Integrand size = 15, antiderivative size = 71 \[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=-a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )+(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-b \sqrt {a+b \tanh ^2(x)} \]
-a^(3/2)*arctanh((a+b*tanh(x)^2)^(1/2)/a^(1/2))+(a+b)^(3/2)*arctanh((a+b*t anh(x)^2)^(1/2)/(a+b)^(1/2))-b*(a+b*tanh(x)^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=-a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )+(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-b \sqrt {a+b \tanh ^2(x)} \]
-(a^(3/2)*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]]) + (a + b)^(3/2)*ArcTanh[ Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - b*Sqrt[a + b*Tanh[x]^2]
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 26, 4153, 26, 354, 95, 25, 174, 73, 221}
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.
\(\displaystyle \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \left (a-b \tan (i x)^2\right )^{3/2}}{\tan (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\left (a-b \tan (i x)^2\right )^{3/2}}{\tan (i x)}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle i \int -\frac {i \coth (x) \left (b \tanh ^2(x)+a\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\coth (x) \left (a+b \tanh ^2(x)\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\coth (x) \left (b \tanh ^2(x)+a\right )^{3/2}}{1-\tanh ^2(x)}d\tanh ^2(x)\) |
\(\Big \downarrow \) 95 |
\(\displaystyle \frac {1}{2} \left (-\int -\frac {\coth (x) \left (a^2+b (2 a+b) \tanh ^2(x)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-2 b \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\int \frac {\coth (x) \left (a^2+b (2 a+b) \tanh ^2(x)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-2 b \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{2} \left (a^2 \int \frac {\coth (x)}{\sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)+(a+b)^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-2 b \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 a^2 \int \frac {1}{\frac {\tanh ^4(x)}{b}-\frac {a}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}+\frac {2 (a+b)^2 \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}-2 b \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )+2 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-2 b \sqrt {a+b \tanh ^2(x)}\right )\) |
(-2*a^(3/2)*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]] + 2*(a + b)^(3/2)*ArcTa nh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - 2*b*Sqrt[a + b*Tanh[x]^2])/2
3.3.23.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d) Int[(b *d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
\[\int \coth \left (x \right ) \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 695 vs. \(2 (57) = 114\).
Time = 0.42 (sec) , antiderivative size = 4039, normalized size of antiderivative = 56.89 \[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/4*(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a + b)*sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*cosh(x) *sinh(x)^7 + (a^3 + a^2*b)*sinh(x)^8 + 2*(2*a^3 + a^2*b)*cosh(x)^6 + 2*(2* a^3 + a^2*b + 14*(a^3 + a^2*b)*cosh(x)^2)*sinh(x)^6 + 4*(14*(a^3 + a^2*b)* cosh(x)^3 + 3*(2*a^3 + a^2*b)*cosh(x))*sinh(x)^5 + (6*a^3 + 4*a^2*b - a*b^ 2 + b^3)*cosh(x)^4 + (70*(a^3 + a^2*b)*cosh(x)^4 + 6*a^3 + 4*a^2*b - a*b^2 + b^3 + 30*(2*a^3 + a^2*b)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a^3 + a^2*b)*cos h(x)^5 + 10*(2*a^3 + a^2*b)*cosh(x)^3 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*co sh(x))*sinh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^3 + 3*a^2*b - b^ 3)*cosh(x)^2 + 2*(14*(a^3 + a^2*b)*cosh(x)^6 + 15*(2*a^3 + a^2*b)*cosh(x)^ 4 + 2*a^3 + 3*a^2*b - b^3 + 3*(6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^2)*s inh(x)^2 + sqrt(2)*(a^2*cosh(x)^6 + 6*a^2*cosh(x)*sinh(x)^5 + a^2*sinh(x)^ 6 + 3*a^2*cosh(x)^4 + 3*(5*a^2*cosh(x)^2 + a^2)*sinh(x)^4 + 4*(5*a^2*cosh( x)^3 + 3*a^2*cosh(x))*sinh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x)^2 + (15*a^ 2*cosh(x)^4 + 18*a^2*cosh(x)^2 + 3*a^2 + 2*a*b - b^2)*sinh(x)^2 + a^2 + 2* a*b + b^2 + 2*(3*a^2*cosh(x)^5 + 6*a^2*cosh(x)^3 + (3*a^2 + 2*a*b - b^2)*c osh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + a^2*b)* cosh(x)^7 + 3*(2*a^3 + a^2*b)*cosh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)* cosh(x)^3 + (2*a^3 + 3*a^2*b - b^3)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*co...
\[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\int \left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}} \coth {\left (x \right )}\, dx \]
\[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\int { {\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}} \coth \left (x\right ) \,d x } \]
\[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\int { {\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}} \coth \left (x\right ) \,d x } \]
Timed out. \[ \int \coth (x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\int \mathrm {coth}\left (x\right )\,{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2} \,d x \]