\(\int \sqrt [3]{a+b x^3} (A+B x+C x^2+D x^3) \, dx\) [107]

Optimal result

Integrand size = 27, antiderivative size = 219 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a C \sqrt [3]{a+b x^3}}{4 b}+\frac {a D x \sqrt [3]{a+b x^3}}{10 b}+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )-\frac {a B \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}+\frac {a (5 A b-a D) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}-\frac {a B \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3}} \] Output:

1/4*a*C*(b*x^3+a)^(1/3)/b+1/10*a*D*x*(b*x^3+a)^(1/3)/b+1/60*(b*x^3+a)^(1/3 
)*(12*D*x^4+15*C*x^3+20*B*x^2+30*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 = 8.84 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.66 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt [3]{a+b x^3} \left (a C \sqrt [3]{1+\frac {b x^3}{a}}+b C x^3 \sqrt [3]{1+\frac {b x^3}{a}}+4 A b x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+2 b B x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+b D x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )}{4 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:

Integrate[(a + b*x^3)^(1/3)*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

((a + b*x^3)^(1/3)*(a*C*(1 + (b*x^3)/a)^(1/3) + b*C*x^3*(1 + (b*x^3)/a)^(1 
/3) + 4*A*b*x*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)] + 2*b*B*x^2* 
Hypergeometric2F1[-1/3, 2/3, 5/3, -((b*x^3)/a)] + b*D*x^4*Hypergeometric2F 
1[-1/3, 4/3, 7/3, -((b*x^3)/a)]))/(4*b*(1 + (b*x^3)/a)^(1/3))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2392, 27, 2427, 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.

\(\displaystyle \int \sqrt [3]{a+b x^3} \left (A+B x+C x^2+D x^3\right ) \, dx\)

\(\Big \downarrow \) 2392

\(\displaystyle a \int \frac {12 D x^3+15 C x^2+20 B x+30 A}{60 \left (b x^3+a\right )^{2/3}}dx+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{60} a \int \frac {12 D x^3+15 C x^2+20 B x+30 A}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 2427

\(\displaystyle \frac {1}{60} a \left (\frac {\int \frac {2 \left (15 b C x^2+20 b B x+6 (5 A b-a D)\right )}{\left (b x^3+a\right )^{2/3}}dx}{2 b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{60} a \left (\frac {\int \frac {15 b C x^2+20 b B x+6 (5 A b-a D)}{\left (b x^3+a\right )^{2/3}}dx}{b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 2425

\(\displaystyle \frac {1}{60} a \left (\frac {\int \frac {6 (5 A b-a D)+20 b B x}{\left (b x^3+a\right )^{2/3}}dx+15 b C \int \frac {x^2}{\left (b x^3+a\right )^{2/3}}dx}{b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 793

\(\displaystyle \frac {1}{60} a \left (\frac {\int \frac {6 (5 A b-a D)+20 b B x}{\left (b x^3+a\right )^{2/3}}dx+15 C \sqrt [3]{a+b x^3}}{b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {1}{60} a \left (\frac {\int \left (\frac {6 (5 A b-a D)}{\left (b x^3+a\right )^{2/3}}+\frac {20 b B x}{\left (b x^3+a\right )^{2/3}}\right )dx+15 C \sqrt [3]{a+b x^3}}{b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{60} a \left (\frac {\frac {6 x \left (\frac {b x^3}{a}+1\right )^{2/3} (5 A b-a D) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {20 \sqrt [3]{b} B \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-10 \sqrt [3]{b} B \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )+15 C \sqrt [3]{a+b x^3}}{b}+\frac {6 D x \sqrt [3]{a+b x^3}}{b}\right )+\frac {1}{60} \sqrt [3]{a+b x^3} \left (30 A x+20 B x^2+15 C x^3+12 D x^4\right )\)

Input:

Int[(a + b*x^3)^(1/3)*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

((a + b*x^3)^(1/3)*(30*A*x + 20*B*x^2 + 15*C*x^3 + 12*D*x^4))/60 + (a*((6* 
D*x*(a + b*x^3)^(1/3))/b + (15*C*(a + b*x^3)^(1/3) - (20*b^(1/3)*B*ArcTan[ 
(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/Sqrt[3] + (6*(5*A*b - a*D) 
*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) - 10*b^(1/3)*B*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/b))/60
 

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 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]

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

Input:

int((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x)
 

Output:

int((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x)
 

Fricas [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Sympy [A] (verification not implemented)

Time = 2.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.68 \[ \int \sqrt [3]{a+b x^3} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A \sqrt [3]{a} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {B \sqrt [3]{a} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + C \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) + \frac {D \sqrt [3]{a} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:

integrate((b*x**3+a)**(1/3)*(D*x**3+C*x**2+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)) + C*Piecewise((a**(1/3)*x**3/3, E 
q(b, 0)), ((a + b*x**3)**(4/3)/(4*b), True)) + D*a**(1/3)*x**4*gamma(4/3)* 
hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))
 

Maxima [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Giac [F]

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

integrate((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(1/3)*(A + B*x + C*x^2 + x^3*D),x)
 

Output:

int((a + b*x^3)^(1/3)*(A + B*x + C*x^2 + x^3*D), x)
 

Reduce [F]

\[ \int \sqrt [3]{a+b x^3} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {30 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b x +15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a c +6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a d x +20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{2}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{3}+12 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{4}+30 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b -6 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} d +20 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \,b^{2}}{60 b} \] Input:

int((b*x^3+a)^(1/3)*(D*x^3+C*x^2+B*x+A),x)
 

Output:

(30*(a + b*x**3)**(1/3)*a*b*x + 15*(a + b*x**3)**(1/3)*a*c + 6*(a + b*x**3 
)**(1/3)*a*d*x + 20*(a + b*x**3)**(1/3)*b**2*x**2 + 15*(a + b*x**3)**(1/3) 
*b*c*x**3 + 12*(a + b*x**3)**(1/3)*b*d*x**4 + 30*int((a + b*x**3)**(1/3)/( 
a + b*x**3),x)*a**2*b - 6*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**2*d + 
 20*int(((a + b*x**3)**(1/3)*x)/(a + b*x**3),x)*a*b**2)/(60*b)