3.1.23 \(\int \frac {(e x)^m (a+b x^n)^2 (A+B x^n)}{c+d x^n} \, dx\) [23]

Optimal. Leaf size=187 \[ -\frac {b (b B c-A b d-2 a B d) x^{1+n} (e x)^m}{d^2 (1+m+n)}+\frac {b^2 B x^{1+2 n} (e x)^m}{d (1+m+2 n)}+\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}-\frac {(b c-a d)^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c d^3 e (1+m)} \]

[Out]

-b*(-A*b*d-2*B*a*d+B*b*c)*x^(1+n)*(e*x)^m/d^2/(1+m+n)+b^2*B*x^(1+2*n)*(e*x)^m/d/(1+m+2*n)+(a^2*B*d^2+b^2*c*(-A
*d+B*c)-2*a*b*d*(-A*d+B*c))*(e*x)^(1+m)/d^3/e/(1+m)-(-a*d+b*c)^2*(-A*d+B*c)*(e*x)^(1+m)*hypergeom([1, (1+m)/n]
,[(1+m+n)/n],-d*x^n/c)/c/d^3/e/(1+m)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {584, 20, 30, 371} \begin {gather*} \frac {(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c d^3 e (m+1)}-\frac {b x^{n+1} (e x)^m (-2 a B d-A b d+b B c)}{d^2 (m+n+1)}+\frac {b^2 B x^{2 n+1} (e x)^m}{d (m+2 n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(a + b*x^n)^2*(A + B*x^n))/(c + d*x^n),x]

[Out]

-((b*(b*B*c - A*b*d - 2*a*B*d)*x^(1 + n)*(e*x)^m)/(d^2*(1 + m + n))) + (b^2*B*x^(1 + 2*n)*(e*x)^m)/(d*(1 + m +
 2*n)) + ((a^2*B*d^2 + b^2*c*(B*c - A*d) - 2*a*b*d*(B*c - A*d))*(e*x)^(1 + m))/(d^3*e*(1 + m)) - ((b*c - a*d)^
2*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*d^3*e*(1 + m))

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 584

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

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx &=\int \left (\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^m}{d^3}+\frac {b (-b B c+A b d+2 a B d) x^n (e x)^m}{d^2}+\frac {b^2 B x^{2 n} (e x)^m}{d}+\frac {(-b c+a d)^2 (-B c+A d) (e x)^m}{d^3 \left (c+d x^n\right )}\right ) \, dx\\ &=\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}+\frac {\left (b^2 B\right ) \int x^{2 n} (e x)^m \, dx}{d}-\frac {\left ((b c-a d)^2 (B c-A d)\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{d^3}-\frac {(b (b B c-A b d-2 a B d)) \int x^n (e x)^m \, dx}{d^2}\\ &=\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}-\frac {(b c-a d)^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c d^3 e (1+m)}+\frac {\left (b^2 B x^{-m} (e x)^m\right ) \int x^{m+2 n} \, dx}{d}-\frac {\left (b (b B c-A b d-2 a B d) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{d^2}\\ &=-\frac {b (b B c-A b d-2 a B d) x^{1+n} (e x)^m}{d^2 (1+m+n)}+\frac {b^2 B x^{1+2 n} (e x)^m}{d (1+m+2 n)}+\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}-\frac {(b c-a d)^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c d^3 e (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.33, size = 156, normalized size = 0.83 \begin {gather*} x (e x)^m \left (\frac {(b c-a d)^2 (B c-A d)}{c d^3 (1+m)}+\frac {a^2 A}{c+c m}+\frac {b (-b B c+A b d+2 a B d) x^n}{d^2 (1+m+n)}+\frac {b^2 B x^{2 n}}{d+d m+2 d n}+\frac {(b c-a d)^2 (-B c+A d) \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c d^3 (1+m)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(a + b*x^n)^2*(A + B*x^n))/(c + d*x^n),x]

[Out]

x*(e*x)^m*(((b*c - a*d)^2*(B*c - A*d))/(c*d^3*(1 + m)) + (a^2*A)/(c + c*m) + (b*(-(b*B*c) + A*b*d + 2*a*B*d)*x
^n)/(d^2*(1 + m + n)) + (b^2*B*x^(2*n))/(d + d*m + 2*d*n) + ((b*c - a*d)^2*(-(B*c) + A*d)*Hypergeometric2F1[1,
 (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*d^3*(1 + m)))

________________________________________________________________________________________

Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{c +d \,x^{n}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)/(c+d*x^n),x)

[Out]

int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)/(c+d*x^n),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)/(c+d*x^n),x, algorithm="maxima")

[Out]

((b^2*c^2*d*e^m - 2*a*b*c*d^2*e^m + a^2*d^3*e^m)*A - (b^2*c^3*e^m - 2*a*b*c^2*d*e^m + a^2*c*d^2*e^m)*B)*integr
ate(x^m/(d^4*x^n + c*d^3), x) + ((m^2*e^m + m*(n + 2)*e^m + (n + 1)*e^m)*B*b^2*d^2*x*e^(m*log(x) + 2*n*log(x))
 - (((m^2*e^m + m*(3*n + 2)*e^m + (2*n^2 + 3*n + 1)*e^m)*b^2*c*d - 2*(m^2*e^m + m*(3*n + 2)*e^m + (2*n^2 + 3*n
 + 1)*e^m)*a*b*d^2)*A - ((m^2*e^m + m*(3*n + 2)*e^m + (2*n^2 + 3*n + 1)*e^m)*b^2*c^2 - 2*(m^2*e^m + m*(3*n + 2
)*e^m + (2*n^2 + 3*n + 1)*e^m)*a*b*c*d + (m^2*e^m + m*(3*n + 2)*e^m + (2*n^2 + 3*n + 1)*e^m)*a^2*d^2)*B)*x*x^m
 + ((m^2*e^m + 2*m*(n + 1)*e^m + (2*n + 1)*e^m)*A*b^2*d^2 - ((m^2*e^m + 2*m*(n + 1)*e^m + (2*n + 1)*e^m)*b^2*c
*d - 2*(m^2*e^m + 2*m*(n + 1)*e^m + (2*n + 1)*e^m)*a*b*d^2)*B)*x*e^(m*log(x) + n*log(x)))/((m^3 + 3*m^2*(n + 1
) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*d^3)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)/(c+d*x^n),x, algorithm="fricas")

[Out]

integral((B*b^2*x^(3*n) + A*a^2 + (2*B*a*b + A*b^2)*x^(2*n) + (B*a^2 + 2*A*a*b)*x^n)*(x*e)^m/(d*x^n + c), x)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 12.16, size = 1085, normalized size = 5.80 \begin {gather*} \frac {A a^{2} e^{m} m x x^{m} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )} + \frac {A a^{2} e^{m} x x^{m} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )} + \frac {2 A a b e^{m} m x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {2 A a b e^{m} x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {2 A a b e^{m} x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {A b^{2} e^{m} m x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {2 A b^{2} e^{m} x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {A b^{2} e^{m} x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {B a^{2} e^{m} m x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {B a^{2} e^{m} x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {B a^{2} e^{m} x x^{m} x^{n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 1 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 1 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )} + \frac {2 B a b e^{m} m x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {4 B a b e^{m} x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {2 B a b e^{m} x x^{m} x^{2 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 2 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 2 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )} + \frac {B b^{2} e^{m} m x x^{m} x^{3 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 3 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 4 + \frac {1}{n}\right )} + \frac {3 B b^{2} e^{m} x x^{m} x^{3 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 3 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )}{c n \Gamma \left (\frac {m}{n} + 4 + \frac {1}{n}\right )} + \frac {B b^{2} e^{m} x x^{m} x^{3 n} \Phi \left (\frac {d x^{n} e^{i \pi }}{c}, 1, \frac {m}{n} + 3 + \frac {1}{n}\right ) \Gamma \left (\frac {m}{n} + 3 + \frac {1}{n}\right )}{c n^{2} \Gamma \left (\frac {m}{n} + 4 + \frac {1}{n}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(a+b*x**n)**2*(A+B*x**n)/(c+d*x**n),x)

[Out]

A*a**2*e**m*m*x*x**m*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(c*n**2*gamma(m/n + 1 +
 1/n)) + A*a**2*e**m*x*x**m*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(c*n**2*gamma(m/
n + 1 + 1/n)) + 2*A*a*b*e**m*m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1
+ 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 2*A*a*b*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1
+ 1/n)*gamma(m/n + 1 + 1/n)/(c*n*gamma(m/n + 2 + 1/n)) + 2*A*a*b*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*
pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + A*b**2*e**m*m*x*x**m*x**(2*n)*le
rchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n)) + 2*A*b*
*2*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n*gamma(m
/n + 3 + 1/n)) + A*b**2*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n +
2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n)) + B*a**2*e**m*m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n +
 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + B*a**2*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_pol
ar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n*gamma(m/n + 2 + 1/n)) + B*a**2*e**m*x*x**m*x**n*lerchp
hi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 2*B*a*b*e*
*m*m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m
/n + 3 + 1/n)) + 4*B*a*b*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n +
 2 + 1/n)/(c*n*gamma(m/n + 3 + 1/n)) + 2*B*a*b*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n
+ 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n)) + B*b**2*e**m*m*x*x**m*x**(3*n)*lerchphi(d*x**n*
exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n**2*gamma(m/n + 4 + 1/n)) + 3*B*b**2*e**m*x*x**m
*x**(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n*gamma(m/n + 4 + 1/n))
 + B*b**2*e**m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n*
*2*gamma(m/n + 4 + 1/n))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)/(c+d*x^n),x, algorithm="giac")

[Out]

integrate((B*x^n + A)*(b*x^n + a)^2*(x*e)^m/(d*x^n + c), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^2}{c+d\,x^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^m*(A + B*x^n)*(a + b*x^n)^2)/(c + d*x^n),x)

[Out]

int(((e*x)^m*(A + B*x^n)*(a + b*x^n)^2)/(c + d*x^n), x)

________________________________________________________________________________________