Integrand size = 17, antiderivative size = 63 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\sqrt {a+b \tanh ^2(x)}-\frac {\left (a+b \tanh ^2(x)\right )^{3/2}}{3 b} \]
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)-(a+b*tanh(x)^2)^(1/ 2)-1/3*(a+b*tanh(x)^2)^(3/2)/b
Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\frac {\sqrt {a+b \tanh ^2(x)} \left (a+3 b+b \tanh ^2(x)\right )}{3 b} \]
Sqrt[a + b]*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - (Sqrt[a + b*Tanh[ x]^2]*(a + 3*b + b*Tanh[x]^2))/(3*b)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 26, 4153, 26, 354, 90, 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.
| \(\displaystyle \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int i \tan (i x)^3 \sqrt {a-b \tan (i x)^2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \tan (i x)^3 \sqrt {a-b \tan (i x)^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle i \int -\frac {i \tanh ^3(x) \sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh (x)\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \int \frac {\tanh ^3(x) \sqrt {a+b \tanh ^2(x)}}{1-\tanh ^2(x)}d\tanh (x)\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\tanh ^2(x) \sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh ^2(x)\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh ^2(x)-\frac {2 \left (a+b \tanh ^2(x)\right )^{3/2}}{3 b}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left ((a+b) \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-\frac {2 \left (a+b \tanh ^2(x)\right )^{3/2}}{3 b}-2 \sqrt {a+b \tanh ^2(x)}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}-\frac {2 \left (a+b \tanh ^2(x)\right )^{3/2}}{3 b}-2 \sqrt {a+b \tanh ^2(x)}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\frac {2 \left (a+b \tanh ^2(x)\right )^{3/2}}{3 b}-2 \sqrt {a+b \tanh ^2(x)}\right )\) |
(2*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*T anh[x]^2] - (2*(a + b*Tanh[x]^2)^(3/2))/(3*b))/2
3.3.10.3.1 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. \(252\) vs. \(2(51)=102\).
Time = 0.09 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.02
| method | result | size |
| derivativedivides | \(-\frac {\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}{3 b}-\frac {\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{2}+\frac {\sqrt {b}\, \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2}-\frac {\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{2}-\frac {\sqrt {b}\, \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2}\) | \(253\) |
| default | \(-\frac {\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}{3 b}-\frac {\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{2}+\frac {\sqrt {b}\, \ln \left (\frac {b \left (1+\tanh \left (x \right )\right )-b}{\sqrt {b}}+\sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}\right )}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2}-\frac {\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{2}-\frac {\sqrt {b}\, \ln \left (\frac {b \left (\tanh \left (x \right )-1\right )+b}{\sqrt {b}}+\sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}\right )}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2}\) | \(253\) |
-1/3*(a+b*tanh(x)^2)^(3/2)/b-1/2*(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/ 2)+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))+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)))-1/2*(b*(tanh(x)-1 )^2+2*b*(tanh(x)-1)+a+b)^(1/2)-1/2*b^(1/2)*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)^(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))/(t anh(x)-1))
Leaf count of result is larger than twice the leaf count of optimal. 882 vs. \(2 (51) = 102\).
Time = 0.34 (sec) , antiderivative size = 2329, normalized size of antiderivative = 36.97 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=\text {Too large to display} \]
[1/12*(3*(b*cosh(x)^6 + 6*b*cosh(x)*sinh(x)^5 + b*sinh(x)^6 + 3*b*cosh(x)^ 4 + 3*(5*b*cosh(x)^2 + b)*sinh(x)^4 + 4*(5*b*cosh(x)^3 + 3*b*cosh(x))*sinh (x)^3 + 3*b*cosh(x)^2 + 3*(5*b*cosh(x)^4 + 6*b*cosh(x)^2 + b)*sinh(x)^2 + 6*(b*cosh(x)^5 + 2*b*cosh(x)^3 + b*cosh(x))*sinh(x) + 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)*cosh(x)^5 + 10*(2*a^3 + a^2* b)*cosh(x)^3 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(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)*sinh(x)^2 + sqrt(2)*(a^2*c osh(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))*sin h(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)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2...
Time = 1.59 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.49 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=- \begin {cases} \frac {2 \left (\frac {b^{2} \sqrt {a + b \tanh ^{2}{\left (x \right )}}}{2} + \frac {b^{2} \left (a + b\right ) \operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{2 \sqrt {- a - b}} + \frac {b \left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}{6}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {\log {\left (\tanh ^{2}{\left (x \right )} - 1 \right )}}{2} + \frac {\tanh ^{2}{\left (x \right )}}{2}\right ) & \text {otherwise} \end {cases} \]
-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), True ))
\[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=\int { \sqrt {b \tanh \left (x\right )^{2} + a} \tanh \left (x\right )^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (51) = 102\).
Time = 0.77 (sec) , antiderivative size = 630, normalized size of antiderivative = 10.00 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {a + b} \log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right ) + \frac {1}{2} \, \sqrt {a + b} \log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right ) - \frac {1}{2} \, \sqrt {a + b} \log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right ) - \frac {4 \, {\left (3 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{5} {\left (a + 2 \, b\right )} + 3 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{4} {\left (3 \, a + 2 \, b\right )} \sqrt {a + b} + 2 \, {\left (3 \, a^{2} - 3 \, a b - 10 \, b^{2}\right )} {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{3} - 6 \, {\left (a^{2} + 3 \, a b + 6 \, b^{2}\right )} {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{2} \sqrt {a + b} - 3 \, {\left (3 \, a^{3} + 4 \, a^{2} b - 9 \, a b^{2} - 26 \, b^{3}\right )} {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} - {\left (3 \, a^{3} - 17 \, a b^{2} + 34 \, b^{3}\right )} \sqrt {a + b}\right )}}{3 \, {\left ({\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{2} + 2 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} \sqrt {a + b} + a - 3 \, b\right )}^{3}} \]
-1/2*sqrt(a + b)*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))) + 1/2*sqrt(a + b)*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*sqrt(a + b)*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))) - 4/3*(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))^5*(a + 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))^4*(3*a + 2*b)*sqrt(a + b) + 2*(3*a^2 - 3*a*b - 10*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))^3 - 6*(a^2 + 3*a*b + 6*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))^2*sqrt(a + b) - 3*(3*a^3 + 4*a^2*b - 9*a*b^2 - 26*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)) - (3*a^3 - 17*a*b^2 + 34*b^3) *sqrt(a + b))/((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))^2 + 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))*sqrt(a + b) + a - 3*b)^3
Time = 3.68 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.05 \[ \int \tanh ^3(x) \sqrt {a+b \tanh ^2(x)} \, dx=-\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}-\frac {{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}}{3\,b}-2\,\mathrm {atan}\left (\frac {2\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,\sqrt {-\frac {a}{4}-\frac {b}{4}}}{a+b}\right )\,\sqrt {-\frac {a}{4}-\frac {b}{4}} \]