\(\int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx\) [215]

Optimal result

Integrand size = 21, antiderivative size = 132 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \] 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^(1/2)+1/2*(a+b+c)^(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))-1/2*(a+b*tanh(x)^2+c*ta 
nh(x)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.99 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\frac {1}{4} \left (\frac {(b+2 c) \text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {c}}+2 \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-2 \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}\right ) \] Input:

Integrate[Tanh[x]*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])])/Sqrt[c] + 2*Sqrt[a + b + 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*Sq 
rt[a + b*Tanh[x]^2 + c*Tanh[x]^4])/4
 

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 26, 4183, 1576, 1162, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4183

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

\(\Big \downarrow \) 1576

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

\(\Big \downarrow \) 1162

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1162

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p/(e*(m + 2*p + 1))   Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - 
b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x 
] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && 
!ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

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 1576

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

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 = 2.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(-\frac {\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\tanh \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{\tanh \left (x \right )^{2}-1}\right )}{2}\)	\(165\)
default	\(-\frac {\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\tanh \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{\tanh \left (x \right )^{2}-1}\right )}{2}\)	\(165\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1827 vs. \(2 (108) = 216\).

Time = 1.35 (sec) , antiderivative size = 7896, normalized size of antiderivative = 59.82 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(tanh(x)**4*c + tanh(x)**2*b + a)*b - int((sqrt(tanh(x)**4*c + tan 
h(x)**2*b + a)*tanh(x)**5)/(tanh(x)**4*b*c - 2*tanh(x)**4*c**2 + tanh(x)** 
2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*b**2*c + 4*int((sqrt(tanh(x)** 
4*c + tanh(x)**2*b + a)*tanh(x)**5)/(tanh(x)**4*b*c - 2*tanh(x)**4*c**2 + 
tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*c**3 + int((sqrt(tanh 
(x)**4*c + tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4*b*c - 2*tanh(x)**4*c**2 
+ tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*a*b**2 - 4*int((sqr 
t(tanh(x)**4*c + tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4*b*c - 2*tanh(x)**4 
*c**2 + tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*a*b*c + 4*int 
((sqrt(tanh(x)**4*c + tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4*b*c - 2*tanh( 
x)**4*c**2 + tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*a*c**2 + 
 int((sqrt(tanh(x)**4*c + tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4*b*c - 2*t 
anh(x)**4*c**2 + tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)*b**3 
 - 2*int((sqrt(tanh(x)**4*c + tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4*b*c - 
 2*tanh(x)**4*c**2 + tanh(x)**2*b**2 - 2*tanh(x)**2*b*c + a*b - 2*a*c),x)* 
b**2*c)/(b - 2*c)