\(\int \sqrt [3]{a+b x^n} (c+d x^n) \, dx\) [108]

Optimal result

Integrand size = 19, antiderivative size = 90 \[ \int \sqrt [3]{a+b x^n} \left (c+d x^n\right ) \, dx=\frac {3 d x \left (a+b x^n\right )^{4/3}}{b (3+4 n)}+\frac {\left (c-\frac {3 a d}{3 b+4 b n}\right ) x \sqrt [3]{a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{\sqrt [3]{1+\frac {b x^n}{a}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08 \[ \int \sqrt [3]{a+b x^n} \left (c+d x^n\right ) \, dx=\frac {x \sqrt [3]{a+b x^n} \left (3 d \left (a+b x^n\right ) \sqrt [3]{1+\frac {b x^n}{a}}+(-3 a d+b c (3+4 n)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )\right )}{b (3+4 n) \sqrt [3]{1+\frac {b x^n}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

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

\(\Big \downarrow \) 913

\(\displaystyle \left (c-\frac {3 a d}{4 b n+3 b}\right ) \int \sqrt [3]{b x^n+a}dx+\frac {3 d x \left (a+b x^n\right )^{4/3}}{b (4 n+3)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\sqrt [3]{a+b x^n} \left (c-\frac {3 a d}{4 b n+3 b}\right ) \int \sqrt [3]{\frac {b x^n}{a}+1}dx}{\sqrt [3]{\frac {b x^n}{a}+1}}+\frac {3 d x \left (a+b x^n\right )^{4/3}}{b (4 n+3)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {x \sqrt [3]{a+b x^n} \left (c-\frac {3 a d}{4 b n+3 b}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{\sqrt [3]{\frac {b x^n}{a}+1}}+\frac {3 d x \left (a+b x^n\right )^{4/3}}{b (4 n+3)}\)

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate((a+b*x^n)^(1/3)*(c+d*x^n),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 = 1.49 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27 \[ \int \sqrt [3]{a+b x^n} \left (c+d x^n\right ) \, dx=\frac {a^{\frac {1}{n}} a^{\frac {1}{3} - \frac {1}{n}} c x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{n} \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {a^{- \frac {2}{3} - \frac {1}{n}} a^{1 + \frac {1}{n}} d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} \] Input:

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

Output:

a**(1/n)*a**(1/3 - 1/n)*c*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**(-2/3 - 1/n)*a**(1 + 1/n)*d* 
x**(n + 1)*gamma(1 + 1/n)*hyper((-1/3, 1 + 1/n), (2 + 1/n,), b*x**n*exp_po 
lar(I*pi)/a)/(n*gamma(2 + 1/n))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{a+b x^n} \left (c+d x^n\right ) \, dx=\frac {3 x^{n} \left (x^{n} b +a \right )^{\frac {1}{3}} b d n x +9 x^{n} \left (x^{n} b +a \right )^{\frac {1}{3}} b d x +3 \left (x^{n} b +a \right )^{\frac {1}{3}} a d n x +12 \left (x^{n} b +a \right )^{\frac {1}{3}} b c n x +9 \left (x^{n} b +a \right )^{\frac {1}{3}} b c x -12 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a^{2} d \,n^{3}-45 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a^{2} d \,n^{2}-27 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a^{2} d n +16 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a b c \,n^{4}+72 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a b c \,n^{3}+81 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a b c \,n^{2}+27 \left (\int \frac {\left (x^{n} b +a \right )^{\frac {1}{3}}}{4 x^{n} b \,n^{2}+15 x^{n} b n +9 x^{n} b +4 a \,n^{2}+15 a n +9 a}d x \right ) a b c n}{b \left (4 n^{2}+15 n +9\right )} \] Input:

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

Output:

(3*x**n*(x**n*b + a)**(1/3)*b*d*n*x + 9*x**n*(x**n*b + a)**(1/3)*b*d*x + 3 
*(x**n*b + a)**(1/3)*a*d*n*x + 12*(x**n*b + a)**(1/3)*b*c*n*x + 9*(x**n*b 
+ a)**(1/3)*b*c*x - 12*int((x**n*b + a)**(1/3)/(4*x**n*b*n**2 + 15*x**n*b* 
n + 9*x**n*b + 4*a*n**2 + 15*a*n + 9*a),x)*a**2*d*n**3 - 45*int((x**n*b + 
a)**(1/3)/(4*x**n*b*n**2 + 15*x**n*b*n + 9*x**n*b + 4*a*n**2 + 15*a*n + 9* 
a),x)*a**2*d*n**2 - 27*int((x**n*b + a)**(1/3)/(4*x**n*b*n**2 + 15*x**n*b* 
n + 9*x**n*b + 4*a*n**2 + 15*a*n + 9*a),x)*a**2*d*n + 16*int((x**n*b + a)* 
*(1/3)/(4*x**n*b*n**2 + 15*x**n*b*n + 9*x**n*b + 4*a*n**2 + 15*a*n + 9*a), 
x)*a*b*c*n**4 + 72*int((x**n*b + a)**(1/3)/(4*x**n*b*n**2 + 15*x**n*b*n + 
9*x**n*b + 4*a*n**2 + 15*a*n + 9*a),x)*a*b*c*n**3 + 81*int((x**n*b + a)**( 
1/3)/(4*x**n*b*n**2 + 15*x**n*b*n + 9*x**n*b + 4*a*n**2 + 15*a*n + 9*a),x) 
*a*b*c*n**2 + 27*int((x**n*b + a)**(1/3)/(4*x**n*b*n**2 + 15*x**n*b*n + 9* 
x**n*b + 4*a*n**2 + 15*a*n + 9*a),x)*a*b*c*n)/(b*(4*n**2 + 15*n + 9))