\(\int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx\) [149]

Optimal result

Integrand size = 19, antiderivative size = 238 \[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\frac {3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac {4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac {d^4 x^2 \left (a+b x^3\right )^{2/3}}{4 b}+\frac {c \left (3 b c^3-4 a d^3\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {d \left (8 b c^3-a d^3\right ) x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{4 b \sqrt [3]{a+b x^3}}-\frac {c \left (3 b c^3-4 a d^3\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}} \] Output:

3*c^2*d^2*(b*x^3+a)^(2/3)/b+4/3*c*d^3*x*(b*x^3+a)^(2/3)/b+1/4*d^4*x^2*(b*x 
^3+a)^(2/3)/b+1/9*c*(-4*a*d^3+3*b*c^3)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a) 
^(1/3))*3^(1/2))*3^(1/2)/b^(4/3)+1/4*d*(-a*d^3+8*b*c^3)*x^2*(1+b*x^3/a)^(1 
/3)*hypergeom([1/3, 2/3],[5/3],-b*x^3/a)/b/(b*x^3+a)^(1/3)-1/6*c*(-4*a*d^3 
+3*b*c^3)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.53 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.65 \[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\frac {180 b^{4/3} c^3 d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+18 b^{4/3} d^4 x^5 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )+5 c \left (54 a \sqrt [3]{b} c d^2+24 a \sqrt [3]{b} d^3 x+54 b^{4/3} c d^2 x^3+24 b^{4/3} d^3 x^4+2 \sqrt {3} \left (3 b c^3-4 a d^3\right ) \sqrt [3]{a+b x^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )+2 \left (-3 b c^3+4 a d^3\right ) \sqrt [3]{a+b x^3} \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+3 b c^3 \sqrt [3]{a+b x^3} \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )-4 a d^3 \sqrt [3]{a+b x^3} \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )\right )}{90 b^{4/3} \sqrt [3]{a+b x^3}} \] Input:

Integrate[(c + d*x)^4/(a + b*x^3)^(1/3),x]
 

Output:

(180*b^(4/3)*c^3*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5 
/3, -((b*x^3)/a)] + 18*b^(4/3)*d^4*x^5*(1 + (b*x^3)/a)^(1/3)*Hypergeometri 
c2F1[1/3, 5/3, 8/3, -((b*x^3)/a)] + 5*c*(54*a*b^(1/3)*c*d^2 + 24*a*b^(1/3) 
*d^3*x + 54*b^(4/3)*c*d^2*x^3 + 24*b^(4/3)*d^3*x^4 + 2*Sqrt[3]*(3*b*c^3 - 
4*a*d^3)*(a + b*x^3)^(1/3)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sq 
rt[3]] + 2*(-3*b*c^3 + 4*a*d^3)*(a + b*x^3)^(1/3)*Log[1 - (b^(1/3)*x)/(a + 
 b*x^3)^(1/3)] + 3*b*c^3*(a + b*x^3)^(1/3)*Log[1 + (b^(2/3)*x^2)/(a + b*x^ 
3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)] - 4*a*d^3*(a + b*x^3)^(1/3)*Log[ 
1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]))/(90 
*b^(4/3)*(a + b*x^3)^(1/3))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.30, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

\(\displaystyle \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx\)

\(\Big \downarrow \) 2432

\(\displaystyle \int \left (\frac {c^4}{\sqrt [3]{a+b x^3}}+\frac {4 c^3 d x}{\sqrt [3]{a+b x^3}}+\frac {6 c^2 d^2 x^2}{\sqrt [3]{a+b x^3}}+\frac {4 c d^3 x^3}{\sqrt [3]{a+b x^3}}+\frac {d^4 x^4}{\sqrt [3]{a+b x^3}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 a c d^3 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {c^4 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {2 a c d^3 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{4/3}}-\frac {c^4 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac {2 c^3 d x^2 \sqrt [3]{\frac {b x^3}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac {3 c^2 d^2 \left (a+b x^3\right )^{2/3}}{b}+\frac {4 c d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac {d^4 x^5 \sqrt [3]{\frac {b x^3}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{5 \sqrt [3]{a+b x^3}}\)

Input:

Int[(c + d*x)^4/(a + b*x^3)^(1/3),x]
 

Output:

(3*c^2*d^2*(a + b*x^3)^(2/3))/b + (4*c*d^3*x*(a + b*x^3)^(2/3))/(3*b) + (c 
^4*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3) 
) - (4*a*c*d^3*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*S 
qrt[3]*b^(4/3)) + (2*c^3*d*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) + (d^4*x^5*(1 + (b*x^3)/a)^(1 
/3)*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)])/(5*(a + b*x^3)^(1/3)) 
- (c^4*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(2*b^(1/3)) + (2*a*c*d^3*Log 
[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(3*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]

\[\int \frac {\left (d x +c \right )^{4}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]

Input:

int((d*x+c)^4/(b*x^3+a)^(1/3),x)
 

Output:

int((d*x+c)^4/(b*x^3+a)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)^4/(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 2.73 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=6 c^{2} d^{2} \left (\begin {cases} \frac {x^{3}}{3 \sqrt [3]{a}} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c^{4} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {4 c^{3} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} + \frac {4 c d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{4} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {8}{3}\right )} \] Input:

integrate((d*x+c)**4/(b*x**3+a)**(1/3),x)
 

Output:

6*c**2*d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/ 
(2*b), True)) + c**4*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*x**3*exp_pol 
ar(I*pi)/a)/(3*a**(1/3)*gamma(4/3)) + 4*c**3*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)) + 4*c*d** 
3*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)) + d**4*x**5*gamma(5/3)*hyper((1/3, 5/3), (8/3,), b*x** 
3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(8/3))
 

Maxima [F]

\[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (d x + c\right )}^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*x+c)^4/(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^4 + inte 
grate((d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x)/(b*x^3 + a)^(1/3 
), x)
 

Giac [F]

\[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (d x + c\right )}^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*x+c)^4/(b*x^3+a)^(1/3),x, algorithm="giac")
 

Output:

integrate((d*x + c)^4/(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {{\left (c+d\,x\right )}^4}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:

int((c + d*x)^4/(a + b*x^3)^(1/3),x)
 

Output:

int((c + d*x)^4/(a + b*x^3)^(1/3), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^4}{\sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) d^{4}+4 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) c \,d^{3}+6 \left (\int \frac {x^{2}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) c^{2} d^{2}+4 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) c^{3} d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) c^{4} \] Input:

int((d*x+c)^4/(b*x^3+a)^(1/3),x)
 

Output:

int(x**4/(a + b*x**3)**(1/3),x)*d**4 + 4*int(x**3/(a + b*x**3)**(1/3),x)*c 
*d**3 + 6*int(x**2/(a + b*x**3)**(1/3),x)*c**2*d**2 + 4*int(x/(a + b*x**3) 
**(1/3),x)*c**3*d + int(1/(a + b*x**3)**(1/3),x)*c**4