\(\int \frac {x^{-1+3 n} (a+b x^n)^3}{c+d x^n} \, dx\) [458]

Optimal result

Integrand size = 26, antiderivative size = 158 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\frac {c (b c-a d)^3 x^n}{d^5 n}-\frac {(b c-a d)^3 x^{2 n}}{2 d^4 n}+\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x^{3 n}}{3 d^3 n}-\frac {b^2 (b c-3 a d) x^{4 n}}{4 d^2 n}+\frac {b^3 x^{5 n}}{5 d n}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6 n} \] Output:

c*(-a*d+b*c)^3*x^n/d^5/n-1/2*(-a*d+b*c)^3*x^(2*n)/d^4/n+1/3*b*(3*a^2*d^2-3 
*a*b*c*d+b^2*c^2)*x^(3*n)/d^3/n-1/4*b^2*(-3*a*d+b*c)*x^(4*n)/d^2/n+1/5*b^3 
*x^(5*n)/d/n-c^2*(-a*d+b*c)^3*ln(c+d*x^n)/d^6/n
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.17 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\frac {d x^n \left (30 a^3 d^3 \left (-2 c+d x^n\right )+30 a^2 b d^2 \left (6 c^2-3 c d x^n+2 d^2 x^{2 n}\right )+15 a b^2 d \left (-12 c^3+6 c^2 d x^n-4 c d^2 x^{2 n}+3 d^3 x^{3 n}\right )+b^3 \left (60 c^4-30 c^3 d x^n+20 c^2 d^2 x^{2 n}-15 c d^3 x^{3 n}+12 d^4 x^{4 n}\right )\right )-60 c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{60 d^6 n} \] Input:

Integrate[(x^(-1 + 3*n)*(a + b*x^n)^3)/(c + d*x^n),x]
 

Output:

(d*x^n*(30*a^3*d^3*(-2*c + d*x^n) + 30*a^2*b*d^2*(6*c^2 - 3*c*d*x^n + 2*d^ 
2*x^(2*n)) + 15*a*b^2*d*(-12*c^3 + 6*c^2*d*x^n - 4*c*d^2*x^(2*n) + 3*d^3*x 
^(3*n)) + b^3*(60*c^4 - 30*c^3*d*x^n + 20*c^2*d^2*x^(2*n) - 15*c*d^3*x^(3* 
n) + 12*d^4*x^(4*n))) - 60*c^2*(b*c - a*d)^3*Log[c + d*x^n])/(60*d^6*n)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {948, 99, 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.

\(\displaystyle \int \frac {x^{3 n-1} \left (a+b x^n\right )^3}{c+d x^n} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {(a d-b c)^3 x^n}{d^4}+\frac {b \left (b^2 c^2-3 a b d c+3 a^2 d^2\right ) x^{2 n}}{d^3}-\frac {b^2 (b c-3 a d) x^{3 n}}{d^2}+\frac {b^3 x^{4 n}}{d}+\frac {c (b c-a d)^3}{d^5}-\frac {c^2 (b c-a d)^3}{d^5 \left (d x^n+c\right )}\right )dx^n}{n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b x^{3 n} \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{3 d^3}-\frac {b^2 x^{4 n} (b c-3 a d)}{4 d^2}-\frac {c^2 (b c-a d)^3 \log \left (c+d x^n\right )}{d^6}+\frac {c x^n (b c-a d)^3}{d^5}-\frac {x^{2 n} (b c-a d)^3}{2 d^4}+\frac {b^3 x^{5 n}}{5 d}}{n}\)

Input:

Int[(x^(-1 + 3*n)*(a + b*x^n)^3)/(c + d*x^n),x]
 

Output:

((c*(b*c - a*d)^3*x^n)/d^5 - ((b*c - a*d)^3*x^(2*n))/(2*d^4) + (b*(b^2*c^2 
 - 3*a*b*c*d + 3*a^2*d^2)*x^(3*n))/(3*d^3) - (b^2*(b*c - 3*a*d)*x^(4*n))/( 
4*d^2) + (b^3*x^(5*n))/(5*d) - (c^2*(b*c - a*d)^3*Log[c + d*x^n])/d^6)/n
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(150)=300\).

Time = 0.97 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.16

method	result	size
risch	\(\frac {b^{3} x^{5 n}}{5 d n}+\frac {3 b^{2} x^{4 n} a}{4 d n}-\frac {b^{3} x^{4 n} c}{4 d^{2} n}+\frac {b \,x^{3 n} a^{2}}{d n}-\frac {b^{2} x^{3 n} a c}{d^{2} n}+\frac {b^{3} x^{3 n} c^{2}}{3 d^{3} n}+\frac {x^{2 n} a^{3}}{2 d n}-\frac {3 x^{2 n} a^{2} b c}{2 d^{2} n}+\frac {3 x^{2 n} a \,b^{2} c^{2}}{2 d^{3} n}-\frac {x^{2 n} b^{3} c^{3}}{2 d^{4} n}-\frac {c \,x^{n} a^{3}}{d^{2} n}+\frac {3 c^{2} x^{n} a^{2} b}{d^{3} n}-\frac {3 c^{3} x^{n} a \,b^{2}}{d^{4} n}+\frac {c^{4} x^{n} b^{3}}{d^{5} n}+\frac {c^{2} \ln \left (x^{n}+\frac {c}{d}\right ) a^{3}}{d^{3} n}-\frac {3 c^{3} \ln \left (x^{n}+\frac {c}{d}\right ) a^{2} b}{d^{4} n}+\frac {3 c^{4} \ln \left (x^{n}+\frac {c}{d}\right ) a \,b^{2}}{d^{5} n}-\frac {c^{5} \ln \left (x^{n}+\frac {c}{d}\right ) b^{3}}{d^{6} n}\)	\(342\)

Input:

int(x^(-1+3*n)*(a+b*x^n)^3/(c+d*x^n),x,method=_RETURNVERBOSE)
 

Output:

1/5*b^3/d/n*(x^n)^5+3/4*b^2/d/n*(x^n)^4*a-1/4*b^3/d^2/n*(x^n)^4*c+b/d/n*(x 
^n)^3*a^2-b^2/d^2/n*(x^n)^3*a*c+1/3*b^3/d^3/n*(x^n)^3*c^2+1/2/d/n*(x^n)^2* 
a^3-3/2/d^2/n*(x^n)^2*a^2*b*c+3/2/d^3/n*(x^n)^2*a*b^2*c^2-1/2/d^4/n*(x^n)^ 
2*b^3*c^3-c/d^2/n*x^n*a^3+3*c^2/d^3/n*x^n*a^2*b-3*c^3/d^4/n*x^n*a*b^2+c^4/ 
d^5/n*x^n*b^3+c^2/d^3/n*ln(x^n+c/d)*a^3-3*c^3/d^4/n*ln(x^n+c/d)*a^2*b+3*c^ 
4/d^5/n*ln(x^n+c/d)*a*b^2-c^5/d^6/n*ln(x^n+c/d)*b^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.46 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\frac {12 \, b^{3} d^{5} x^{5 \, n} - 15 \, {\left (b^{3} c d^{4} - 3 \, a b^{2} d^{5}\right )} x^{4 \, n} + 20 \, {\left (b^{3} c^{2} d^{3} - 3 \, a b^{2} c d^{4} + 3 \, a^{2} b d^{5}\right )} x^{3 \, n} - 30 \, {\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{2 \, n} + 60 \, {\left (b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} + 3 \, a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{n} - 60 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \log \left (d x^{n} + c\right )}{60 \, d^{6} n} \] Input:

integrate(x^(-1+3*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="fricas")
 

Output:

1/60*(12*b^3*d^5*x^(5*n) - 15*(b^3*c*d^4 - 3*a*b^2*d^5)*x^(4*n) + 20*(b^3* 
c^2*d^3 - 3*a*b^2*c*d^4 + 3*a^2*b*d^5)*x^(3*n) - 30*(b^3*c^3*d^2 - 3*a*b^2 
*c^2*d^3 + 3*a^2*b*c*d^4 - a^3*d^5)*x^(2*n) + 60*(b^3*c^4*d - 3*a*b^2*c^3* 
d^2 + 3*a^2*b*c^2*d^3 - a^3*c*d^4)*x^n - 60*(b^3*c^5 - 3*a*b^2*c^4*d + 3*a 
^2*b*c^3*d^2 - a^3*c^2*d^3)*log(d*x^n + c))/(d^6*n)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (138) = 276\).

Time = 7.75 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.71 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\begin {cases} \frac {\left (a + b\right )^{3} \log {\left (x \right )}}{c} & \text {for}\: d = 0 \wedge n = 0 \\\frac {\frac {a^{3} x x^{3 n - 1}}{3 n} + \frac {3 a^{2} b x x^{n} x^{3 n - 1}}{4 n} + \frac {3 a b^{2} x x^{2 n} x^{3 n - 1}}{5 n} + \frac {b^{3} x x^{3 n} x^{3 n - 1}}{6 n}}{c} & \text {for}\: d = 0 \\\frac {\left (a + b\right )^{3} \log {\left (x \right )}}{c + d} & \text {for}\: n = 0 \\\frac {a^{3} c^{2} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{3} n} - \frac {a^{3} c x^{n}}{d^{2} n} + \frac {a^{3} x^{2 n}}{2 d n} - \frac {3 a^{2} b c^{3} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{4} n} + \frac {3 a^{2} b c^{2} x^{n}}{d^{3} n} - \frac {3 a^{2} b c x^{2 n}}{2 d^{2} n} + \frac {a^{2} b x^{3 n}}{d n} + \frac {3 a b^{2} c^{4} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{5} n} - \frac {3 a b^{2} c^{3} x^{n}}{d^{4} n} + \frac {3 a b^{2} c^{2} x^{2 n}}{2 d^{3} n} - \frac {a b^{2} c x^{3 n}}{d^{2} n} + \frac {3 a b^{2} x^{4 n}}{4 d n} - \frac {b^{3} c^{5} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{6} n} + \frac {b^{3} c^{4} x^{n}}{d^{5} n} - \frac {b^{3} c^{3} x^{2 n}}{2 d^{4} n} + \frac {b^{3} c^{2} x^{3 n}}{3 d^{3} n} - \frac {b^{3} c x^{4 n}}{4 d^{2} n} + \frac {b^{3} x^{5 n}}{5 d n} & \text {otherwise} \end {cases} \] Input:

integrate(x**(-1+3*n)*(a+b*x**n)**3/(c+d*x**n),x)
 

Output:

Piecewise(((a + b)**3*log(x)/c, Eq(d, 0) & Eq(n, 0)), ((a**3*x*x**(3*n - 1 
)/(3*n) + 3*a**2*b*x*x**n*x**(3*n - 1)/(4*n) + 3*a*b**2*x*x**(2*n)*x**(3*n 
 - 1)/(5*n) + b**3*x*x**(3*n)*x**(3*n - 1)/(6*n))/c, Eq(d, 0)), ((a + b)** 
3*log(x)/(c + d), Eq(n, 0)), (a**3*c**2*log(c/d + x**n)/(d**3*n) - a**3*c* 
x**n/(d**2*n) + a**3*x**(2*n)/(2*d*n) - 3*a**2*b*c**3*log(c/d + x**n)/(d** 
4*n) + 3*a**2*b*c**2*x**n/(d**3*n) - 3*a**2*b*c*x**(2*n)/(2*d**2*n) + a**2 
*b*x**(3*n)/(d*n) + 3*a*b**2*c**4*log(c/d + x**n)/(d**5*n) - 3*a*b**2*c**3 
*x**n/(d**4*n) + 3*a*b**2*c**2*x**(2*n)/(2*d**3*n) - a*b**2*c*x**(3*n)/(d* 
*2*n) + 3*a*b**2*x**(4*n)/(4*d*n) - b**3*c**5*log(c/d + x**n)/(d**6*n) + b 
**3*c**4*x**n/(d**5*n) - b**3*c**3*x**(2*n)/(2*d**4*n) + b**3*c**2*x**(3*n 
)/(3*d**3*n) - b**3*c*x**(4*n)/(4*d**2*n) + b**3*x**(5*n)/(5*d*n), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.81 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=-\frac {1}{60} \, b^{3} {\left (\frac {60 \, c^{5} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{6} n} - \frac {12 \, d^{4} x^{5 \, n} - 15 \, c d^{3} x^{4 \, n} + 20 \, c^{2} d^{2} x^{3 \, n} - 30 \, c^{3} d x^{2 \, n} + 60 \, c^{4} x^{n}}{d^{5} n}\right )} + \frac {1}{4} \, a b^{2} {\left (\frac {12 \, c^{4} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{5} n} + \frac {3 \, d^{3} x^{4 \, n} - 4 \, c d^{2} x^{3 \, n} + 6 \, c^{2} d x^{2 \, n} - 12 \, c^{3} x^{n}}{d^{4} n}\right )} - \frac {1}{2} \, a^{2} b {\left (\frac {6 \, c^{3} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{4} n} - \frac {2 \, d^{2} x^{3 \, n} - 3 \, c d x^{2 \, n} + 6 \, c^{2} x^{n}}{d^{3} n}\right )} + \frac {1}{2} \, a^{3} {\left (\frac {2 \, c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{3} n} + \frac {d x^{2 \, n} - 2 \, c x^{n}}{d^{2} n}\right )} \] Input:

integrate(x^(-1+3*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="maxima")
 

Output:

-1/60*b^3*(60*c^5*log((d*x^n + c)/d)/(d^6*n) - (12*d^4*x^(5*n) - 15*c*d^3* 
x^(4*n) + 20*c^2*d^2*x^(3*n) - 30*c^3*d*x^(2*n) + 60*c^4*x^n)/(d^5*n)) + 1 
/4*a*b^2*(12*c^4*log((d*x^n + c)/d)/(d^5*n) + (3*d^3*x^(4*n) - 4*c*d^2*x^( 
3*n) + 6*c^2*d*x^(2*n) - 12*c^3*x^n)/(d^4*n)) - 1/2*a^2*b*(6*c^3*log((d*x^ 
n + c)/d)/(d^4*n) - (2*d^2*x^(3*n) - 3*c*d*x^(2*n) + 6*c^2*x^n)/(d^3*n)) + 
 1/2*a^3*(2*c^2*log((d*x^n + c)/d)/(d^3*n) + (d*x^(2*n) - 2*c*x^n)/(d^2*n) 
)
 

Giac [F]

\[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{3} x^{3 \, n - 1}}{d x^{n} + c} \,d x } \] Input:

integrate(x^(-1+3*n)*(a+b*x^n)^3/(c+d*x^n),x, algorithm="giac")
 

Output:

integrate((b*x^n + a)^3*x^(3*n - 1)/(d*x^n + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\int \frac {x^{3\,n-1}\,{\left (a+b\,x^n\right )}^3}{c+d\,x^n} \,d x \] Input:

int((x^(3*n - 1)*(a + b*x^n)^3)/(c + d*x^n),x)
 

Output:

int((x^(3*n - 1)*(a + b*x^n)^3)/(c + d*x^n), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.79 \[ \int \frac {x^{-1+3 n} \left (a+b x^n\right )^3}{c+d x^n} \, dx=\frac {12 x^{5 n} b^{3} d^{5}+45 x^{4 n} a \,b^{2} d^{5}-15 x^{4 n} b^{3} c \,d^{4}+60 x^{3 n} a^{2} b \,d^{5}-60 x^{3 n} a \,b^{2} c \,d^{4}+20 x^{3 n} b^{3} c^{2} d^{3}+30 x^{2 n} a^{3} d^{5}-90 x^{2 n} a^{2} b c \,d^{4}+90 x^{2 n} a \,b^{2} c^{2} d^{3}-30 x^{2 n} b^{3} c^{3} d^{2}-60 x^{n} a^{3} c \,d^{4}+180 x^{n} a^{2} b \,c^{2} d^{3}-180 x^{n} a \,b^{2} c^{3} d^{2}+60 x^{n} b^{3} c^{4} d +60 \,\mathrm {log}\left (x^{n} d +c \right ) a^{3} c^{2} d^{3}-180 \,\mathrm {log}\left (x^{n} d +c \right ) a^{2} b \,c^{3} d^{2}+180 \,\mathrm {log}\left (x^{n} d +c \right ) a \,b^{2} c^{4} d -60 \,\mathrm {log}\left (x^{n} d +c \right ) b^{3} c^{5}}{60 d^{6} n} \] Input:

int(x^(-1+3*n)*(a+b*x^n)^3/(c+d*x^n),x)
 

Output:

(12*x**(5*n)*b**3*d**5 + 45*x**(4*n)*a*b**2*d**5 - 15*x**(4*n)*b**3*c*d**4 
 + 60*x**(3*n)*a**2*b*d**5 - 60*x**(3*n)*a*b**2*c*d**4 + 20*x**(3*n)*b**3* 
c**2*d**3 + 30*x**(2*n)*a**3*d**5 - 90*x**(2*n)*a**2*b*c*d**4 + 90*x**(2*n 
)*a*b**2*c**2*d**3 - 30*x**(2*n)*b**3*c**3*d**2 - 60*x**n*a**3*c*d**4 + 18 
0*x**n*a**2*b*c**2*d**3 - 180*x**n*a*b**2*c**3*d**2 + 60*x**n*b**3*c**4*d 
+ 60*log(x**n*d + c)*a**3*c**2*d**3 - 180*log(x**n*d + c)*a**2*b*c**3*d**2 
 + 180*log(x**n*d + c)*a*b**2*c**4*d - 60*log(x**n*d + c)*b**3*c**5)/(60*d 
**6*n)