\(\int \frac {(e x)^{-1+2 n}}{(a+b \text {sech}(c+d x^n))^2} \, dx\) [90]

Optimal result

Integrand size = 24, antiderivative size = 717 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\frac {(e x)^{2 n}}{2 a^2 e n}+\frac {b^3 x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d e n}-\frac {2 b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d e n}-\frac {b^3 x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d e n}+\frac {2 b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d e n}-\frac {b^2 x^{-2 n} (e x)^{2 n} \log \left (b+a \cosh \left (c+d x^n\right )\right )}{a^2 \left (a^2-b^2\right ) d^2 e n}+\frac {b^3 x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2 e n}-\frac {2 b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2 e n}-\frac {b^3 x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2 e n}+\frac {2 b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2 e n}+\frac {b^2 x^{-n} (e x)^{2 n} \sinh \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (b+a \cosh \left (c+d x^n\right )\right )} \] Output:

1/2*(e*x)^(2*n)/a^2/e/n+b^3*(e*x)^(2*n)*ln(1+a*exp(c+d*x^n)/(b-(-a^2+b^2)^ 
(1/2)))/a^2/(-a^2+b^2)^(3/2)/d/e/n/(x^n)-2*b*(e*x)^(2*n)*ln(1+a*exp(c+d*x^ 
n)/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d/e/n/(x^n)-b^3*(e*x)^(2*n)* 
ln(1+a*exp(c+d*x^n)/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d/e/n/(x^n) 
+2*b*(e*x)^(2*n)*ln(1+a*exp(c+d*x^n)/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^ 
(1/2)/d/e/n/(x^n)-b^2*(e*x)^(2*n)*ln(b+a*cosh(c+d*x^n))/a^2/(a^2-b^2)/d^2/ 
e/n/(x^(2*n))+b^3*(e*x)^(2*n)*polylog(2,-a*exp(c+d*x^n)/(b-(-a^2+b^2)^(1/2 
)))/a^2/(-a^2+b^2)^(3/2)/d^2/e/n/(x^(2*n))-2*b*(e*x)^(2*n)*polylog(2,-a*ex 
p(c+d*x^n)/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2/e/n/(x^(2*n))-b^ 
3*(e*x)^(2*n)*polylog(2,-a*exp(c+d*x^n)/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^ 
2)^(3/2)/d^2/e/n/(x^(2*n))+2*b*(e*x)^(2*n)*polylog(2,-a*exp(c+d*x^n)/(b+(- 
a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2/e/n/(x^(2*n))+b^2*(e*x)^(2*n)*si 
nh(c+d*x^n)/a/(a^2-b^2)/d/e/n/(x^n)/(b+a*cosh(c+d*x^n))
 

Mathematica [A] (warning: unable to verify)

Time = 4.16 (sec) , antiderivative size = 469, normalized size of antiderivative = 0.65 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\frac {x^{-2 n} (e x)^{2 n} \left (b+a \cosh \left (c+d x^n\right )\right ) \text {sech}^2\left (c+d x^n\right ) \left (\frac {4 b^2 d e^{2 c} x^n \left (b+a \cosh \left (c+d x^n\right )\right )}{\left (a^2-b^2\right ) \left (1+e^{2 c}\right )}+\frac {2 b \left (b+a \cosh \left (c+d x^n\right )\right ) \left (b \sqrt {-a^2+b^2} \log \left (a+2 b e^{c+d x^n}+a e^{2 \left (c+d x^n\right )}\right )+\left (2 a^2-b^2\right ) \left (d x^n \log \left (1+\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )+\operatorname {PolyLog}\left (2,\frac {a e^{c+d x^n}}{-b+\sqrt {-a^2+b^2}}\right )\right )-\left (2 a^2-b^2\right ) \left (d x^n \log \left (1+\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )+\operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )\right )\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {2 b^2 d x^n \text {sech}(c) \left (b \sinh (c)-a \sinh \left (d x^n\right )\right )}{(-a+b) (a+b)}+\frac {2 b^2 d x^n \left (b+a \cosh \left (c+d x^n\right )\right ) \tanh (c)}{-a^2+b^2}+\frac {d x^n \left (b+a \cosh \left (c+d x^n\right )\right ) \left (\left (a^2-b^2\right ) d x^n+2 b^2 \tanh (c)\right )}{(a-b) (a+b)}\right )}{2 a^2 d^2 e n \left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \] Input:

Integrate[(e*x)^(-1 + 2*n)/(a + b*Sech[c + d*x^n])^2,x]
 

Output:

((e*x)^(2*n)*(b + a*Cosh[c + d*x^n])*Sech[c + d*x^n]^2*((4*b^2*d*E^(2*c)*x 
^n*(b + a*Cosh[c + d*x^n]))/((a^2 - b^2)*(1 + E^(2*c))) + (2*b*(b + a*Cosh 
[c + d*x^n])*(b*Sqrt[-a^2 + b^2]*Log[a + 2*b*E^(c + d*x^n) + a*E^(2*(c + d 
*x^n))] + (2*a^2 - b^2)*(d*x^n*Log[1 + (a*E^(c + d*x^n))/(b - Sqrt[-a^2 + 
b^2])] + PolyLog[2, (a*E^(c + d*x^n))/(-b + Sqrt[-a^2 + b^2])]) - (2*a^2 - 
 b^2)*(d*x^n*Log[1 + (a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2])] + PolyLog[2 
, -((a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2]))])))/(-a^2 + b^2)^(3/2) + (2* 
b^2*d*x^n*Sech[c]*(b*Sinh[c] - a*Sinh[d*x^n]))/((-a + b)*(a + b)) + (2*b^2 
*d*x^n*(b + a*Cosh[c + d*x^n])*Tanh[c])/(-a^2 + b^2) + (d*x^n*(b + a*Cosh[ 
c + d*x^n])*((a^2 - b^2)*d*x^n + 2*b^2*Tanh[c]))/((a - b)*(a + b))))/(2*a^ 
2*d^2*e*n*x^(2*n)*(a + b*Sech[c + d*x^n])^2)
 

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 563, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5963, 5959, 3042, 4679, 2009}

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.

\(\displaystyle \int \frac {(e x)^{2 n-1}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx\)

\(\Big \downarrow \) 5963

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^{2 n-1}}{\left (a+b \text {sech}\left (d x^n+c\right )\right )^2}dx}{e}\)

\(\Big \downarrow \) 5959

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{\left (a+b \text {sech}\left (d x^n+c\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{\left (a+b \csc \left (i d x^n+i c+\frac {\pi }{2}\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \left (-\frac {2 b x^n}{a^2 \left (b+a \cosh \left (d x^n+c\right )\right )}+\frac {x^n}{a^2}+\frac {b^2 x^n}{a^2 \left (b+a \cosh \left (d x^n+c\right )\right )^2}\right )dx^n}{e n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (-\frac {2 b \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}+\frac {2 b \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}-\frac {b^2 \log \left (a \cosh \left (c+d x^n\right )+b\right )}{a^2 d^2 \left (a^2-b^2\right )}-\frac {2 b x^n \log \left (\frac {a e^{c+d x^n}}{b-\sqrt {b^2-a^2}}+1\right )}{a^2 d \sqrt {b^2-a^2}}+\frac {2 b x^n \log \left (\frac {a e^{c+d x^n}}{\sqrt {b^2-a^2}+b}+1\right )}{a^2 d \sqrt {b^2-a^2}}+\frac {b^2 x^n \sinh \left (c+d x^n\right )}{a d \left (a^2-b^2\right ) \left (a \cosh \left (c+d x^n\right )+b\right )}+\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}-\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}+\frac {b^3 x^n \log \left (\frac {a e^{c+d x^n}}{b-\sqrt {b^2-a^2}}+1\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}-\frac {b^3 x^n \log \left (\frac {a e^{c+d x^n}}{\sqrt {b^2-a^2}+b}+1\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}+\frac {x^{2 n}}{2 a^2}\right )}{e n}\)

Input:

Int[(e*x)^(-1 + 2*n)/(a + b*Sech[c + d*x^n])^2,x]
 

Output:

((e*x)^(2*n)*(x^(2*n)/(2*a^2) + (b^3*x^n*Log[1 + (a*E^(c + d*x^n))/(b - Sq 
rt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) - (2*b*x^n*Log[1 + (a*E^(c + 
d*x^n))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) - (b^3*x^n*Log[1 
 + (a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + 
 (2*b*x^n*Log[1 + (a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^ 
2 + b^2]*d) - (b^2*Log[b + a*Cosh[c + d*x^n]])/(a^2*(a^2 - b^2)*d^2) + (b^ 
3*PolyLog[2, -((a*E^(c + d*x^n))/(b - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^ 
2)^(3/2)*d^2) - (2*b*PolyLog[2, -((a*E^(c + d*x^n))/(b - Sqrt[-a^2 + b^2]) 
)])/(a^2*Sqrt[-a^2 + b^2]*d^2) - (b^3*PolyLog[2, -((a*E^(c + d*x^n))/(b + 
Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (2*b*PolyLog[2, -((a*E 
^(c + d*x^n))/(b + Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (b^2* 
x^n*Sinh[c + d*x^n])/(a*(a^2 - b^2)*d*(b + a*Cosh[c + d*x^n]))))/(e*n*x^(2 
*n))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 5959

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] 
)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m 
 + 1)/n], 0] && IntegerQ[p]
 

rule 5963

Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), 
x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m* 
(a + b*Sech[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
 

Maple [F]

\[\int \frac {\left (e x \right )^{-1+2 n}}{{\left (a +b \,\operatorname {sech}\left (c +d \,x^{n}\right )\right )}^{2}}d x\]

Input:

int((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x)
 

Output:

int((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9020 vs. \(2 (681) = 1362\).

Time = 0.26 (sec) , antiderivative size = 9020, normalized size of antiderivative = 12.58 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\int \frac {\left (e x\right )^{2 n - 1}}{\left (a + b \operatorname {sech}{\left (c + d x^{n} \right )}\right )^{2}}\, dx \] Input:

integrate((e*x)**(-1+2*n)/(a+b*sech(c+d*x**n))**2,x)
 

Output:

Integral((e*x)**(2*n - 1)/(a + b*sech(c + d*x**n))**2, x)
 

Maxima [F]

\[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{{\left (b \operatorname {sech}\left (d x^{n} + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x, algorithm="maxima")
 

Output:

-1/2*(4*a*b^2*e^(2*n)*x^n - (a^3*d*e^(2*n) - a*b^2*d*e^(2*n))*x^(2*n) - (a 
^3*d*e^(2*n)*e^(2*c) - a*b^2*d*e^(2*n)*e^(2*c))*e^(2*d*x^n + 2*n*log(x)) + 
 2*(2*b^3*e^(2*n)*e^(n*log(x) + c) - (a^2*b*d*e^(2*n)*e^c - b^3*d*e^(2*n)* 
e^c)*x^(2*n))*e^(d*x^n))/(a^5*d*e*n - a^3*b^2*d*e*n + (a^5*d*e*n*e^(2*c) - 
 a^3*b^2*d*e*n*e^(2*c))*e^(2*d*x^n) + 2*(a^4*b*d*e*n*e^c - a^2*b^3*d*e*n*e 
^c)*e^(d*x^n)) - integrate(-2*(a*b^2*e^(2*n)*x^n + (b^3*e^(2*n)*e^(n*log(x 
) + c) - (2*a^2*b*d*e^(2*n)*e^c - b^3*d*e^(2*n)*e^c)*x^(2*n))*e^(d*x^n))/( 
(a^5*d*e*e^(2*c) - a^3*b^2*d*e*e^(2*c))*x*e^(2*d*x^n) + 2*(a^4*b*d*e*e^c - 
 a^2*b^3*d*e*e^c)*x*e^(d*x^n) + (a^5*d*e - a^3*b^2*d*e)*x), x)
 

Giac [F]

\[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{{\left (b \operatorname {sech}\left (d x^{n} + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((e*x)^(2*n - 1)/(b*sech(d*x^n + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\int \frac {{\left (e\,x\right )}^{2\,n-1}}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x^n\right )}\right )}^2} \,d x \] Input:

int((e*x)^(2*n - 1)/(a + b/cosh(c + d*x^n))^2,x)
 

Output:

int((e*x)^(2*n - 1)/(a + b/cosh(c + d*x^n))^2, x)
 

Reduce [F]

\[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \text {sech}\left (c+d x^n\right )\right )^2} \, dx=\text {too large to display} \] Input:

int((e*x)^(-1+2*n)/(a+b*sech(c+d*x^n))^2,x)
 

Output:

(e**(2*n)*( - 8*e**(2*x**n*d + 2*c)*sqrt(a**2 - b**2)*atan((e**(x**n*d + c 
)*a + b)/sqrt(a**2 - b**2))*a**3*b + 4*e**(2*x**n*d + 2*c)*sqrt(a**2 - b** 
2)*atan((e**(x**n*d + c)*a + b)/sqrt(a**2 - b**2))*a*b**3 - 16*e**(x**n*d 
+ c)*sqrt(a**2 - b**2)*atan((e**(x**n*d + c)*a + b)/sqrt(a**2 - b**2))*a** 
2*b**2 + 8*e**(x**n*d + c)*sqrt(a**2 - b**2)*atan((e**(x**n*d + c)*a + b)/ 
sqrt(a**2 - b**2))*b**4 - 8*sqrt(a**2 - b**2)*atan((e**(x**n*d + c)*a + b) 
/sqrt(a**2 - b**2))*a**3*b + 4*sqrt(a**2 - b**2)*atan((e**(x**n*d + c)*a + 
 b)/sqrt(a**2 - b**2))*a*b**3 - 16*e**(2*x**n*d + 3*c)*int((x**(2*n)*e**(x 
**n*d))/(e**(4*x**n*d + 4*c)*a**2*x + 4*e**(3*x**n*d + 3*c)*a*b*x + 2*e**( 
2*x**n*d + 2*c)*a**2*x + 4*e**(2*x**n*d + 2*c)*b**2*x + 4*e**(x**n*d + c)* 
a*b*x + a**2*x),x)*a**6*b*d**2*n + 24*e**(2*x**n*d + 3*c)*int((x**(2*n)*e* 
*(x**n*d))/(e**(4*x**n*d + 4*c)*a**2*x + 4*e**(3*x**n*d + 3*c)*a*b*x + 2*e 
**(2*x**n*d + 2*c)*a**2*x + 4*e**(2*x**n*d + 2*c)*b**2*x + 4*e**(x**n*d + 
c)*a*b*x + a**2*x),x)*a**4*b**3*d**2*n - 8*e**(2*x**n*d + 3*c)*int((x**(2* 
n)*e**(x**n*d))/(e**(4*x**n*d + 4*c)*a**2*x + 4*e**(3*x**n*d + 3*c)*a*b*x 
+ 2*e**(2*x**n*d + 2*c)*a**2*x + 4*e**(2*x**n*d + 2*c)*b**2*x + 4*e**(x**n 
*d + c)*a*b*x + a**2*x),x)*a**2*b**5*d**2*n + x**(2*n)*e**(2*x**n*d + 2*c) 
*a**5*d**2 - x**(2*n)*e**(2*x**n*d + 2*c)*a**3*b**2*d**2 - 4*x**n*e**(2*x* 
*n*d + 2*c)*a**3*b**2*d + 4*x**n*e**(2*x**n*d + 2*c)*a*b**4*d + 2*e**(2*x* 
*n*d + 2*c)*log(e**(2*x**n*d + 2*c)*a + 2*e**(x**n*d + c)*b + a)*a**3*b...