3.1.22 \(\int \frac {(e x)^m (a+b x^n)^3 (A+B x^n)}{c+d x^n} \, dx\) [22]
Optimal. Leaf size=272 \[ \frac {b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{1+n} (e x)^m}{d^3 (1+m+n)}-\frac {b^2 (b B c-A b d-3 a B d) x^{1+2 n} (e x)^m}{d^2 (1+m+2 n)}+\frac {b^3 B x^{1+3 n} (e x)^m}{d (1+m+3 n)}+\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {(b c-a d)^3 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c d^4 e (1+m)} \]
[Out]
b*(3*a^2*B*d^2+b^2*c*(-A*d+B*c)-3*a*b*d*(-A*d+B*c))*x^(1+n)*(e*x)^m/d^3/(1+m+n)-b^2*(-A*b*d-3*B*a*d+B*b*c)*x^(
1+2*n)*(e*x)^m/d^2/(1+m+2*n)+b^3*B*x^(1+3*n)*(e*x)^m/d/(1+m+3*n)+(a^3*B*d^3-b^3*c^2*(-A*d+B*c)+3*a*b^2*c*d*(-A
*d+B*c)-3*a^2*b*d^2*(-A*d+B*c))*(e*x)^(1+m)/d^4/e/(1+m)+(-a*d+b*c)^3*(-A*d+B*c)*(e*x)^(1+m)*hypergeom([1, (1+m
)/n],[(1+m+n)/n],-d*x^n/c)/c/d^4/e/(1+m)
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Rubi [A]
time = 0.27, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps
used = 9, 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 {b x^{n+1} (e x)^m \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 (m+n+1)}+\frac {(e x)^{m+1} \left (a^3 B d^3-3 a^2 b d^2 (B c-A d)+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}-\frac {b^2 x^{2 n+1} (e x)^m (-3 a B d-A b d+b B c)}{d^2 (m+2 n+1)}+\frac {(e x)^{m+1} (b c-a d)^3 (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^4 e (m+1)}+\frac {b^3 B x^{3 n+1} (e x)^m}{d (m+3 n+1)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n),x]
[Out]
(b*(3*a^2*B*d^2 + b^2*c*(B*c - A*d) - 3*a*b*d*(B*c - A*d))*x^(1 + n)*(e*x)^m)/(d^3*(1 + m + n)) - (b^2*(b*B*c
- A*b*d - 3*a*B*d)*x^(1 + 2*n)*(e*x)^m)/(d^2*(1 + m + 2*n)) + (b^3*B*x^(1 + 3*n)*(e*x)^m)/(d*(1 + m + 3*n)) +
((a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 3*a*b^2*c*d*(B*c - A*d) - 3*a^2*b*d^2*(B*c - A*d))*(e*x)^(1 + m))/(d^4*e*(
1 + m)) + ((b*c - a*d)^3*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)
])/(c*d^4*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 )^3 \left (A+B x^n\right )}{c+d x^n} \, dx &=\int \left (\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^m}{d^4}+\frac {b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^n (e x)^m}{d^3}+\frac {b^2 (-b B c+A b d+3 a B d) x^{2 n} (e x)^m}{d^2}+\frac {b^3 B x^{3 n} (e x)^m}{d}+\frac {(-b c+a d)^3 (-B c+A d) (e x)^m}{d^4 \left (c+d x^n\right )}\right ) \, dx\\ &=\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {\left (b^3 B\right ) \int x^{3 n} (e x)^m \, dx}{d}+\frac {\left ((b c-a d)^3 (B c-A d)\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{d^4}-\frac {\left (b^2 (b B c-A b d-3 a B d)\right ) \int x^{2 n} (e x)^m \, dx}{d^2}+\frac {\left (b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right )\right ) \int x^n (e x)^m \, dx}{d^3}\\ &=\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {(b c-a d)^3 (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^4 e (1+m)}+\frac {\left (b^3 B x^{-m} (e x)^m\right ) \int x^{m+3 n} \, dx}{d}-\frac {\left (b^2 (b B c-A b d-3 a B d) x^{-m} (e x)^m\right ) \int x^{m+2 n} \, dx}{d^2}+\frac {\left (b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{d^3}\\ &=\frac {b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{1+n} (e x)^m}{d^3 (1+m+n)}-\frac {b^2 (b B c-A b d-3 a B d) x^{1+2 n} (e x)^m}{d^2 (1+m+2 n)}+\frac {b^3 B x^{1+3 n} (e x)^m}{d (1+m+3 n)}+\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {(b c-a d)^3 (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^4 e (1+m)}\\ \end {align*}
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Mathematica [A]
time = 0.62, size = 213, normalized size = 0.78 \begin {gather*} x (e x)^m \left (-\frac {(b c-a d)^3 (B c-A d)}{c d^4 (1+m)}+\frac {a^3 A}{c+c m}+\frac {b \left (3 a^2 B d^2+b^2 c (B c-A d)+3 a b d (-B c+A d)\right ) x^n}{d^3 (1+m+n)}+\frac {b^2 (-b B c+A b d+3 a B d) x^{2 n}}{d^2 (1+m+2 n)}+\frac {b^3 B x^{3 n}}{d+d m+3 d n}+\frac {(b c-a d)^3 (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^4 (1+m)}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n),x]
[Out]
x*(e*x)^m*(-(((b*c - a*d)^3*(B*c - A*d))/(c*d^4*(1 + m))) + (a^3*A)/(c + c*m) + (b*(3*a^2*B*d^2 + b^2*c*(B*c -
A*d) + 3*a*b*d*(-(B*c) + A*d))*x^n)/(d^3*(1 + m + n)) + (b^2*(-(b*B*c) + A*b*d + 3*a*B*d)*x^(2*n))/(d^2*(1 +
m + 2*n)) + (b^3*B*x^(3*n))/(d + d*m + 3*d*n) + ((b*c - a*d)^3*(B*c - A*d)*Hypergeometric2F1[1, (1 + m)/n, (1
+ m + n)/n, -((d*x^n)/c)])/(c*d^4*(1 + m)))
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{3} \left (A +B \,x^{n}\right )}{c +d \,x^{n}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x)
[Out]
int((e*x)^m*(a+b*x^n)^3*(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)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="maxima")
[Out]
-((b^3*c^3*d*e^m - 3*a*b^2*c^2*d^2*e^m + 3*a^2*b*c*d^3*e^m - a^3*d^4*e^m)*A - (b^3*c^4*e^m - 3*a*b^2*c^3*d*e^m
+ 3*a^2*b*c^2*d^2*e^m - a^3*c*d^3*e^m)*B)*integrate(x^m/(d^5*x^n + c*d^4), x) + ((m^3*e^m + 3*m^2*(n + 1)*e^m
+ (2*n^2 + 6*n + 3)*m*e^m + (2*n^2 + 3*n + 1)*e^m)*B*b^3*d^3*x*e^(m*log(x) + 3*n*log(x)) + (((m^3*e^m + 3*m^2
*(2*n + 1)*e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*b^3*c^2*d - 3*(m^3*e^m + 3*m^2*(2
*n + 1)*e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*a*b^2*c*d^2 + 3*(m^3*e^m + 3*m^2*(2*
n + 1)*e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*a^2*b*d^3)*A - ((m^3*e^m + 3*m^2*(2*n
+ 1)*e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*b^3*c^3 - 3*(m^3*e^m + 3*m^2*(2*n + 1)
*e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*a*b^2*c^2*d + 3*(m^3*e^m + 3*m^2*(2*n + 1)*
e^m + (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*a^2*b*c*d^2 - (m^3*e^m + 3*m^2*(2*n + 1)*e^m
+ (11*n^2 + 12*n + 3)*m*e^m + (6*n^3 + 11*n^2 + 6*n + 1)*e^m)*a^3*d^3)*B)*x*x^m + ((m^3*e^m + m^2*(4*n + 3)*e
^m + (3*n^2 + 8*n + 3)*m*e^m + (3*n^2 + 4*n + 1)*e^m)*A*b^3*d^3 - ((m^3*e^m + m^2*(4*n + 3)*e^m + (3*n^2 + 8*n
+ 3)*m*e^m + (3*n^2 + 4*n + 1)*e^m)*b^3*c*d^2 - 3*(m^3*e^m + m^2*(4*n + 3)*e^m + (3*n^2 + 8*n + 3)*m*e^m + (3
*n^2 + 4*n + 1)*e^m)*a*b^2*d^3)*B)*x*e^(m*log(x) + 2*n*log(x)) - (((m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*
n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*b^3*c*d^2 - 3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m +
(6*n^2 + 5*n + 1)*e^m)*a*b^2*d^3)*A - ((m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n
+ 1)*e^m)*b^3*c^2*d - 3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*a*b^2
*c*d^2 + 3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*a^2*b*d^3)*B)*x*e^
(m*log(x) + n*log(x)))/((m^4 + 2*m^3*(3*n + 2) + (11*n^2 + 18*n + 6)*m^2 + 6*n^3 + 2*(3*n^3 + 11*n^2 + 9*n + 2
)*m + 11*n^2 + 6*n + 1)*d^4)
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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)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="fricas")
[Out]
integral((B*b^3*x^(4*n) + A*a^3 + (3*B*a*b^2 + A*b^3)*x^(3*n) + 3*(B*a^2*b + A*a*b^2)*x^(2*n) + (B*a^3 + 3*A*a
^2*b)*x^n)*(x*e)^m/(d*x^n + c), x)
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Sympy [C] Result contains complex when optimal does not.
time = 23.44, size = 1503, normalized size = 5.53 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)**m*(a+b*x**n)**3*(A+B*x**n)/(c+d*x**n),x)
[Out]
A*a**3*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**3*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)) + 3*A*a**2*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)) + 3*A*a**2*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)) + 3*A*a**2*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)) + 3*A*a*b**2*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)) + 6*A*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)) + 3*A*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)) + A*b**3*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*A*b**3*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)) + A
*b**3*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*g
amma(m/n + 4 + 1/n)) + B*a**3*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**3*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)) + B*a**3*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_pola
r(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 3*B*a**2*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))
+ 6*B*a**2*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)) + 3*B*a**2*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)) + 3*B*a*b**2*e**m*m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_p
olar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n**2*gamma(m/n + 4 + 1/n)) + 9*B*a*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)) +
3*B*a*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)) + B*b**3*e**m*m*x*x**m*x**(4*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 4 + 1/n)
*gamma(m/n + 4 + 1/n)/(c*n**2*gamma(m/n + 5 + 1/n)) + 4*B*b**3*e**m*x*x**m*x**(4*n)*lerchphi(d*x**n*exp_polar(
I*pi)/c, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(c*n*gamma(m/n + 5 + 1/n)) + B*b**3*e**m*x*x**m*x**(4*n)*lerch
phi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(c*n**2*gamma(m/n + 5 + 1/n))
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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)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="giac")
[Out]
integrate((B*x^n + A)*(b*x^n + a)^3*(x*e)^m/(d*x^n + c), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^3}{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)^3)/(c + d*x^n),x)
[Out]
int(((e*x)^m*(A + B*x^n)*(a + b*x^n)^3)/(c + d*x^n), x)
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