\(\int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx\) [145]

Optimal result

Integrand size = 19, antiderivative size = 183 \[ \int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx=\frac {1}{6} \left (3 c^2 x+4 c d x^2\right ) \sqrt [3]{a+b x^3}+\frac {d^2 \left (a+b x^3\right )^{4/3}}{4 b}-\frac {2 a c d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a c^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a c d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{2/3}} \] Output:

1/6*(4*c*d*x^2+3*c^2*x)*(b*x^3+a)^(1/3)+1/4*d^2*(b*x^3+a)^(4/3)/b-2/9*a*c* 
d*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(2/3)+1/2* 
a*c^2*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)^( 
2/3)-1/3*a*c*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)
 

Mathematica [A] (verified)

Time = 8.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.61 \[ \int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a+b x^3} \left (4 b c^2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (d \left (a+b x^3\right ) \sqrt [3]{1+\frac {b x^3}{a}}+4 b c x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )\right )\right )}{4 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:

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

Output:

((a + b*x^3)^(1/3)*(4*b*c^2*x*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/ 
a)] + d*(d*(a + b*x^3)*(1 + (b*x^3)/a)^(1/3) + 4*b*c*x^2*Hypergeometric2F1 
[-1/3, 2/3, 5/3, -((b*x^3)/a)])))/(4*b*(1 + (b*x^3)/a)^(1/3))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2392, 27, 2425, 793, 2432, 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.

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

\(\Big \downarrow \) 2392

\(\displaystyle a \int \frac {6 c^2+8 d x c+3 d^2 x^2}{12 \left (b x^3+a\right )^{2/3}}dx+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} a \int \frac {6 c^2+8 d x c+3 d^2 x^2}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

\(\Big \downarrow \) 2425

\(\displaystyle \frac {1}{12} a \left (\int \frac {6 c^2+8 d x c}{\left (b x^3+a\right )^{2/3}}dx+3 d^2 \int \frac {x^2}{\left (b x^3+a\right )^{2/3}}dx\right )+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

\(\Big \downarrow \) 793

\(\displaystyle \frac {1}{12} a \left (\int \frac {6 c^2+8 d x c}{\left (b x^3+a\right )^{2/3}}dx+\frac {3 d^2 \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {1}{12} a \left (\int \left (\frac {6 c^2}{\left (b x^3+a\right )^{2/3}}+\frac {8 d x c}{\left (b x^3+a\right )^{2/3}}\right )dx+\frac {3 d^2 \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{12} a \left (-\frac {8 c d \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {4 c d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+\frac {6 c^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {3 d^2 \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{12} \sqrt [3]{a+b x^3} \left (6 c^2 x+8 c d x^2+3 d^2 x^3\right )\)

Input:

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

Output:

((a + b*x^3)^(1/3)*(6*c^2*x + 8*c*d*x^2 + 3*d^2*x^3))/12 + (a*((3*d^2*(a + 
 b*x^3)^(1/3))/b - (8*c*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqr 
t[3]])/(Sqrt[3]*b^(2/3)) + (6*c^2*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F 
1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x^3)^(2/3) - (4*c*d*Log[b^(1/3)*x - 
 (a + b*x^3)^(1/3)])/b^(2/3)))/12
 

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 793

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n) 
^(p + 1)/(b*n*(p + 1)), x] /; FreeQ[{a, b, m, n, p}, x] && EqQ[m, n - 1] && 
 NeQ[p, -1]
 

rule 2009

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

rule 2392

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{q = Expon[Pq 
, x], i}, Simp[(a + b*x^n)^p*Sum[Coeff[Pq, x, i]*(x^(i + 1)/(n*p + i + 1)), 
 {i, 0, q}], x] + Simp[a*n*p   Int[(a + b*x^n)^(p - 1)*Sum[Coeff[Pq, x, i]* 
(x^i/(n*p + i + 1)), {i, 0, q}], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x 
] && IGtQ[(n - 1)/2, 0] && GtQ[p, 0]
 

rule 2425

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Pq, x, n - 
 1]   Int[x^(n - 1)*(a + b*x^n)^p, x], x] + Int[ExpandToSum[Pq - Coeff[Pq, 
x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && PolyQ[P 
q, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.62 \[ \int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 \sqrt [3]{a} c d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + d^{2} \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**(1/3)*c**2*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*p 
i)/a)/(3*gamma(4/3)) + 2*a**(1/3)*c*d*x**2*gamma(2/3)*hyper((-1/3, 2/3), ( 
5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3)) + d**2*Piecewise((a**(1/3) 
*x**3/3, Eq(b, 0)), ((a + b*x**3)**(4/3)/(4*b), True))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^2 \sqrt [3]{a+b x^3} \, dx=\frac {3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,d^{2}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{2} x +8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c d \,x^{2}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,d^{2} x^{3}+6 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a b \,c^{2}+8 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a b c d}{12 b} \] Input:

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

Output:

(3*(a + b*x**3)**(1/3)*a*d**2 + 6*(a + b*x**3)**(1/3)*b*c**2*x + 8*(a + b* 
x**3)**(1/3)*b*c*d*x**2 + 3*(a + b*x**3)**(1/3)*b*d**2*x**3 + 6*int((a + b 
*x**3)**(1/3)/(a + b*x**3),x)*a*b*c**2 + 8*int(((a + b*x**3)**(1/3)*x)/(a 
+ b*x**3),x)*a*b*c*d)/(12*b)