\(\int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx\) [210]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c} \] Output:

1/4*(b-2*c)*arctanh(1/2*(b+2*c*tanh(x)^2)/c^(1/2)/(a+b*tanh(x)^2+c*tanh(x) 
^4)^(1/2))/c^(3/2)+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)-1/2*(a+b*tanh(x)^2+c*tan 
h(x)^4)^(1/2)/c
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {1}{4} \left (\frac {(-b+2 c) \text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{c^{3/2}}+\frac {2 \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {a+b+c}}-\frac {2 \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{c}\right ) \] Input:

Integrate[Tanh[x]^5/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]
 

Output:

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

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.68, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 26, 4183, 1578, 1267, 27, 1269, 1092, 219, 1154, 219}

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.

\(\displaystyle \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \tan (i x)^5}{\sqrt {a-b \tan (i x)^2+c \tan (i x)^4}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\tan (i x)^5}{\sqrt {c \tan (i x)^4-b \tan (i x)^2+a}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle -\int \frac {i \tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {c \tanh ^4(x)+b \tanh ^2(x)+a}}d(i \tanh (x))\)

\(\Big \downarrow \) 1578

\(\displaystyle -\frac {1}{2} \int -\frac {\tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )\)

\(\Big \downarrow \) 1267

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {b-(b-2 c) \tanh ^2(x)}{2 \left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {b-(b-2 c) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{2} \left (-\frac {(b-2 c) \int \frac {1}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )+2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (-\frac {2 (b-2 c) \int \frac {1}{\tanh ^2(x)+4 c}d\left (-\frac {b-2 i c \tanh (x)}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}\right )+2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {2 c \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}d\left (-\tanh ^2(x)\right )+\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}-4 c \int \frac {1}{\tanh ^2(x)+4 (a+b+c)}d\frac {2 a+b-i (b+2 c) \tanh (x)}{\sqrt {-c \tanh ^2(x)-i b \tanh (x)+a}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {i (b-2 c) \arctan \left (\frac {\tanh (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}-\frac {2 i c \arctan \left (\frac {\tanh (x)}{2 \sqrt {a+b+c}}\right )}{\sqrt {a+b+c}}}{2 c}-\frac {\sqrt {a-i b \tanh (x)-c \tanh ^2(x)}}{c}\right )\)

Input:

Int[Tanh[x]^5/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]
 

Output:

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

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

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

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-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]
 

rule 1267

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

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(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] &&  !IGtQ[m, 0]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[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] && Int 
egerQ[(m - 1)/2]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4183

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( 
x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] 
 :> Simp[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[n 
2, 2*n] && NeQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(149\)
default	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(149\)

Input:

int(tanh(x)^5/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-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*tanh(x)^2+c*tanh(x)^4)^(1/2)/c+1/4*b/c^(3/2)*ln((1/2*b+c*tan 
h(x)^2)/c^(1/2)+(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2))+1/2/(a+b+c)^(1/2)*arcta 
nh(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*ta 
nh(x)^4)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2075 vs. \(2 (111) = 222\).

Time = 1.15 (sec) , antiderivative size = 8891, normalized size of antiderivative = 65.86 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(x)^5/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="fricas 
")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(tanh(x)**5/(a+b*tanh(x)**2+c*tanh(x)**4)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(tanh(x)^5/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="maxima 
")
 

Output:

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

Giac [F]

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

integrate(tanh(x)^5/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(tanh(x)^5/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2),x)
 

Output:

int(tanh(x)^5/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2), x)
 

Reduce [F]

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

int(tanh(x)^5/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x)
 

Output:

int((sqrt(tanh(x)**4*c + tanh(x)**2*b + a)*tanh(x)**5)/(tanh(x)**4*c + tan 
h(x)**2*b + a),x)