\(\int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx\) [294]

Optimal result

Integrand size = 19, antiderivative size = 139 \[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \text {arctanh}(1-d-d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3} \] Output:

1/12*b*x^4-1/3*x^3*arctanh(-1+d+d*tanh(b*x+a))-1/6*x^3*ln(1+(1-d)*exp(2*b* 
x+2*a))-1/4*x^2*polylog(2,-(1-d)*exp(2*b*x+2*a))/b+1/4*x*polylog(3,-(1-d)* 
exp(2*b*x+2*a))/b^2-1/8*polylog(4,-(1-d)*exp(2*b*x+2*a))/b^3
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88 \[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {8 b^3 x^3 \text {arctanh}(1-d-d \tanh (a+b x))-4 b^3 x^3 \log \left (1-\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {e^{-2 (a+b x)}}{-1+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {e^{-2 (a+b x)}}{-1+d}\right )+3 \operatorname {PolyLog}\left (4,\frac {e^{-2 (a+b x)}}{-1+d}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTanh[1 - d - d*Tanh[a + b*x]],x]
 

Output:

(8*b^3*x^3*ArcTanh[1 - d - d*Tanh[a + b*x]] - 4*b^3*x^3*Log[1 - 1/((-1 + d 
)*E^(2*(a + b*x)))] + 6*b^2*x^2*PolyLog[2, 1/((-1 + d)*E^(2*(a + b*x)))] + 
 6*b*x*PolyLog[3, 1/((-1 + d)*E^(2*(a + b*x)))] + 3*PolyLog[4, 1/((-1 + d) 
*E^(2*(a + b*x)))])/(24*b^3)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6793, 2615, 2620, 3011, 7163, 2720, 7143}

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.

\(\displaystyle \int x^2 \text {arctanh}(d (-\tanh (a+b x))-d+1) \, dx\)

\(\Big \downarrow \) 6793

\(\displaystyle \frac {1}{3} b \int \frac {x^3}{e^{2 a+2 b x} (1-d)+1}dx+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \int \frac {e^{2 a+2 b x} x^3}{e^{2 a+2 b x} (1-d)+1}dx\right )+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \left (\frac {x^3 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {3 \int x^2 \log \left (e^{2 a+2 b x} (1-d)+1\right )dx}{2 b (1-d)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \left (\frac {x^3 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{2 b (1-d)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \left (\frac {x^3 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{2 b (1-d)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \left (\frac {x^3 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )de^{2 a+2 b x}}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{2 b (1-d)}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{3} x^3 \text {arctanh}(d (-\tanh (a+b x))-d+1)+\frac {1}{3} b \left (\frac {x^4}{4}-(1-d) \left (\frac {x^3 \log \left ((1-d) e^{2 a+2 b x}+1\right )}{2 b (1-d)}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{2 b}\right )}{2 b (1-d)}\right )\right )\)

Input:

Int[x^2*ArcTanh[1 - d - d*Tanh[a + b*x]],x]
 

Output:

(x^3*ArcTanh[1 - d - d*Tanh[a + b*x]])/3 + (b*(x^4/4 - (1 - d)*((x^3*Log[1 
 + (1 - d)*E^(2*a + 2*b*x)])/(2*b*(1 - d)) - (3*(-1/2*(x^2*PolyLog[2, -((1 
 - d)*E^(2*a + 2*b*x))])/b + ((x*PolyLog[3, -((1 - d)*E^(2*a + 2*b*x))])/( 
2*b) - PolyLog[4, -((1 - d)*E^(2*a + 2*b*x))]/(4*b^2))/b))/(2*b*(1 - d)))) 
)/3
 

Defintions of rubi rules used

rule 2615

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

rule 2620

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

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 6793

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tanh[a + b*x]]/(f*( 
m + 1))), x] + Simp[b/(f*(m + 1))   Int[(e + f*x)^(m + 1)/(c - d + c*E^(2*a 
 + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c 
- d)^2, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 3.18 (sec) , antiderivative size = 1668, normalized size of antiderivative = 12.00

method	result	size
risch	\(\text {Expression too large to display}\)	\(1668\)

Input:

int(-x^2*arctanh(-1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/12*b*x^4-1/3*x^3*ln(exp(b*x+a))+1/8/b^3/(d-1)*polylog(4,(d-1)*exp(2*b*x+ 
2*a))+1/6/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x^3-1/6*d/(d-1)*ln(1-(d-1)*exp( 
2*b*x+2*a))*x^3-1/6/b^3*a^3/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)-1/ 
8/b^3*d/(d-1)*polylog(4,(d-1)*exp(2*b*x+2*a))+1/4/b/(d-1)*polylog(2,(d-1)* 
exp(2*b*x+2*a))*x^2-1/3/b^3/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*a^3-1/4/b^3/( 
d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*a^2-1/4/b^2/(d-1)*polylog(3,(d-1)*exp 
(2*b*x+2*a))*x+1/2/b^3*a^3/(d-1)*ln(1+exp(b*x+a)*(d-1)^(1/2))+1/2/b^3*a^3/ 
(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))+1/2/b^3*a^2/(d-1)*dilog(1-exp(b*x+a)*(d 
-1)^(1/2))+1/2/b^3*a^2/(d-1)*dilog(1+exp(b*x+a)*(d-1)^(1/2))+1/6/b^3*d*a^3 
/(d-1)*ln(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1)-1/2/b^3*d*a^3/(d-1)*ln(1+exp( 
b*x+a)*(d-1)^(1/2))-1/2/b^3*d*a^3/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))-1/2/b 
^3*d*a^2/(d-1)*dilog(1-exp(b*x+a)*(d-1)^(1/2))-1/2/b^3*d*a^2/(d-1)*dilog(1 
+exp(b*x+a)*(d-1)^(1/2))-1/4/b*d/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*x^2 
+1/3/b^3*d/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*a^3+1/4/b^3*d/(d-1)*polylog(2, 
(d-1)*exp(2*b*x+2*a))*a^2+1/4/b^2*d/(d-1)*polylog(3,(d-1)*exp(2*b*x+2*a))* 
x-1/2/b^2/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*a^2*x+1/2/b^2*a^2/(d-1)*x*ln(1+ 
exp(b*x+a)*(d-1)^(1/2))+1/2/b^2*a^2/(d-1)*x*ln(1-exp(b*x+a)*(d-1)^(1/2))-1 
/12*(-I*Pi*csgn(I*(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1))*csgn(I/(exp(2*b*x+2 
*a)+1)*(d*exp(2*b*x+2*a)-exp(2*b*x+2*a)-1))^2+2*ln(d)-2*I*Pi+I*Pi*csgn(I/( 
exp(2*b*x+2*a)+1)*d*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (109) = 218\).

Time = 0.10 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.59 \[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\frac {b^{4} x^{4} - 2 \, b^{3} x^{3} \log \left (-\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) + 12 \, b x {\rm polylog}\left (3, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}} \] Input:

integrate(-x^2*arctanh(-1+d+d*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/12*(b^4*x^4 - 2*b^3*x^3*log(-(d*cosh(b*x + a) + d*sinh(b*x + a))/((d - 2 
)*cosh(b*x + a) + d*sinh(b*x + a))) - 6*b^2*x^2*dilog(sqrt(d - 1)*(cosh(b* 
x + a) + sinh(b*x + a))) - 6*b^2*x^2*dilog(-sqrt(d - 1)*(cosh(b*x + a) + s 
inh(b*x + a))) + 2*a^3*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 1)*sinh(b*x + 
a) + 2*sqrt(d - 1)) + 2*a^3*log(2*(d - 1)*cosh(b*x + a) + 2*(d - 1)*sinh(b 
*x + a) - 2*sqrt(d - 1)) + 12*b*x*polylog(3, sqrt(d - 1)*(cosh(b*x + a) + 
sinh(b*x + a))) + 12*b*x*polylog(3, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x 
 + a))) - 2*(b^3*x^3 + a^3)*log(sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a) 
) + 1) - 2*(b^3*x^3 + a^3)*log(-sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a) 
) + 1) - 12*polylog(4, sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - 12*p 
olylog(4, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))))/b^3
 

Sympy [F]

\[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=- \int x^{2} \operatorname {atanh}{\left (d \tanh {\left (a + b x \right )} + d - 1 \right )}\, dx \] Input:

integrate(-x**2*atanh(-1+d+d*tanh(b*x+a)),x)
 

Output:

-Integral(x**2*atanh(d*tanh(a + b*x) + d - 1), x)
 

Maxima [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88 \[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=-\frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d - 1\right ) + \frac {1}{36} \, {\left (\frac {3 \, x^{4}}{d} - \frac {2 \, {\left (4 \, b^{3} x^{3} \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d \] Input:

integrate(-x^2*arctanh(-1+d+d*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

-1/3*x^3*arctanh(d*tanh(b*x + a) + d - 1) + 1/36*(3*x^4/d - 2*(4*b^3*x^3*l 
og(-(d - 1)*e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog((d - 1)*e^(2*b*x + 2*a) 
) - 6*b*x*polylog(3, (d - 1)*e^(2*b*x + 2*a)) + 3*polylog(4, (d - 1)*e^(2* 
b*x + 2*a)))/(b^4*d))*b*d
 

Giac [F]

\[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\int { -x^{2} \operatorname {artanh}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \,d x } \] Input:

integrate(-x^2*arctanh(-1+d+d*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(-x^2*arctanh(d*tanh(b*x + a) + d - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=\int -x^2\,\mathrm {atanh}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \] Input:

int(-x^2*atanh(d + d*tanh(a + b*x) - 1),x)
 

Output:

int(-x^2*atanh(d + d*tanh(a + b*x) - 1), x)
 

Reduce [F]

\[ \int x^2 \text {arctanh}(1-d-d \tanh (a+b x)) \, dx=-\left (\int \mathit {atanh} \left (\tanh \left (b x +a \right ) d +d -1\right ) x^{2}d x \right ) \] Input:

int(-x^2*atanh(-1+d+d*tanh(b*x+a)),x)
 

Output:

 - int(atanh(tanh(a + b*x)*d + d - 1)*x**2,x)