\(\int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx\) [201]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {c}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}} \]

[Out]

-1/2*arctanh(1/2*(b+2*c*tanh(x)^2)/c^(1/2)/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))/c^(1/2)+1/2*arctanh(1/2*(2*a+b+(
b+2*c)*tanh(x)^2)/(a+b+c)^(1/2)/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))/(a+b+c)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3781, 1265, 857, 635, 212, 738} \[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {\text {arctanh}\left (\frac {2 a+(b+2 c) \tanh ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {c}} \]

[In]

Int[Tanh[x]^3/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]

[Out]

-1/2*ArcTanh[(b + 2*c*Tanh[x]^2)/(2*Sqrt[c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])]/Sqrt[c] + ArcTanh[(2*a + b +
(b + 2*c)*Tanh[x]^2)/(2*Sqrt[a + b + c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])]/(2*Sqrt[a + b + c])

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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1265

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

Rule 3781

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^3}{\left (1+x^2\right ) \sqrt {a-b x^2+c x^4}} \, dx,x,i \tanh (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right ) \\ & = \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {-b-2 c \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )+\text {Subst}\left (\int \frac {1}{4 a+4 b+4 c-x^2} \, dx,x,\frac {2 a+b+(b+2 c) \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right ) \\ & = \frac {\text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {c}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {c}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {a+b+c}}\right ) \]

[In]

Integrate[Tanh[x]^3/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]

[Out]

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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(90\)
default	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(90\)

[In]

int(tanh(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln((1/2*b+c*tanh(x)^2)/c^(1/2)+(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))/c^(1/2)+1/2/(a+b+c)^(1/2)*arctanh(1/2*(
b*tanh(x)^2+2*c*tanh(x)^2+2*a+b)/(a+b+c)^(1/2)/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1518 vs. \(2 (85) = 170\).

Time = 0.94 (sec) , antiderivative size = 6663, normalized size of antiderivative = 63.46 \[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )} + c \tanh ^{4}{\left (x \right )}}}\, dx \]

[In]

integrate(tanh(x)**3/(a+b*tanh(x)**2+c*tanh(x)**4)**(1/2),x)

[Out]

Integral(tanh(x)**3/sqrt(a + b*tanh(x)**2 + c*tanh(x)**4), x)

Maxima [F]

\[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)^3/sqrt(c*tanh(x)^4 + b*tanh(x)^2 + a), x)

Giac [F]

\[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]

[In]

integrate(tanh(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(tanh(x)^3/sqrt(c*tanh(x)^4 + b*tanh(x)^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^3}{\sqrt {c\,{\mathrm {tanh}\left (x\right )}^4+b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \]

[In]

int(tanh(x)^3/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2),x)

[Out]

int(tanh(x)^3/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2), x)