\(\int \frac {A+B x+D x^3}{\sqrt [3]{a+b x^3}} \, dx\) [100]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {A+B x+D x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {D x \left (a+b x^3\right )^{2/3}}{3 b}+\frac {(3 A b-a D) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {B x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac {(3 A b-a D) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 10.38 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.84 \[ \int \frac {A+B x+D x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {1}{18} \left (\frac {6 a D x}{b \sqrt [3]{a+b x^3}}+\frac {6 D x^4}{\sqrt [3]{a+b x^3}}+\frac {2 \sqrt {3} (3 A b-a D) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{b^{4/3}}+\frac {9 B x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac {6 A \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}+\frac {2 a D \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{4/3}}+\frac {3 A \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b}}-\frac {a D \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{4/3}}\right ) \] Input:

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

Output:

((6*a*D*x)/(b*(a + b*x^3)^(1/3)) + (6*D*x^4)/(a + b*x^3)^(1/3) + (2*Sqrt[3 
]*(3*A*b - a*D)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/b^( 
4/3) + (9*B*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -(( 
b*x^3)/a)])/(a + b*x^3)^(1/3) - (6*A*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3) 
])/b^(1/3) + (2*a*D*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/b^(4/3) + (3*A 
*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]) 
/b^(1/3) - (a*D*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + 
 b*x^3)^(1/3)])/b^(4/3))/18
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.35, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {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.

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

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(D*x*(a + b*x^3)^(2/3))/(3*b) + (A*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^( 
1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) - (a*D*ArcTan[(1 + (2*b^(1/3)*x)/(a + b* 
x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*b^(4/3)) + (B*x^2*(1 + (b*x^3)/a)^(1/3)*H 
ypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3)^(1/3)) - (A* 
Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (a*D*Log[-(b^(1/3)*x) 
 + (a + b*x^3)^(1/3)])/(6*b^(4/3))
 

Defintions of rubi rules used

rule 2009

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

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]

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

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x+D x^3}{\sqrt [3]{a+b x^3}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.74 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x+D x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {A x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {B x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} + \frac {D x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \] Input:

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

Output:

A*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**( 
1/3)*gamma(4/3)) + B*x**2*gamma(2/3)*hyper((1/3, 2/3), (5/3,), b*x**3*exp_ 
polar(I*pi)/a)/(3*a**(1/3)*gamma(5/3)) + D*x**4*gamma(4/3)*hyper((1/3, 4/3 
), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3))
 

Maxima [F]

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

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

Output:

-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/ 
3))/b^(1/3) - log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3 
)/x^2)/b^(1/3) + 2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3))*A + integr 
ate((D*x^3 + B*x)/(b*x^3 + a)^(1/3), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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