3.1.22 \(\int \frac {(e x)^m (a+b x^n)^3 (A+B x^n)}{c+d x^n} \, dx\) [22]

3.1.22.1 Optimal result

Integrand size = 31, antiderivative size = 272 \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{1+n} (e x)^m}{d^3 (1+m+n)}-\frac {b^2 (b B c-A b d-3 a B d) x^{1+2 n} (e x)^m}{d^2 (1+m+2 n)}+\frac {b^3 B x^{1+3 n} (e x)^m}{d (1+m+3 n)}+\frac {\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac {(b c-a d)^3 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c d^4 e (1+m)} \]

output

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)
 

3.1.22.2 Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {x (e x)^m \left (\frac {a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 c^2 (-B c+A d)+3 a^2 b d^2 (-B c+A d)}{1+m}+\frac {b d \left (3 a^2 B d^2+b^2 c (B c-A d)+3 a b d (-B c+A d)\right ) x^n}{1+m+n}+\frac {b^2 d^2 (-b B c+A b d+3 a B d) x^{2 n}}{1+m+2 n}+\frac {b^3 B d^3 x^{3 n}}{1+m+3 n}+\frac {(b c-a d)^3 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c (1+m)}\right )}{d^4} \]

input

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

output

(x*(e*x)^m*((a^3*B*d^3 + 3*a*b^2*c*d*(B*c - A*d) + b^3*c^2*(-(B*c) + A*d) 
+ 3*a^2*b*d^2*(-(B*c) + A*d))/(1 + m) + (b*d*(3*a^2*B*d^2 + b^2*c*(B*c - A 
*d) + 3*a*b*d*(-(B*c) + A*d))*x^n)/(1 + m + n) + (b^2*d^2*(-(b*B*c) + A*b* 
d + 3*a*B*d)*x^(2*n))/(1 + m + 2*n) + (b^3*B*d^3*x^(3*n))/(1 + m + 3*n) + 
((b*c - a*d)^3*(B*c - A*d)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, 
-((d*x^n)/c)])/(c*(1 + m))))/d^4
 

3.1.22.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 1040

\(\displaystyle \int \left (\frac {b x^n (e x)^m \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3}+\frac {(e x)^m \left (a^3 B d^3-3 a^2 b d^2 (B c-A d)+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4}+\frac {b^2 x^{2 n} (e x)^m (3 a B d+A b d-b B c)}{d^2}+\frac {(e x)^m (a d-b c)^3 (A d-B c)}{d^4 \left (c+d x^n\right )}+\frac {b^3 B x^{3 n} (e x)^m}{d}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b x^{n+1} (e x)^m \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 (m+n+1)}+\frac {(e x)^{m+1} \left (a^3 B d^3-3 a^2 b d^2 (B c-A d)+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}-\frac {b^2 x^{2 n+1} (e x)^m (-3 a B d-A b d+b B c)}{d^2 (m+2 n+1)}+\frac {(e x)^{m+1} (b c-a d)^3 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right )}{c d^4 e (m+1)}+\frac {b^3 B x^{3 n+1} (e x)^m}{d (m+3 n+1)}\)

input

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

output

(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))
 

3.1.22.3.1 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]
 

3.1.22.4 Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{3} \left (A +B \,x^{n}\right )}{c +d \,x^{n}}d x\]

input

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

output

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

3.1.22.5 Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{3} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]

input

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="fricas")
 

output

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

3.1.22.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 16.64 (sec) , antiderivative size = 1933, normalized size of antiderivative = 7.11 \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\text {Too large to display} \]

input

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

output

A*a**3*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**3*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)) + 3*A*a**2*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)) + 3*A*a**2*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)) + 3*A*a**2* 
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**2*gamma(m 
/n + 2 + 1/n)) + 3*A*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)*gamm 
a(m/n + 2 + 1/n)/(n**2*gamma(m/n + 3 + 1/n)) + 6*A*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*p 
i)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(n*gamma(m/n + 3 + 1/n)) + 3* 
A*a*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m*x**(m + 2*n + 1)*lerc 
hphi(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)) + A*b**3*c**(-m/n - 4 - 1/n)*c**(m/n + 3 + 1/n)*e* 
*m*m*x**(m + 3*n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1...
 

3.1.22.7 Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{3} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]

input

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="maxima")
 

output

-((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 + 3*m^2*(n + 1) + (2*n^2 
+ 6*n + 3)*m + 2*n^2 + 3*n + 1)*B*b^3*d^3*e^m*x*e^(m*log(x) + 3*n*log(x)) 
+ (((m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n 
+ 1)*b^3*c^2*d*e^m - 3*(m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3 
)*m + 11*n^2 + 6*n + 1)*a*b^2*c*d^2*e^m + 3*(m^3 + 3*m^2*(2*n + 1) + 6*n^3 
 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*a^2*b*d^3*e^m)*A - ((m^3 + 3* 
m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*b^3*c^3* 
e^m - 3*(m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 
6*n + 1)*a*b^2*c^2*d*e^m + 3*(m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12 
*n + 3)*m + 11*n^2 + 6*n + 1)*a^2*b*c*d^2*e^m - (m^3 + 3*m^2*(2*n + 1) + 6 
*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*a^3*d^3*e^m)*B)*x*x^m + ( 
(m^3 + m^2*(4*n + 3) + (3*n^2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*A*b^3*d^3*e^ 
m - ((m^3 + m^2*(4*n + 3) + (3*n^2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*b^3*c*d 
^2*e^m - 3*(m^3 + m^2*(4*n + 3) + (3*n^2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*a 
*b^2*d^3*e^m)*B)*x*e^(m*log(x) + 2*n*log(x)) - (((m^3 + m^2*(5*n + 3) + (6 
*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*b^3*c*d^2*e^m - 3*(m^3 + m^2*(5*n + 
3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*a*b^2*d^3*e^m)*A - ((m^3 + m^ 
2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*b^3*c^2*d*e^m - 3...
 

3.1.22.8 Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{3} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]

input

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="giac")
 

output

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

3.1.22.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^3}{c+d\,x^n} \,d x \]

input

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

output

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