\(\int \frac {(e x)^m (a+b x^n)^2 (A+B x^n)}{c+d x^n} \, dx\) [41]

Optimal result

Integrand size = 31, antiderivative size = 193 \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {\left (a^2 B d^2+b^2 c (B c-A d)-2 a b d (B c-A d)\right ) (e x)^{1+m}}{d^3 e (1+m)}-\frac {b (b B c-A b d-2 a B d) x^n (e x)^{1+m}}{d^2 e (1+m+n)}+\frac {b^2 B x^{2 n} (e x)^{1+m}}{d e (1+m+2 n)}-\frac {(b c-a d)^2 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c d^3 e (1+m)} \] Output:

(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)-b* 
(-A*b*d-2*B*a*d+B*b*c)*x^n*(e*x)^(1+m)/d^2/e/(1+m+n)+b^2*B*x^(2*n)*(e*x)^( 
1+m)/d/e/(1+m+2*n)-(-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)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {x (e x)^m \left (\frac {a^2 B d^2+b^2 c (B c-A d)+2 a b d (-B c+A d)}{1+m}+\frac {b d (-b B c+A b d+2 a B d) x^n}{1+m+n}+\frac {b^2 B d^2 x^{2 n}}{1+m+2 n}-\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c (1+m)}\right )}{d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1040, 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.

\(\displaystyle \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx\)

\(\Big \downarrow \) 1040

\(\displaystyle \int \left (\frac {(e x)^m \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3}+\frac {(e x)^m (a d-b c)^2 (A d-B c)}{d^3 \left (c+d x^n\right )}+\frac {b x^n (e x)^m (2 a B d+A b d-b B c)}{d^2}+\frac {b^2 B x^{2 n} (e x)^m}{d}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac {(e x)^{m+1} (b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right )}{c d^3 e (m+1)}-\frac {b x^{n+1} (e x)^m (-2 a B d-A b d+b B c)}{d^2 (m+n+1)}+\frac {b^2 B x^{2 n+1} (e x)^m}{d (m+2 n+1)}\)

Input:

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

Output:

-((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))
 

Defintions of rubi rules used

rule 1040

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]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \] Input:

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

Output:

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)*(e*x)^m/(d*x^n + c), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.73 (sec) , antiderivative size = 1402, normalized size of antiderivative = 7.26 \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\text {Too large to display} \] Input:

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

Output:

A*a**2*c**(m/n + 1/n)*c**(-m/n - 1 - 1/n)*e**m*m*x**(m + 1)*lerchphi(d*x** 
n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(n**2*gamma(m/n + 1 + 
1/n)) + A*a**2*c**(m/n + 1/n)*c**(-m/n - 1 - 1/n)*e**m*x**(m + 1)*lerchphi 
(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(n**2*gamma(m/n 
+ 1 + 1/n)) + 2*A*a*b*c**(-m/n - 2 - 1/n)*c**(m/n + 1 + 1/n)*e**m*m*x**(m 
+ n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 
1 + 1/n)/(n**2*gamma(m/n + 2 + 1/n)) + 2*A*a*b*c**(-m/n - 2 - 1/n)*c**(m/n 
 + 1 + 1/n)*e**m*x**(m + n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n 
+ 1 + 1/n)*gamma(m/n + 1 + 1/n)/(n*gamma(m/n + 2 + 1/n)) + 2*A*a*b*c**(-m/ 
n - 2 - 1/n)*c**(m/n + 1 + 1/n)*e**m*x**(m + n + 1)*lerchphi(d*x**n*exp_po 
lar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(n**2*gamma(m/n + 2 + 
1/n)) + A*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m*m*x**(m + 2*n + 
 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1 
/n)/(n**2*gamma(m/n + 3 + 1/n)) + 2*A*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 
 + 1/n)*e**m*x**(m + 2*n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 
2 + 1/n)*gamma(m/n + 2 + 1/n)/(n*gamma(m/n + 3 + 1/n)) + A*b**2*c**(-m/n - 
 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m*x**(m + 2*n + 1)*lerchphi(d*x**n*exp_pol 
ar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(n**2*gamma(m/n + 3 + 1 
/n)) + B*a**2*c**(-m/n - 2 - 1/n)*c**(m/n + 1 + 1/n)*e**m*m*x**(m + n + 1) 
*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1...
 

Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{2} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^2}{c+d\,x^n} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{c+d x^n} \, dx =\text {Too large to display} \] Input:

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

Output:

(e**m*(x**(m + 2*n)*b**3*d**2*m**2*x + x**(m + 2*n)*b**3*d**2*m*n*x + 2*x* 
*(m + 2*n)*b**3*d**2*m*x + x**(m + 2*n)*b**3*d**2*n*x + x**(m + 2*n)*b**3* 
d**2*x + 3*x**(m + n)*a*b**2*d**2*m**2*x + 6*x**(m + n)*a*b**2*d**2*m*n*x 
+ 6*x**(m + n)*a*b**2*d**2*m*x + 6*x**(m + n)*a*b**2*d**2*n*x + 3*x**(m + 
n)*a*b**2*d**2*x - x**(m + n)*b**3*c*d*m**2*x - 2*x**(m + n)*b**3*c*d*m*n* 
x - 2*x**(m + n)*b**3*c*d*m*x - 2*x**(m + n)*b**3*c*d*n*x - x**(m + n)*b** 
3*c*d*x + 3*x**m*a**2*b*d**2*m**2*x + 9*x**m*a**2*b*d**2*m*n*x + 6*x**m*a* 
*2*b*d**2*m*x + 6*x**m*a**2*b*d**2*n**2*x + 9*x**m*a**2*b*d**2*n*x + 3*x** 
m*a**2*b*d**2*x - 3*x**m*a*b**2*c*d*m**2*x - 9*x**m*a*b**2*c*d*m*n*x - 6*x 
**m*a*b**2*c*d*m*x - 6*x**m*a*b**2*c*d*n**2*x - 9*x**m*a*b**2*c*d*n*x - 3* 
x**m*a*b**2*c*d*x + x**m*b**3*c**2*m**2*x + 3*x**m*b**3*c**2*m*n*x + 2*x** 
m*b**3*c**2*m*x + 2*x**m*b**3*c**2*n**2*x + 3*x**m*b**3*c**2*n*x + x**m*b* 
*3*c**2*x + int(x**m/(x**n*d + c),x)*a**3*d**3*m**3 + 3*int(x**m/(x**n*d + 
 c),x)*a**3*d**3*m**2*n + 3*int(x**m/(x**n*d + c),x)*a**3*d**3*m**2 + 2*in 
t(x**m/(x**n*d + c),x)*a**3*d**3*m*n**2 + 6*int(x**m/(x**n*d + c),x)*a**3* 
d**3*m*n + 3*int(x**m/(x**n*d + c),x)*a**3*d**3*m + 2*int(x**m/(x**n*d + c 
),x)*a**3*d**3*n**2 + 3*int(x**m/(x**n*d + c),x)*a**3*d**3*n + int(x**m/(x 
**n*d + c),x)*a**3*d**3 - 3*int(x**m/(x**n*d + c),x)*a**2*b*c*d**2*m**3 - 
9*int(x**m/(x**n*d + c),x)*a**2*b*c*d**2*m**2*n - 9*int(x**m/(x**n*d + c), 
x)*a**2*b*c*d**2*m**2 - 6*int(x**m/(x**n*d + c),x)*a**2*b*c*d**2*m*n**2...