\(\int \sqrt [3]{a+b x^3} (A+B x+D x^3) \, dx\) [99]

Optimal result

Integrand size = 22, antiderivative size = 193 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+D x^3\right ) \, dx=\frac {a D x \sqrt [3]{a+b x^3}}{10 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )-\frac {a B \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a (5 A b-a D) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}-\frac {a B \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \] Output:

1/10*a*D*x*(b*x^3+a)^(1/3)/b+1/30*(b*x^3+a)^(1/3)*(6*D*x^4+10*B*x^2+15*A*x 
)-1/9*a*B*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(2 
/3)+1/10*a*(5*A*b-D*a)*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x 
^3/a)/b/(b*x^3+a)^(2/3)-1/6*a*B*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(2/3)
 

Mathematica [A] (verified)

Time = 6.59 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.52 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+D x^3\right ) \, dx=\frac {x \sqrt [3]{a+b x^3} \left (4 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+2 B x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+D x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )}{4 \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:

Integrate[(a + b*x^3)^(1/3)*(A + B*x + D*x^3),x]
 

Output:

(x*(a + b*x^3)^(1/3)*(4*A*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)] 
+ 2*B*x*Hypergeometric2F1[-1/3, 2/3, 5/3, -((b*x^3)/a)] + D*x^3*Hypergeome 
tric2F1[-1/3, 4/3, 7/3, -((b*x^3)/a)]))/(4*(1 + (b*x^3)/a)^(1/3))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2392, 27, 2427, 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.

\(\displaystyle \int \sqrt [3]{a+b x^3} \left (A+B x+D x^3\right ) \, dx\)

\(\Big \downarrow \) 2392

\(\displaystyle a \int \frac {6 D x^3+10 B x+15 A}{30 \left (b x^3+a\right )^{2/3}}dx+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{30} a \int \frac {6 D x^3+10 B x+15 A}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

\(\Big \downarrow \) 2427

\(\displaystyle \frac {1}{30} a \left (\frac {\int \frac {2 (3 (5 A b-a D)+10 b B x)}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {3 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{30} a \left (\frac {\int \frac {3 (5 A b-a D)+10 b B x}{\left (b x^3+a\right )^{2/3}}dx}{b}+\frac {3 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {1}{30} a \left (\frac {\int \left (\frac {3 (5 A b-a D)}{\left (b x^3+a\right )^{2/3}}+\frac {10 b B x}{\left (b x^3+a\right )^{2/3}}\right )dx}{b}+\frac {3 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{30} a \left (\frac {\frac {3 x \left (\frac {b x^3}{a}+1\right )^{2/3} (5 A b-a D) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {10 \sqrt [3]{b} B \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-5 \sqrt [3]{b} B \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b}+\frac {3 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 A x+10 B x^2+6 D x^4\right )\)

Input:

Int[(a + b*x^3)^(1/3)*(A + B*x + D*x^3),x]
 

Output:

((a + b*x^3)^(1/3)*(15*A*x + 10*B*x^2 + 6*D*x^4))/30 + (a*((3*D*x*(a + b*x 
^3)^(1/3))/b + ((-10*b^(1/3)*B*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3) 
)/Sqrt[3]])/Sqrt[3] + (3*(5*A*b - a*D)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeomet 
ric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x^3)^(2/3) - 5*b^(1/3)*B*Log[b 
^(1/3)*x - (a + b*x^3)^(1/3)])/b))/30
 

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 2427

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

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

Input:

int((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x)
 

Output:

int((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x)
 

Fricas [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x, algorithm="fricas")
 

Output:

integral((D*x^3 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.91 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.64 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+D x^3\right ) \, dx=\frac {A \sqrt [3]{a} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {B \sqrt [3]{a} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {D \sqrt [3]{a} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:

integrate((b*x**3+a)**(1/3)*(D*x**3+B*x+A),x)
 

Output:

A*a**(1/3)*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*gamma(4/3)) + B*a**(1/3)*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*a**(1/3)*x**4*gamma(4/3)*hype 
r((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))
 

Maxima [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x, algorithm="maxima")
 

Output:

integrate((D*x^3 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Giac [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((D*x^3 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(1/3)*(A + B*x + x^3*D),x)
 

Output:

int((a + b*x^3)^(1/3)*(A + B*x + x^3*D), x)
 

Reduce [F]

\[ \int \sqrt [3]{a+b x^3} \left (A+B x+D x^3\right ) \, dx=\frac {15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b x +3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a d x +10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{2}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{4}+15 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b -3 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} d +10 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \,b^{2}}{30 b} \] Input:

int((b*x^3+a)^(1/3)*(D*x^3+B*x+A),x)
 

Output:

(15*(a + b*x**3)**(1/3)*a*b*x + 3*(a + b*x**3)**(1/3)*a*d*x + 10*(a + b*x* 
*3)**(1/3)*b**2*x**2 + 6*(a + b*x**3)**(1/3)*b*d*x**4 + 15*int((a + b*x**3 
)**(1/3)/(a + b*x**3),x)*a**2*b - 3*int((a + b*x**3)**(1/3)/(a + b*x**3),x 
)*a**2*d + 10*int(((a + b*x**3)**(1/3)*x)/(a + b*x**3),x)*a*b**2)/(30*b)