\(\int \frac {(c+d x^n)^3}{a+b x^n} \, dx\) [82]

Optimal result

Integrand size = 19, antiderivative size = 155 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=-\frac {d \left (3 a b c d (1+2 n)-a^2 d^2 (1+2 n)-2 b^2 c^2 (1+3 n)\right ) x}{b^3 (1+2 n)}-\frac {d^2 (a d (1+2 n)-b (c+4 c n)) x^{1+n}}{b^2 (1+n) (1+2 n)}+\frac {d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac {(b c-a d)^3 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^3} \] Output:

-d*(3*a*b*c*d*(1+2*n)-a^2*d^2*(1+2*n)-2*b^2*c^2*(1+3*n))*x/b^3/(1+2*n)-d^2 
*(a*d*(1+2*n)-b*(4*c*n+c))*x^(1+n)/b^2/(1+n)/(1+2*n)+d*x*(c+d*x^n)^2/b/(1+ 
2*n)+(-a*d+b*c)^3*x*hypergeom([1, 1/n],[1+1/n],-b*x^n/a)/a/b^3
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 1.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {x \left (3 c^2 d x^n \Phi \left (-\frac {b x^n}{a},1,1+\frac {1}{n}\right )+3 c d^2 x^{2 n} \Phi \left (-\frac {b x^n}{a},1,2+\frac {1}{n}\right )+d^3 x^{3 n} \Phi \left (-\frac {b x^n}{a},1,3+\frac {1}{n}\right )+c^3 \Phi \left (-\frac {b x^n}{a},1,\frac {1}{n}\right )\right )}{a n} \] Input:

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

Output:

(x*(3*c^2*d*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, 1 + n^(-1)] + 3*c*d^2*x^( 
2*n)*HurwitzLerchPhi[-((b*x^n)/a), 1, 2 + n^(-1)] + d^3*x^(3*n)*HurwitzLer 
chPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + c^3*HurwitzLerchPhi[-((b*x^n)/a), 1, 
n^(-1)]))/(a*n)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {933, 25, 1025, 25, 913, 778}

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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\frac {\left (d x^n+c\right ) \left (d (a d (2 n+1)-b (4 n c+c)) x^n+c (a d-b (2 n c+c))\right )}{b x^n+a}dx}{b (2 n+1)}+\frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}-\frac {\int \frac {\left (d x^n+c\right ) \left (d (a d (2 n+1)-b (4 n c+c)) x^n+c (a d-b (2 n c+c))\right )}{b x^n+a}dx}{b (2 n+1)}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}-\frac {\frac {\int -\frac {d \left (b^2 \left (6 n^2+4 n+1\right ) c^2-a b d \left (6 n^2+7 n+2\right ) c+a^2 d^2 \left (2 n^2+3 n+1\right )\right ) x^n+c \left (b^2 \left (2 n^2+3 n+1\right ) c^2-a b d (5 n+2) c+a^2 d^2 (2 n+1)\right )}{b x^n+a}dx}{b (n+1)}+\frac {d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b (n+1)}}{b (2 n+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}-\frac {\frac {d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b (n+1)}-\frac {\int \frac {d \left (b^2 \left (6 n^2+4 n+1\right ) c^2-a b d \left (6 n^2+7 n+2\right ) c+a^2 d^2 \left (2 n^2+3 n+1\right )\right ) x^n+c \left (b^2 \left (2 n^2+3 n+1\right ) c^2-a b d (5 n+2) c+a^2 d^2 (2 n+1)\right )}{b x^n+a}dx}{b (n+1)}}{b (2 n+1)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}-\frac {\frac {d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b (n+1)}-\frac {\frac {(n+1) (2 n+1) (b c-a d)^3 \int \frac {1}{b x^n+a}dx}{b}+\frac {d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (6 n^2+7 n+2\right )+b^2 c^2 \left (6 n^2+4 n+1\right )\right )}{b}}{b (n+1)}}{b (2 n+1)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^n\right )^2}{b (2 n+1)}-\frac {\frac {d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b (n+1)}-\frac {\frac {d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (6 n^2+7 n+2\right )+b^2 c^2 \left (6 n^2+4 n+1\right )\right )}{b}+\frac {(n+1) (2 n+1) x (b c-a d)^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b}}{b (n+1)}}{b (2 n+1)}\)

Input:

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

Output:

(d*x*(c + d*x^n)^2)/(b*(1 + 2*n)) - ((d*(a*d*(1 + 2*n) - b*(c + 4*c*n))*x* 
(c + d*x^n))/(b*(1 + n)) - ((d*(a^2*d^2*(1 + 3*n + 2*n^2) + b^2*c^2*(1 + 4 
*n + 6*n^2) - a*b*c*d*(2 + 7*n + 6*n^2))*x)/b + ((b*c - a*d)^3*(1 + n)*(1 
+ 2*n)*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b))/(b 
*(1 + n)))/(b*(1 + 2*n))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 913

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

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

rule 1025

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3)/(b*x^n + a), 
x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.69 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.32 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {a^{\frac {1}{n}} a^{-1 - \frac {1}{n}} c^{3} x \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{n^{2} \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 a^{-4 - \frac {1}{n}} a^{3 + \frac {1}{n}} d^{3} x^{3 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 3 + \frac {1}{n}\right ) \Gamma \left (3 + \frac {1}{n}\right )}{n \Gamma \left (4 + \frac {1}{n}\right )} + \frac {a^{-4 - \frac {1}{n}} a^{3 + \frac {1}{n}} d^{3} x^{3 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 3 + \frac {1}{n}\right ) \Gamma \left (3 + \frac {1}{n}\right )}{n^{2} \Gamma \left (4 + \frac {1}{n}\right )} + \frac {6 a^{-3 - \frac {1}{n}} a^{2 + \frac {1}{n}} c d^{2} x^{2 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 2 + \frac {1}{n}\right ) \Gamma \left (2 + \frac {1}{n}\right )}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {3 a^{-3 - \frac {1}{n}} a^{2 + \frac {1}{n}} c d^{2} x^{2 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 2 + \frac {1}{n}\right ) \Gamma \left (2 + \frac {1}{n}\right )}{n^{2} \Gamma \left (3 + \frac {1}{n}\right )} - \frac {3 a^{- \frac {1}{n}} a^{1 + \frac {1}{n}} b^{\frac {1}{n}} b^{-1 - \frac {1}{n}} c^{2} d x \Phi \left (\frac {a x^{- n} e^{i \pi }}{b}, 1, \frac {e^{i \pi }}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a n^{2} \Gamma \left (1 + \frac {1}{n}\right )} \] Input:

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

Output:

a**(1/n)*a**(-1 - 1/n)*c**3*x*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*g 
amma(1/n)/(n**2*gamma(1 + 1/n)) + 3*a**(-4 - 1/n)*a**(3 + 1/n)*d**3*x**(3* 
n + 1)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 3 + 1/n)*gamma(3 + 1/n)/(n*ga 
mma(4 + 1/n)) + a**(-4 - 1/n)*a**(3 + 1/n)*d**3*x**(3*n + 1)*lerchphi(b*x* 
*n*exp_polar(I*pi)/a, 1, 3 + 1/n)*gamma(3 + 1/n)/(n**2*gamma(4 + 1/n)) + 6 
*a**(-3 - 1/n)*a**(2 + 1/n)*c*d**2*x**(2*n + 1)*lerchphi(b*x**n*exp_polar( 
I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(n*gamma(3 + 1/n)) + 3*a**(-3 - 1/n)*a 
**(2 + 1/n)*c*d**2*x**(2*n + 1)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 2 + 
1/n)*gamma(2 + 1/n)/(n**2*gamma(3 + 1/n)) - 3*a**(1 + 1/n)*b**(1/n)*b**(-1 
 - 1/n)*c**2*d*x*lerchphi(a*exp_polar(I*pi)/(b*x**n), 1, exp_polar(I*pi)/n 
)*gamma(1/n)/(a*a**(1/n)*n**2*gamma(1 + 1/n))
 

Maxima [F]

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

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

Output:

(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*integrate(1/(b^4*x^n + 
 a*b^3), x) + (b^2*d^3*(n + 1)*x*x^(2*n) + (3*b^2*c*d^2*(2*n + 1) - a*b*d^ 
3*(2*n + 1))*x*x^n + (3*(2*n^2 + 3*n + 1)*b^2*c^2*d - 3*(2*n^2 + 3*n + 1)* 
a*b*c*d^2 + (2*n^2 + 3*n + 1)*a^2*d^3)*x)/((2*n^2 + 3*n + 1)*b^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {3 b^{2} c^{2} d x +3 \left (\int \frac {1}{x^{n} b +a}d x \right ) a^{2} b c \,d^{2}-3 \left (\int \frac {1}{x^{n} b +a}d x \right ) a \,b^{2} c^{2} d -2 x^{n} a b \,d^{3} n x +6 x^{n} b^{2} c \,d^{2} n x -6 a b c \,d^{2} n^{2} x -9 a b c \,d^{2} n x +x^{2 n} b^{2} d^{3} x +a^{2} d^{3} x +2 a^{2} d^{3} n^{2} x +3 a^{2} d^{3} n x -2 \left (\int \frac {1}{x^{n} b +a}d x \right ) a^{3} d^{3} n^{2}-3 \left (\int \frac {1}{x^{n} b +a}d x \right ) a^{3} d^{3} n +2 \left (\int \frac {1}{x^{n} b +a}d x \right ) b^{3} c^{3} n^{2}+3 \left (\int \frac {1}{x^{n} b +a}d x \right ) b^{3} c^{3} n +3 x^{n} b^{2} c \,d^{2} x +6 b^{2} c^{2} d \,n^{2} x +9 b^{2} c^{2} d n x -3 a b c \,d^{2} x +6 \left (\int \frac {1}{x^{n} b +a}d x \right ) a^{2} b c \,d^{2} n^{2}+9 \left (\int \frac {1}{x^{n} b +a}d x \right ) a^{2} b c \,d^{2} n -6 \left (\int \frac {1}{x^{n} b +a}d x \right ) a \,b^{2} c^{2} d \,n^{2}-9 \left (\int \frac {1}{x^{n} b +a}d x \right ) a \,b^{2} c^{2} d n +x^{2 n} b^{2} d^{3} n x -x^{n} a b \,d^{3} x -\left (\int \frac {1}{x^{n} b +a}d x \right ) a^{3} d^{3}+\left (\int \frac {1}{x^{n} b +a}d x \right ) b^{3} c^{3}}{b^{3} \left (2 n^{2}+3 n +1\right )} \] Input:

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

Output:

(x**(2*n)*b**2*d**3*n*x + x**(2*n)*b**2*d**3*x - 2*x**n*a*b*d**3*n*x - x** 
n*a*b*d**3*x + 6*x**n*b**2*c*d**2*n*x + 3*x**n*b**2*c*d**2*x - 2*int(1/(x* 
*n*b + a),x)*a**3*d**3*n**2 - 3*int(1/(x**n*b + a),x)*a**3*d**3*n - int(1/ 
(x**n*b + a),x)*a**3*d**3 + 6*int(1/(x**n*b + a),x)*a**2*b*c*d**2*n**2 + 9 
*int(1/(x**n*b + a),x)*a**2*b*c*d**2*n + 3*int(1/(x**n*b + a),x)*a**2*b*c* 
d**2 - 6*int(1/(x**n*b + a),x)*a*b**2*c**2*d*n**2 - 9*int(1/(x**n*b + a),x 
)*a*b**2*c**2*d*n - 3*int(1/(x**n*b + a),x)*a*b**2*c**2*d + 2*int(1/(x**n* 
b + a),x)*b**3*c**3*n**2 + 3*int(1/(x**n*b + a),x)*b**3*c**3*n + int(1/(x* 
*n*b + a),x)*b**3*c**3 + 2*a**2*d**3*n**2*x + 3*a**2*d**3*n*x + a**2*d**3* 
x - 6*a*b*c*d**2*n**2*x - 9*a*b*c*d**2*n*x - 3*a*b*c*d**2*x + 6*b**2*c**2* 
d*n**2*x + 9*b**2*c**2*d*n*x + 3*b**2*c**2*d*x)/(b**3*(2*n**2 + 3*n + 1))