\(\int \frac {(c x)^m (A+B x^n)}{(a+b x^n)^3} \, dx\) [409]

Optimal result

Integrand size = 22, antiderivative size = 112 \[ \int \frac {(c x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3} \, dx=\frac {(A b-a B) (c x)^{1+m}}{2 a b c n \left (a+b x^n\right )^2}+\frac {(a B (1+m)-A b (1+m-2 n)) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{2 a^3 b c (1+m) n} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.74 \[ \int \frac {(c x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3} \, dx=\frac {x (c x)^m \left (a B \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )+(A b-a B) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )\right )}{a^3 b (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {957, 888}

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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {(a B (m+1)-A b (m-2 n+1)) \int \frac {(c x)^m}{\left (b x^n+a\right )^2}dx}{2 a b n}+\frac {(c x)^{m+1} (A b-a B)}{2 a b c n \left (a+b x^n\right )^2}\)

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 112.94 (sec) , antiderivative size = 8303, normalized size of antiderivative = 74.13 \[ \int \frac {(c x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

A*(a**2*a**(m/n + 1/n)*a**(-m/n - 3 - 1/n)*c**m*m**3*x**(m + 1)*lerchphi(b 
*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(2*a**2*n**4*gamma 
(m/n + 1 + 1/n) + 4*a*b*n**4*x**n*gamma(m/n + 1 + 1/n) + 2*b**2*n**4*x**(2 
*n)*gamma(m/n + 1 + 1/n)) - 3*a**2*a**(m/n + 1/n)*a**(-m/n - 3 - 1/n)*c**m 
*m**2*n*x**(m + 1)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma( 
m/n + 1/n)/(2*a**2*n**4*gamma(m/n + 1 + 1/n) + 4*a*b*n**4*x**n*gamma(m/n + 
 1 + 1/n) + 2*b**2*n**4*x**(2*n)*gamma(m/n + 1 + 1/n)) - a**2*a**(m/n + 1/ 
n)*a**(-m/n - 3 - 1/n)*c**m*m**2*n*x**(m + 1)*gamma(m/n + 1/n)/(2*a**2*n** 
4*gamma(m/n + 1 + 1/n) + 4*a*b*n**4*x**n*gamma(m/n + 1 + 1/n) + 2*b**2*n** 
4*x**(2*n)*gamma(m/n + 1 + 1/n)) + 3*a**2*a**(m/n + 1/n)*a**(-m/n - 3 - 1/ 
n)*c**m*m**2*x**(m + 1)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*g 
amma(m/n + 1/n)/(2*a**2*n**4*gamma(m/n + 1 + 1/n) + 4*a*b*n**4*x**n*gamma( 
m/n + 1 + 1/n) + 2*b**2*n**4*x**(2*n)*gamma(m/n + 1 + 1/n)) + 2*a**2*a**(m 
/n + 1/n)*a**(-m/n - 3 - 1/n)*c**m*m*n**2*x**(m + 1)*lerchphi(b*x**n*exp_p 
olar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(2*a**2*n**4*gamma(m/n + 1 + 
1/n) + 4*a*b*n**4*x**n*gamma(m/n + 1 + 1/n) + 2*b**2*n**4*x**(2*n)*gamma(m 
/n + 1 + 1/n)) + 3*a**2*a**(m/n + 1/n)*a**(-m/n - 3 - 1/n)*c**m*m*n**2*x** 
(m + 1)*gamma(m/n + 1/n)/(2*a**2*n**4*gamma(m/n + 1 + 1/n) + 4*a*b*n**4*x* 
*n*gamma(m/n + 1 + 1/n) + 2*b**2*n**4*x**(2*n)*gamma(m/n + 1 + 1/n)) - 6*a 
**2*a**(m/n + 1/n)*a**(-m/n - 3 - 1/n)*c**m*m*n*x**(m + 1)*lerchphi(b*x...
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3} \, dx=c^{m} \left (\int \frac {x^{m}}{x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) \] Input:

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

Output:

c**m*int(x**m/(x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)