\(\int \frac {(c+d x)^2}{(a+b x^3)^{2/3}} \, dx\) [157]

Optimal result

Integrand size = 19, antiderivative size = 141 \[ \int \frac {(c+d x)^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {d^2 \sqrt [3]{a+b x^3}}{b}-\frac {2 c d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {c^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {c d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.67 \[ \int \frac {(c+d x)^2}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {b c^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (d \left (a+b x^3\right )+b c x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {b x^3}{a+b x^3}\right )\right )}{b \left (a+b x^3\right )^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 141, 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.

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

\(\Big \downarrow \) 2425

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

\(\Big \downarrow \) 793

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

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^2*(a + b*x^3)^(1/3))/b - (2*c*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^( 
1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) + (c^2*x*(1 + (b*x^3)/a)^(2/3)*Hypergeom 
etric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x^3)^(2/3) - (c*d*Log[b^(1/3 
)*x - (a + b*x^3)^(1/3)])/b^(2/3)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.77 \[ \int \frac {(c+d x)^2}{\left (a+b x^3\right )^{2/3}} \, dx=d^{2} \left (\begin {cases} \frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {\sqrt [3]{a + b x^{3}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \] Input:

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

Output:

d**2*Piecewise((x**3/(3*a**(2/3)), Eq(b, 0)), ((a + b*x**3)**(1/3)/b, True 
)) + c**2*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a) 
/(3*a**(2/3)*gamma(4/3)) + 2*c*d*x**2*gamma(2/3)*hyper((2/3, 2/3), (5/3,), 
 b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(5/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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