Integrand size = 17, antiderivative size = 82 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-(a+b) \sqrt {a+b \tanh ^2(x)}-\frac {1}{3} \left (a+b \tanh ^2(x)\right )^{3/2}-\frac {\left (a+b \tanh ^2(x)\right )^{5/2}}{5 b} \]
(a+b)^(3/2)*arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))-(a+b)*(a+b*tanh(x)^ 2)^(1/2)-1/3*(a+b*tanh(x)^2)^(3/2)-1/5*(a+b*tanh(x)^2)^(5/2)/b
Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\frac {\sqrt {a+b \tanh ^2(x)} \left (3 a^2+20 a b+15 b^2+b (6 a+5 b) \tanh ^2(x)+3 b^2 \tanh ^4(x)\right )}{15 b} \]
(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - (Sqrt[a + b*Tan h[x]^2]*(3*a^2 + 20*a*b + 15*b^2 + b*(6*a + 5*b)*Tanh[x]^2 + 3*b^2*Tanh[x] ^4))/(15*b)
Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {3042, 26, 4153, 26, 354, 90, 60, 60, 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.
| \(\displaystyle \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int i \tan (i x)^3 \left (a-b \tan (i x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \tan (i x)^3 \left (a-b \tan (i x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle i \int -\frac {i \tanh ^3(x) \left (b \tanh ^2(x)+a\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \int \frac {\tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\tanh ^2(x) \left (b \tanh ^2(x)+a\right )^{3/2}}{1-\tanh ^2(x)}d\tanh ^2(x)\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (b \tanh ^2(x)+a\right )^{3/2}}{1-\tanh ^2(x)}d\tanh ^2(x)-\frac {2 \left (a+b \tanh ^2(x)\right )^{5/2}}{5 b}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left ((a+b) \int \frac {\sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh ^2(x)-\frac {2 \left (a+b \tanh ^2(x)\right )^{5/2}}{5 b}-\frac {2}{3} \left (a+b \tanh ^2(x)\right )^{3/2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left ((a+b) \left ((a+b) \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-2 \sqrt {a+b \tanh ^2(x)}\right )-\frac {2 \left (a+b \tanh ^2(x)\right )^{5/2}}{5 b}-\frac {2}{3} \left (a+b \tanh ^2(x)\right )^{3/2}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left ((a+b) \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}-2 \sqrt {a+b \tanh ^2(x)}\right )-\frac {2 \left (a+b \tanh ^2(x)\right )^{5/2}}{5 b}-\frac {2}{3} \left (a+b \tanh ^2(x)\right )^{3/2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left ((a+b) \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \tanh ^2(x)}\right )-\frac {2 \left (a+b \tanh ^2(x)\right )^{5/2}}{5 b}-\frac {2}{3} \left (a+b \tanh ^2(x)\right )^{3/2}\right )\) |
((-2*(a + b*Tanh[x]^2)^(3/2))/3 - (2*(a + b*Tanh[x]^2)^(5/2))/(5*b) + (a + b)*(2*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*Tanh[x]^2]))/2
3.3.19.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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]))
Leaf count of result is larger than twice the leaf count of optimal. \(487\) vs. \(2(66)=132\).
Time = 0.08 (sec) , antiderivative size = 488, normalized size of antiderivative = 5.95
| method | result | size |
| derivativedivides | \(-\frac {\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {5}{2}}}{5 b}-\frac {\left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}{6}+\frac {b \left (\frac {\left (2 b \left (1+\tanh \left (x \right )\right )-2 b \right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{4 b}+\frac {\left (4 \left (a +b \right ) b -4 b^{2}\right ) \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )}{8 b^{\frac {3}{2}}}\right )}{2}-\frac {\left (a +b \right ) \left (\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}-\sqrt {b}\, \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )-\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )\right )}{2}-\frac {\left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}{6}-\frac {b \left (\frac {\left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{4 b}+\frac {\left (4 \left (a +b \right ) b -4 b^{2}\right ) \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )}{8 b^{\frac {3}{2}}}\right )}{2}-\frac {\left (a +b \right ) \left (\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}+\sqrt {b}\, \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )-\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )\right )}{2}\) | \(488\) |
| default | \(-\frac {\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {5}{2}}}{5 b}-\frac {\left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}{6}+\frac {b \left (\frac {\left (2 b \left (1+\tanh \left (x \right )\right )-2 b \right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{4 b}+\frac {\left (4 \left (a +b \right ) b -4 b^{2}\right ) \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )}{8 b^{\frac {3}{2}}}\right )}{2}-\frac {\left (a +b \right ) \left (\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}-\sqrt {b}\, \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )-\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )\right )}{2}-\frac {\left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}{6}-\frac {b \left (\frac {\left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{4 b}+\frac {\left (4 \left (a +b \right ) b -4 b^{2}\right ) \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )}{8 b^{\frac {3}{2}}}\right )}{2}-\frac {\left (a +b \right ) \left (\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}+\sqrt {b}\, \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )-\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )\right )}{2}\) | \(488\) |
-1/5*(a+b*tanh(x)^2)^(5/2)/b-1/6*(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(3/ 2)+1/2*b*(1/4*(2*b*(1+tanh(x))-2*b)/b*(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b )^(1/2)+1/8*(4*(a+b)*b-4*b^2)/b^(3/2)*ln((b*(1+tanh(x))-b)/b^(1/2)+(b*(1+t anh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)))-1/2*(a+b)*((b*(1+tanh(x))^2-2*b*(1+ tanh(x))+a+b)^(1/2)-b^(1/2)*ln((b*(1+tanh(x))-b)/b^(1/2)+(b*(1+tanh(x))^2- 2*b*(1+tanh(x))+a+b)^(1/2))-(a+b)^(1/2)*ln((2*a+2*b-2*b*(1+tanh(x))+2*(a+b )^(1/2)*(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2))/(1+tanh(x))))-1/6*(b* (tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)-1/2*b*(1/4*(2*b*(tanh(x)-1)+2*b)/ b*(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+1/8*(4*(a+b)*b-4*b^2)/b^(3/2 )*ln((b*(tanh(x)-1)+b)/b^(1/2)+(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2) ))-1/2*(a+b)*((b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+b^(1/2)*ln((b*(t anh(x)-1)+b)/b^(1/2)+(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2))-(a+b)^(1 /2)*ln((2*a+2*b+2*b*(tanh(x)-1)+2*(a+b)^(1/2)*(b*(tanh(x)-1)^2+2*b*(tanh(x )-1)+a+b)^(1/2))/(tanh(x)-1)))
Leaf count of result is larger than twice the leaf count of optimal. 2188 vs. \(2 (66) = 132\).
Time = 0.49 (sec) , antiderivative size = 4941, normalized size of antiderivative = 60.26 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/60*(15*((a*b + b^2)*cosh(x)^10 + 10*(a*b + b^2)*cosh(x)*sinh(x)^9 + (a* b + b^2)*sinh(x)^10 + 5*(a*b + b^2)*cosh(x)^8 + 5*(9*(a*b + b^2)*cosh(x)^2 + a*b + b^2)*sinh(x)^8 + 40*(3*(a*b + b^2)*cosh(x)^3 + (a*b + b^2)*cosh(x ))*sinh(x)^7 + 10*(a*b + b^2)*cosh(x)^6 + 10*(21*(a*b + b^2)*cosh(x)^4 + 1 4*(a*b + b^2)*cosh(x)^2 + a*b + b^2)*sinh(x)^6 + 4*(63*(a*b + b^2)*cosh(x) ^5 + 70*(a*b + b^2)*cosh(x)^3 + 15*(a*b + b^2)*cosh(x))*sinh(x)^5 + 10*(a* b + b^2)*cosh(x)^4 + 10*(21*(a*b + b^2)*cosh(x)^6 + 35*(a*b + b^2)*cosh(x) ^4 + 15*(a*b + b^2)*cosh(x)^2 + a*b + b^2)*sinh(x)^4 + 40*(3*(a*b + b^2)*c osh(x)^7 + 7*(a*b + b^2)*cosh(x)^5 + 5*(a*b + b^2)*cosh(x)^3 + (a*b + b^2) *cosh(x))*sinh(x)^3 + 5*(a*b + b^2)*cosh(x)^2 + 5*(9*(a*b + b^2)*cosh(x)^8 + 28*(a*b + b^2)*cosh(x)^6 + 30*(a*b + b^2)*cosh(x)^4 + 12*(a*b + b^2)*co sh(x)^2 + a*b + b^2)*sinh(x)^2 + a*b + b^2 + 10*((a*b + b^2)*cosh(x)^9 + 4 *(a*b + b^2)*cosh(x)^7 + 6*(a*b + b^2)*cosh(x)^5 + 4*(a*b + b^2)*cosh(x)^3 + (a*b + b^2)*cosh(x))*sinh(x))*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)*cosh(x)^5 + 10*(2*a^3 + a^2*b)*cosh(x)^3 + (6*a^3...
Time = 10.26 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.72 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=- a \left (\begin {cases} \frac {2 \left (\frac {b^{2} \sqrt {a + b \tanh ^{2}{\left (x \right )}}}{2} + \frac {b^{2} \left (a + b\right ) \operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{2 \sqrt {- a - b}} + \frac {b \left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}{6}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {\log {\left (\tanh ^{2}{\left (x \right )} - 1 \right )}}{2} + \frac {\tanh ^{2}{\left (x \right )}}{2}\right ) & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} \frac {2 \left (\frac {b^{3} \sqrt {a + b \tanh ^{2}{\left (x \right )}}}{2} + \frac {b^{3} \left (a + b\right ) \operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{2 \sqrt {- a - b}} + \frac {b \left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}{10} + \frac {\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}} \left (- \frac {a b}{2} + \frac {b^{2}}{2}\right )}{3}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {\log {\left (\tanh ^{2}{\left (x \right )} - 1 \right )}}{2} + \frac {\tanh ^{4}{\left (x \right )}}{4} + \frac {\tanh ^{2}{\left (x \right )}}{2}\right ) & \text {otherwise} \end {cases}\right ) \]
-a*Piecewise((2*(b**2*sqrt(a + b*tanh(x)**2)/2 + b**2*(a + b)*atan(sqrt(a + b*tanh(x)**2)/sqrt(-a - b))/(2*sqrt(-a - b)) + b*(a + b*tanh(x)**2)**(3/ 2)/6)/b**2, Ne(b, 0)), (sqrt(a)*(log(tanh(x)**2 - 1)/2 + tanh(x)**2/2), Tr ue)) - b*Piecewise((2*(b**3*sqrt(a + b*tanh(x)**2)/2 + b**3*(a + b)*atan(s qrt(a + b*tanh(x)**2)/sqrt(-a - b))/(2*sqrt(-a - b)) + b*(a + b*tanh(x)**2 )**(5/2)/10 + (a + b*tanh(x)**2)**(3/2)*(-a*b/2 + b**2/2)/3)/b**3, Ne(b, 0 )), (sqrt(a)*(log(tanh(x)**2 - 1)/2 + tanh(x)**4/4 + tanh(x)**2/2), True))
\[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\int { {\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}} \tanh \left (x\right )^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1063 vs. \(2 (66) = 132\).
Time = 1.46 (sec) , antiderivative size = 1063, normalized size of antiderivative = 12.96 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/2*(a + b)^(3/2)*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x ) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) + sqrt(a + b))) - 1/2*(a + b)^(3/2) *log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) - sqrt(a + b))) - 1/2*(a^2 + 2*a*b + b^2)*log(abs(-( sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2* x) + a + b))*(a + b) - sqrt(a + b)*(a - b)))/sqrt(a + b) - 4/15*(15*(a^2 + 4*a*b + 3*b^2)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^ (2*x) - 2*b*e^(2*x) + a + b))^9 + 15*(7*a^2 + 20*a*b + 9*b^2)*(sqrt(a + b) *e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) )^8*sqrt(a + b) + 20*(15*a^3 + 39*a^2*b + 21*a*b^2 + b^3)*(sqrt(a + b)*e^( 2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^7 + 20*(21*a^3 + 21*a^2*b - 57*a*b^2 - 65*b^3)*(sqrt(a + b)*e^(2*x) - sqrt(a *e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^6*sqrt(a + b) + 2*(105*a^4 - 210*a^3*b - 1860*a^2*b^2 - 1590*a*b^3 + 19*b^4)*(sqrt(a + b) *e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) )^5 - 10*(21*a^4 + 126*a^3*b + 288*a^2*b^2 - 390*a*b^3 - 349*b^4)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^4*sqrt(a + b) - 20*(21*a^5 + 63*a^4*b - 18*a^3*b^2 - 378*a^2*b^3 - 2 35*a*b^4 + 19*b^5)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a *e^(2*x) - 2*b*e^(2*x) + a + b))^3 - 20*(15*a^5 + 21*a^4*b - 126*a^3*b^...
Time = 9.97 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \tanh ^3(x) \left (a+b \tanh ^2(x)\right )^{3/2} \, dx=-\frac {{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{5/2}}{5\,b}-\left (\frac {a+b}{3\,b}-\frac {a}{3\,b}\right )\,{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}-\left (a+b\right )\,\left (\frac {a+b}{b}-\frac {a}{b}\right )\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}-\mathrm {atan}\left (\frac {{\left (a+b\right )}^{3/2}\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,1{}\mathrm {i}}{a^2+2\,a\,b+b^2}\right )\,{\left (a+b\right )}^{3/2}\,1{}\mathrm {i} \]