\(\int \frac {\tanh (x)}{(a+b \tanh ^4(x))^{5/2}} \, dx\) [263]

Optimal result

Integrand size = 15, antiderivative size = 118 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}} \]

[Out]

1/2*arctanh((a+b*tanh(x)^2)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))/(a+b)^(5/2)+1/6*(-3*a^2+b*(5*a+2*b)*tanh(x)^2)/
a^2/(a+b)^2/(a+b*tanh(x)^4)^(1/2)+1/6*(-a+b*tanh(x)^2)/a/(a+b)/(a+b*tanh(x)^4)^(3/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3751, 1262, 755, 837, 12, 739, 212} \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}} \]

[In]

Int[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])]/(2*(a + b)^(5/2)) - (a - b*Tanh[x]^2)/(6*a*(a +
b)*(a + b*Tanh[x]^4)^(3/2)) - (3*a^2 - b*(5*a + 2*b)*Tanh[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

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}{\left (1-x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tanh (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(1-x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh ^2(x)\right ) \\ & = -\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {\text {Subst}\left (\int \frac {-3 a-2 b+2 b x}{(1-x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh ^2(x)\right )}{6 a (a+b)} \\ & = -\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\text {Subst}\left (\int \frac {3 a^2 b}{(1-x) \sqrt {a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{6 a^2 b (a+b)^2} \\ & = -\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x^2}} \, dx,x,\tanh ^2(x)\right )}{2 (a+b)^2} \\ & = -\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}}-\frac {\text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {-a-b \tanh ^2(x)}{\sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^2} \\ & = \frac {\text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {a-b \tanh ^2(x)}{6 a (a+b) \left (a+b \tanh ^4(x)\right )^{3/2}}-\frac {3 a^2-b (5 a+2 b) \tanh ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tanh ^4(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\frac {1}{6} \left (\frac {3 \text {arctanh}\left (\frac {a+b \tanh ^2(x)}{\sqrt {a+b} \sqrt {a+b \tanh ^4(x)}}\right )}{(a+b)^{5/2}}+\frac {-a^2 (4 a+b)+3 a b (2 a+b) \tanh ^2(x)-3 a^2 b \tanh ^4(x)+b^2 (5 a+2 b) \tanh ^6(x)}{a^2 (a+b)^2 \left (a+b \tanh ^4(x)\right )^{3/2}}\right ) \]

[In]

Integrate[Tanh[x]/(a + b*Tanh[x]^4)^(5/2),x]

[Out]

((3*ArcTanh[(a + b*Tanh[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tanh[x]^4])])/(a + b)^(5/2) + (-(a^2*(4*a + b)) + 3*a*b*
(2*a + b)*Tanh[x]^2 - 3*a^2*b*Tanh[x]^4 + b^2*(5*a + 2*b)*Tanh[x]^6)/(a^2*(a + b)^2*(a + b*Tanh[x]^4)^(3/2)))/
6

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.71 (sec) , antiderivative size = 637, normalized size of antiderivative = 5.40

method	result	size
derivativedivides	\(-\frac {\left (-\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}+\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}-\frac {\left (\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}+\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (-\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}-\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}\)	\(637\)
default	\(-\frac {\left (-\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}+\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}-\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}-\frac {\left (\frac {\tanh \left (x \right )^{3}}{6 a \left (a +b \right ) b}-\frac {\tanh \left (x \right )^{2}}{6 a \left (a +b \right ) b}+\frac {\tanh \left (x \right )}{6 a \left (a +b \right ) b}+\frac {1}{6 \left (a +b \right ) b^{2}}\right ) \sqrt {a +b \tanh \left (x \right )^{4}}}{2 \left (\tanh \left (x \right )^{4}+\frac {a}{b}\right )^{2}}+\frac {b \left (-\frac {\left (3 a +b \right ) \tanh \left (x \right )^{3}}{8 a^{2} \left (a +b \right )^{2}}+\frac {\left (5 a +2 b \right ) \tanh \left (x \right )^{2}}{12 a^{2} \left (a +b \right )^{2}}-\frac {\left (11 a +5 b \right ) \tanh \left (x \right )}{24 a^{2} \left (a +b \right )^{2}}-\frac {1}{4 \left (a +b \right )^{2} b}\right )}{\sqrt {\left (\tanh \left (x \right )^{4}+\frac {a}{b}\right ) b}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {2 b \tanh \left (x \right )^{2}+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tanh \left (x \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tanh \left (x \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, -\frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \tanh \left (x \right )^{4}}}}{2 \left (a +b \right )^{2}}\)	\(637\)

[In]

int(tanh(x)/(a+b*tanh(x)^4)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(-1/6/a/(a+b)/b*tanh(x)^3-1/6/a/(a+b)/b*tanh(x)^2-1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1
/2)/(tanh(x)^4+a/b)^2+b*(1/8*(3*a+b)/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2+1/24/a^2*(11*a
+5*b)/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*
tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh(x)^4)^(1/2))-1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^
(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2)
,-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))-1/2*(1/6/a/(a+b)/b*tanh(x)^3-1/6/a/
(a+b)/b*tanh(x)^2+1/6/a/(a+b)/b*tanh(x)+1/6/(a+b)/b^2)*(a+b*tanh(x)^4)^(1/2)/(tanh(x)^4+a/b)^2+b*(-1/8*(3*a+b)
/a^2/(a+b)^2*tanh(x)^3+1/12*(5*a+2*b)/a^2/(a+b)^2*tanh(x)^2-1/24/a^2*(11*a+5*b)/(a+b)^2*tanh(x)-1/4/(a+b)^2/b)
/((tanh(x)^4+a/b)*b)^(1/2)-1/2/(a+b)^2*(-1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*tanh(x)^2+2*a)/(a+b)^(1/2)/(a+b*tanh
(x)^4)^(1/2))+1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tanh(x)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tanh(x)^2
)^(1/2)/(a+b*tanh(x)^4)^(1/2)*EllipticPi(tanh(x)*(I/a^(1/2)*b^(1/2))^(1/2),-I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1
/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8210 vs. \(2 (102) = 204\).

Time = 2.04 (sec) , antiderivative size = 16463, normalized size of antiderivative = 139.52 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int \frac {\tanh {\left (x \right )}}{\left (a + b \tanh ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tanh(x)/(a+b*tanh(x)**4)**(5/2),x)

[Out]

Integral(tanh(x)/(a + b*tanh(x)**4)**(5/2), x)

Maxima [F]

\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)

Giac [F]

\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{4} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tanh(x)/(a+b*tanh(x)^4)^(5/2),x, algorithm="giac")

[Out]

integrate(tanh(x)/(b*tanh(x)^4 + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^4(x)\right )^{5/2}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{{\left (b\,{\mathrm {tanh}\left (x\right )}^4+a\right )}^{5/2}} \,d x \]

[In]

int(tanh(x)/(a + b*tanh(x)^4)^(5/2),x)

[Out]

int(tanh(x)/(a + b*tanh(x)^4)^(5/2), x)