\(\int (c+d x) \sqrt [3]{a+b x^3} \, dx\) [146]

Optimal result

Integrand size = 17, antiderivative size = 155 \[ \int (c+d x) \sqrt [3]{a+b x^3} \, dx=\frac {1}{6} \left (3 c x+2 d x^2\right ) \sqrt [3]{a+b x^3}-\frac {a d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a c x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 \left (a+b x^3\right )^{2/3}}-\frac {a d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 7.67 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.48 \[ \int (c+d x) \sqrt [3]{a+b x^3} \, dx=\frac {x \sqrt [3]{a+b x^3} \left (2 c \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )\right )}{2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2392, 27, 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.

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

\(\Big \downarrow \) 2392

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.51 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int (c+d x) \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {\sqrt [3]{a} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \] Input:

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

Output:

a**(1/3)*c*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*gamma(4/3)) + a**(1/3)*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))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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