\(\int \frac {\tanh ^3(x)}{(a+b \tanh ^2(x))^{5/2}} \, dx\) [249]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}} \]

[Out]

arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)-1/(a+b)^2/(a+b*tanh(x)^2)^(1/2)+1/3*a/b/(a+b)/(a+b*tanh
(x)^2)^(3/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3751, 457, 79, 53, 65, 214} \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}} \]

[In]

Int[Tanh[x]^3/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/(a + b)^(5/2) + a/(3*b*(a + b)*(a + b*Tanh[x]^2)^(3/2)) - 1/((a + b
)^2*Sqrt[a + b*Tanh[x]^2])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

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]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^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] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^3}{\left (1-x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(1-x) (a+b x)^{5/2}} \, dx,x,\tanh ^2(x)\right ) \\ & = \frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{(1-x) (a+b x)^{3/2}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)} \\ & = \frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)^2} \\ & = \frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}}+\frac {\text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tanh ^2(x)}\right )}{b (a+b)^2} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {a}{3 b (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {a (a+b)-3 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right ) \left (a+b \tanh ^2(x)\right )}{3 b (a+b)^2 \left (a+b \tanh ^2(x)\right )^{3/2}} \]

[In]

Integrate[Tanh[x]^3/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

(a*(a + b) - 3*b*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tanh[x]^2)/(a + b)]*(a + b*Tanh[x]^2))/(3*b*(a + b)^2*
(a + b*Tanh[x]^2)^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(434\) vs. \(2(62)=124\).

Time = 0.08 (sec) , antiderivative size = 435, normalized size of antiderivative = 5.88

method	result	size
derivativedivides	\(\frac {1}{3 b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{6 \left (a +b \right ) \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}}}-\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \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}}}-\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\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 \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{6 \left (a +b \right ) \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}}}+\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \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}}}+\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\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 \left (a +b \right )^{\frac {5}{2}}}\)	\(435\)
default	\(\frac {1}{3 b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{6 \left (a +b \right ) \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}}}-\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \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}}}-\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\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 \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{6 \left (a +b \right ) \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}}}+\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \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}}}+\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\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 \left (a +b \right )^{\frac {5}{2}}}\)	\(435\)

[In]

int(tanh(x)^3/(a+b*tanh(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/b/(a+b*tanh(x)^2)^(3/2)-1/6/(a+b)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(3/2)-1/6*b/(a+b)/a/(b*(1+tanh(x))
^2-2*b*(1+tanh(x))+a+b)^(3/2)*tanh(x)-1/3*b/(a+b)/a^2/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)*tanh(x)-1/2/
(a+b)^2/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)-1/2/(a+b)^2/a/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)*
b*tanh(x)+1/2/(a+b)^(5/2)*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/(a+b)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)+1/6*b/(a+b)/a/(b*(tanh(x)-1)^2+2*b*(tanh
(x)-1)+a+b)^(3/2)*tanh(x)+1/3*b/(a+b)/a^2/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)*tanh(x)-1/2/(a+b)^2/(b*(
tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+1/2/(a+b)^2/a/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)*b*tanh(x)+1/
2/(a+b)^(5/2)*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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3028 vs. \(2 (62) = 124\).

Time = 0.66 (sec) , antiderivative size = 6621, normalized size of antiderivative = 89.47 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2)^(5/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tanh(x)**3/(a+b*tanh(x)**2)**(5/2),x)

[Out]

Integral(tanh(x)**3/(a + b*tanh(x)**2)**(5/2), x)

Maxima [F]

\[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)^3/(b*tanh(x)^2 + a)^(5/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 725 vs. \(2 (62) = 124\).

Time = 0.53 (sec) , antiderivative size = 725, normalized size of antiderivative = 9.80 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {{\left ({\left (\frac {{\left (a^{8} b + 2 \, a^{7} b^{2} - 5 \, a^{6} b^{3} - 20 \, a^{5} b^{4} - 25 \, a^{4} b^{5} - 14 \, a^{3} b^{6} - 3 \, a^{2} b^{7}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}} + \frac {3 \, {\left (a^{8} b + 2 \, a^{7} b^{2} - a^{6} b^{3} - 4 \, a^{5} b^{4} - a^{4} b^{5} + 2 \, a^{3} b^{6} + a^{2} b^{7}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} + \frac {3 \, {\left (a^{8} b + 2 \, a^{7} b^{2} - a^{6} b^{3} - 4 \, a^{5} b^{4} - a^{4} b^{5} + 2 \, a^{3} b^{6} + a^{2} b^{7}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} + \frac {a^{8} b + 2 \, a^{7} b^{2} - 5 \, a^{6} b^{3} - 20 \, a^{5} b^{4} - 25 \, a^{4} b^{5} - 14 \, a^{3} b^{6} - 3 \, a^{2} b^{7}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}}{3 \, {\left (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 )}^{\frac {3}{2}}} - \frac {\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 )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} + \frac {\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 )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {\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 )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/3*((((a^8*b + 2*a^7*b^2 - 5*a^6*b^3 - 20*a^5*b^4 - 25*a^4*b^5 - 14*a^3*b^6 - 3*a^2*b^7)*e^(2*x)/(a^8*b^2 + 6
*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8) + 3*(a^8*b + 2*a^7*b^2 - a^6*b^3 - 4*a^
5*b^4 - a^4*b^5 + 2*a^3*b^6 + a^2*b^7)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7
+ a^2*b^8))*e^(2*x) + 3*(a^8*b + 2*a^7*b^2 - a^6*b^3 - 4*a^5*b^4 - a^4*b^5 + 2*a^3*b^6 + a^2*b^7)/(a^8*b^2 +
6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))*e^(2*x) + (a^8*b + 2*a^7*b^2 - 5*a^6*
b^3 - 20*a^5*b^4 - 25*a^4*b^5 - 14*a^3*b^6 - 3*a^2*b^7)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^
4*b^6 + 6*a^3*b^7 + a^2*b^8))/(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b)^(3/2) - 1/2*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)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) + 1/2*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)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) - 1/2*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)))/((a^2 + 2*a*b + b
^2)*sqrt(a + b))

Mupad [B] (verification not implemented)

Time = 4.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,\left (2\,a^2+4\,a\,b+2\,b^2\right )}{2\,{\left (a+b\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}}+\frac {\frac {a}{3\,\left (a+b\right )}-\frac {b\,\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}{{\left (a+b\right )}^2}}{b\,{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}} \]

[In]

int(tanh(x)^3/(a + b*tanh(x)^2)^(5/2),x)

[Out]

atanh(((a + b*tanh(x)^2)^(1/2)*(4*a*b + 2*a^2 + 2*b^2))/(2*(a + b)^(5/2)))/(a + b)^(5/2) + (a/(3*(a + b)) - (b
*(a + b*tanh(x)^2))/(a + b)^2)/(b*(a + b*tanh(x)^2)^(3/2))