-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (2 \int \frac {1}{1-\tanh ^2(x)}d\sqrt {\tanh (x)}-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (2 \left (\frac {1}{2} \int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)}+\frac {1}{2} \int \frac {1}{\tanh (x)+1}d\sqrt {\tanh (x)}\right )-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (2 \left (\frac {1}{2} \int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)}+\frac {1}{2} \arctan \left (\sqrt {\tanh (x)}\right )\right )-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {\tanh ^3(x)} \left (2 \left (\frac {1}{2} \arctan \left (\sqrt {\tanh (x)}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\tanh (x)}\right )\right )-2 \sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\) |
((2*(ArcTan[Sqrt[Tanh[x]]]/2 + ArcTanh[Sqrt[Tanh[x]]]/2) - 2*Sqrt[Tanh[x]] )*Sqrt[Tanh[x]^3])/Tanh[x]^(3/2)
3.1.33.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
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]])
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(-\frac {\sqrt {\tanh \left (x \right )^{3}}\, \left (4 \sqrt {\tanh \left (x \right )}+\ln \left (\sqrt {\tanh \left (x \right )}-1\right )-\ln \left (\sqrt {\tanh \left (x \right )}+1\right )-2 \arctan \left (\sqrt {\tanh \left (x \right )}\right )\right )}{2 \tanh \left (x \right )^{\frac {3}{2}}}\) | \(43\) |
default | \(-\frac {\sqrt {\tanh \left (x \right )^{3}}\, \left (4 \sqrt {\tanh \left (x \right )}+\ln \left (\sqrt {\tanh \left (x \right )}-1\right )-\ln \left (\sqrt {\tanh \left (x \right )}+1\right )-2 \arctan \left (\sqrt {\tanh \left (x \right )}\right )\right )}{2 \tanh \left (x \right )^{\frac {3}{2}}}\) | \(43\) |
-1/2*(tanh(x)^3)^(1/2)*(4*tanh(x)^(1/2)+ln(tanh(x)^(1/2)-1)-ln(tanh(x)^(1/ 2)+1)-2*arctan(tanh(x)^(1/2)))/tanh(x)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (43) = 86\).
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.86 \[ \int \sqrt {\tanh ^3(x)} \, dx=-2 \, \sqrt {\frac {\sinh \left (x\right )}{\cosh \left (x\right )}} + \arctan \left (-\cosh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) \sinh \left (x\right ) - \sinh \left (x\right )^{2} + {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \sqrt {\frac {\sinh \left (x\right )}{\cosh \left (x\right )}}\right ) - \frac {1}{2} \, \log \left (-\cosh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) \sinh \left (x\right ) - \sinh \left (x\right )^{2} + {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \sqrt {\frac {\sinh \left (x\right )}{\cosh \left (x\right )}}\right ) \]
-2*sqrt(sinh(x)/cosh(x)) + arctan(-cosh(x)^2 - 2*cosh(x)*sinh(x) - sinh(x) ^2 + (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(sinh(x)/cosh(x)) ) - 1/2*log(-cosh(x)^2 - 2*cosh(x)*sinh(x) - sinh(x)^2 + (cosh(x)^2 + 2*co sh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(sinh(x)/cosh(x)))
\[ \int \sqrt {\tanh ^3(x)} \, dx=\int \sqrt {\tanh ^{3}{\left (x \right )}}\, dx \]
\[ \int \sqrt {\tanh ^3(x)} \, dx=\int { \sqrt {\tanh \left (x\right )^{3}} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \sqrt {\tanh ^3(x)} \, dx=\frac {4}{\sqrt {e^{\left (4 \, x\right )} - 1} - e^{\left (2 \, x\right )} - 1} + \arctan \left (\sqrt {e^{\left (4 \, x\right )} - 1} - e^{\left (2 \, x\right )}\right ) - \frac {1}{2} \, \log \left (-\sqrt {e^{\left (4 \, x\right )} - 1} + e^{\left (2 \, x\right )}\right ) \]
4/(sqrt(e^(4*x) - 1) - e^(2*x) - 1) + arctan(sqrt(e^(4*x) - 1) - e^(2*x)) - 1/2*log(-sqrt(e^(4*x) - 1) + e^(2*x))
Timed out. \[ \int \sqrt {\tanh ^3(x)} \, dx=\int \sqrt {{\mathrm {tanh}\left (x\right )}^3} \,d x \]