\(\int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx\) [151]

Optimal result

Integrand size = 19, antiderivative size = 147 \[ \int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx=\frac {d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac {c^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {c d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac {c^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}} \] Output:

1/2*d^2*(b*x^3+a)^(2/3)/b+1/3*c^2*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3 
))*3^(1/2))*3^(1/2)/b^(1/3)+c*d*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)-1/2*c^2*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^ 
(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.22 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx=\frac {d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac {c^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {c d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac {c^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}+\frac {c^2 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{6 \sqrt [3]{b}} \] Input:

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

Output:

(d^2*(a + b*x^3)^(2/3))/(2*b) + (c^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3) 
^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (c*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hyper 
geometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(a + b*x^3)^(1/3) - (c^2*Log[1 
- (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(3*b^(1/3)) + (c^2*Log[1 + (b^(2/3)*x^2) 
/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(6*b^(1/3))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.

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

\(\Big \downarrow \) 2425

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

\(\Big \downarrow \) 793

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

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^2*(a + b*x^3)^(2/3))/(2*b) + (c^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3) 
^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (c*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hyper 
geometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(a + b*x^3)^(1/3) - (c^2*Log[-( 
b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3))
 

Defintions of rubi rules used

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 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]

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

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]

\[ \int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] 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 = 1.54 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx=d^{2} \left (\begin {cases} \frac {x^{3}}{3 \sqrt [3]{a}} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} \] Input:

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

Output:

d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/(2*b), 
True)) + c**2*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)) + 2*c*d*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))
 

Maxima [F]

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

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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