\(\int \frac {(c+d x^n)^2}{\sqrt [3]{a+b x^n}} \, dx\) [111]

Optimal result

Integrand size = 21, antiderivative size = 187 \[ \int \frac {\left (c+d x^n\right )^2}{\sqrt [3]{a+b x^n}} \, dx=-\frac {3 d (3 a d (1+n)-b c (3+8 n)) x \left (a+b x^n\right )^{2/3}}{b^2 (3+2 n) (3+5 n)}+\frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (3+5 n)}+\frac {\left (9 a^2 d^2 (1+n)-6 a b c d (3+5 n)+b^2 c^2 \left (9+21 n+10 n^2\right )\right ) x \sqrt [3]{1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^2 (3+2 n) (3+5 n) \sqrt [3]{a+b x^n}} \] Output:

-3*d*(3*a*d*(1+n)-b*c*(3+8*n))*x*(a+b*x^n)^(2/3)/b^2/(3+2*n)/(3+5*n)+3*d*x 
*(a+b*x^n)^(2/3)*(c+d*x^n)/b/(3+5*n)+(9*a^2*d^2*(1+n)-6*a*b*c*d*(3+5*n)+b^ 
2*c^2*(10*n^2+21*n+9))*x*(1+b*x^n/a)^(1/3)*hypergeom([1/3, 1/n],[1+1/n],-b 
*x^n/a)/b^2/(3+2*n)/(3+5*n)/(a+b*x^n)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 5.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.78 \[ \int \frac {\left (c+d x^n\right )^2}{\sqrt [3]{a+b x^n}} \, dx=\frac {3 d x \left (a+b x^n\right ) \left (-3 a d (1+n)+2 b c (3+5 n)+b d (3+2 n) x^n\right )+\left (9 a^2 d^2 (1+n)-6 a b c d (3+5 n)+b^2 c^2 \left (9+21 n+10 n^2\right )\right ) x \sqrt [3]{1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^2 (3+2 n) (3+5 n) \sqrt [3]{a+b x^n}} \] Input:

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

Output:

(3*d*x*(a + b*x^n)*(-3*a*d*(1 + n) + 2*b*c*(3 + 5*n) + b*d*(3 + 2*n)*x^n) 
+ (9*a^2*d^2*(1 + n) - 6*a*b*c*d*(3 + 5*n) + b^2*c^2*(9 + 21*n + 10*n^2))* 
x*(1 + (b*x^n)/a)^(1/3)*Hypergeometric2F1[1/3, n^(-1), 1 + n^(-1), -((b*x^ 
n)/a)])/(b^2*(3 + 2*n)*(3 + 5*n)*(a + b*x^n)^(1/3))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 27, 913, 779, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {3 \int -\frac {d (3 a d (n+1)-b c (8 n+3)) x^n+c (3 a d-b c (5 n+3))}{3 \sqrt [3]{b x^n+a}}dx}{b (5 n+3)}+\frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (5 n+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (5 n+3)}-\frac {\int \frac {d (3 a d (n+1)-b c (8 n+3)) x^n+c (3 a d-b c (5 n+3))}{\sqrt [3]{b x^n+a}}dx}{b (5 n+3)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (5 n+3)}-\frac {\frac {3 d x \left (a+b x^n\right )^{2/3} (3 a d (n+1)-b c (8 n+3))}{b (2 n+3)}-\frac {\left (9 a^2 d^2 (n+1)-6 a b c d (5 n+3)+b^2 c^2 \left (10 n^2+21 n+9\right )\right ) \int \frac {1}{\sqrt [3]{b x^n+a}}dx}{b (2 n+3)}}{b (5 n+3)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (5 n+3)}-\frac {\frac {3 d x \left (a+b x^n\right )^{2/3} (3 a d (n+1)-b c (8 n+3))}{b (2 n+3)}-\frac {\sqrt [3]{\frac {b x^n}{a}+1} \left (9 a^2 d^2 (n+1)-6 a b c d (5 n+3)+b^2 c^2 \left (10 n^2+21 n+9\right )\right ) \int \frac {1}{\sqrt [3]{\frac {b x^n}{a}+1}}dx}{b (2 n+3) \sqrt [3]{a+b x^n}}}{b (5 n+3)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {3 d x \left (a+b x^n\right )^{2/3} \left (c+d x^n\right )}{b (5 n+3)}-\frac {\frac {3 d x \left (a+b x^n\right )^{2/3} (3 a d (n+1)-b c (8 n+3))}{b (2 n+3)}-\frac {x \sqrt [3]{\frac {b x^n}{a}+1} \left (9 a^2 d^2 (n+1)-6 a b c d (5 n+3)+b^2 c^2 \left (10 n^2+21 n+9\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b (2 n+3) \sqrt [3]{a+b x^n}}}{b (5 n+3)}\)

Input:

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

Output:

(3*d*x*(a + b*x^n)^(2/3)*(c + d*x^n))/(b*(3 + 5*n)) - ((3*d*(3*a*d*(1 + n) 
 - b*c*(3 + 8*n))*x*(a + b*x^n)^(2/3))/(b*(3 + 2*n)) - ((9*a^2*d^2*(1 + n) 
 - 6*a*b*c*d*(3 + 5*n) + b^2*c^2*(9 + 21*n + 10*n^2))*x*(1 + (b*x^n)/a)^(1 
/3)*Hypergeometric2F1[1/3, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(b*(3 + 2*n) 
*(a + b*x^n)^(1/3)))/(b*(3 + 5*n))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 779

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

Maple [F]

\[\int \frac {\left (c +d \,x^{n}\right )^{2}}{\left (a +b \,x^{n}\right )^{\frac {1}{3}}}d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (c+d x^n\right )^2}{\sqrt [3]{a+b x^n}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c+d*x^n)^2/(a+b*x^n)^(1/3),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.13 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c+d x^n\right )^2}{\sqrt [3]{a+b x^n}} \, dx=\frac {a^{\frac {1}{n}} a^{- \frac {1}{3} - \frac {1}{n}} c^{2} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{n} \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {a^{- \frac {7}{3} - \frac {1}{n}} a^{2 + \frac {1}{n}} d^{2} x^{2 n + 1} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {2 a^{- \frac {4}{3} - \frac {1}{n}} a^{1 + \frac {1}{n}} c d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} \] Input:

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

Output:

a**(1/n)*a**(-1/3 - 1/n)*c**2*x*gamma(1/n)*hyper((1/3, 1/n), (1 + 1/n,), b 
*x**n*exp_polar(I*pi)/a)/(n*gamma(1 + 1/n)) + a**(-7/3 - 1/n)*a**(2 + 1/n) 
*d**2*x**(2*n + 1)*gamma(2 + 1/n)*hyper((1/3, 2 + 1/n), (3 + 1/n,), b*x**n 
*exp_polar(I*pi)/a)/(n*gamma(3 + 1/n)) + 2*a**(-4/3 - 1/n)*a**(1 + 1/n)*c* 
d*x**(n + 1)*gamma(1 + 1/n)*hyper((1/3, 1 + 1/n), (2 + 1/n,), b*x**n*exp_p 
olar(I*pi)/a)/(n*gamma(2 + 1/n))
 

Maxima [F]

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

integrate((c+d*x^n)^2/(a+b*x^n)^(1/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((c+d*x^n)^2/(a+b*x^n)^(1/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^n\right )^2}{\sqrt [3]{a+b x^n}} \, dx=\int \frac {{\left (c+d\,x^n\right )}^2}{{\left (a+b\,x^n\right )}^{1/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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