\(\int (c+d x)^4 \sqrt [3]{a+b x^3} \, dx\) [143]

Optimal result

Integrand size = 19, antiderivative size = 304 \[ \int (c+d x)^4 \sqrt [3]{a+b x^3} \, dx=\frac {3 a c^2 d^2 \sqrt [3]{a+b x^3}}{2 b}+\frac {2 a c d^3 x \sqrt [3]{a+b x^3}}{5 b}+\frac {a d^4 x^2 \sqrt [3]{a+b x^3}}{18 b}+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )-\frac {a d \left (12 b c^3-a d^3\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3}}+\frac {a c \left (5 b c^3-4 a d^3\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{10 b \left (a+b x^3\right )^{2/3}}-\frac {a d \left (12 b c^3-a d^3\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3}} \] Output:

3/2*a*c^2*d^2*(b*x^3+a)^(1/3)/b+2/5*a*c*d^3*x*(b*x^3+a)^(1/3)/b+1/18*a*d^4 
*x^2*(b*x^3+a)^(1/3)/b+1/30*(b*x^3+a)^(1/3)*(5*d^4*x^5+24*c*d^3*x^4+45*c^2 
*d^2*x^3+40*c^3*d*x^2+15*c^4*x)-1/27*a*d*(-a*d^3+12*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^(5/3)+1/10*a*c*(-4*a*d^3+5* 
b*c^3)*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)-1/18*a*d*(-a*d^3+12*b*c^3)*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)
 

Mathematica [A] (verified)

Time = 9.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.54 \[ \int (c+d x)^4 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a+b x^3} \left (6 b c^4 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (12 b c^3-a d^3\right ) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+d^2 \left (\left (9 c^2+d^2 x^2\right ) \left (a+b x^3\right ) \sqrt [3]{1+\frac {b x^3}{a}}+6 b c d x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )\right )}{6 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:

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

Output:

((a + b*x^3)^(1/3)*(6*b*c^4*x*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/ 
a)] + d*(12*b*c^3 - a*d^3)*x^2*Hypergeometric2F1[-1/3, 2/3, 5/3, -((b*x^3) 
/a)] + d^2*((9*c^2 + d^2*x^2)*(a + b*x^3)*(1 + (b*x^3)/a)^(1/3) + 6*b*c*d* 
x^4*Hypergeometric2F1[-1/3, 4/3, 7/3, -((b*x^3)/a)])))/(6*b*(1 + (b*x^3)/a 
)^(1/3))
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2392, 27, 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.

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

\(\Big \downarrow \) 2392

\(\displaystyle a \int \frac {15 c^4+40 d x c^3+45 d^2 x^2 c^2+24 d^3 x^3 c+5 d^4 x^4}{30 \left (b x^3+a\right )^{2/3}}dx+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{30} a \int \frac {15 c^4+40 d x c^3+45 d^2 x^2 c^2+24 d^3 x^3 c+5 d^4 x^4}{\left (b x^3+a\right )^{2/3}}dx+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {1}{30} a \int \left (\frac {15 c^4}{\left (b x^3+a\right )^{2/3}}+\frac {40 d x c^3}{\left (b x^3+a\right )^{2/3}}+\frac {45 d^2 x^2 c^2}{\left (b x^3+a\right )^{2/3}}+\frac {24 d^3 x^3 c}{\left (b x^3+a\right )^{2/3}}+\frac {5 d^4 x^4}{\left (b x^3+a\right )^{2/3}}\right )dx+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{30} a \left (-\frac {40 c^3 d \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {10 a d^4 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}-\frac {20 c^3 d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+\frac {5 a d^4 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}}+\frac {15 c^4 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {45 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {6 c d^3 x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {5 d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )+\frac {1}{30} \sqrt [3]{a+b x^3} \left (15 c^4 x+40 c^3 d x^2+45 c^2 d^2 x^3+24 c d^3 x^4+5 d^4 x^5\right )\)

Input:

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

Output:

((a + b*x^3)^(1/3)*(15*c^4*x + 40*c^3*d*x^2 + 45*c^2*d^2*x^3 + 24*c*d^3*x^ 
4 + 5*d^4*x^5))/30 + (a*((45*c^2*d^2*(a + b*x^3)^(1/3))/b + (5*d^4*x^2*(a 
+ b*x^3)^(1/3))/(3*b) - (40*c^3*d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1 
/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) + (10*a*d^4*ArcTan[(1 + (2*b^(1/3)*x)/(a 
+ b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*b^(5/3)) + (15*c^4*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) + 
(6*c*d^3*x^4*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x 
^3)/a)])/(a + b*x^3)^(2/3) - (20*c^3*d*Log[b^(1/3)*x - (a + b*x^3)^(1/3)]) 
/b^(2/3) + (5*a*d^4*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(3*b^(5/3))))/30
 

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 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 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]

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

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

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

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.70 \[ \int (c+d x)^4 \sqrt [3]{a+b x^3} \, dx=\frac {\sqrt [3]{a} c^{4} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 \sqrt [3]{a} c^{3} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {4 \sqrt [3]{a} c d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} d^{4} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} + 6 c^{2} d^{2} \left (\begin {cases} \frac {\sqrt [3]{a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{4 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**(1/3)*c**4*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*p 
i)/a)/(3*gamma(4/3)) + 4*a**(1/3)*c**3*d*x**2*gamma(2/3)*hyper((-1/3, 2/3) 
, (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(5/3)) + 4*a**(1/3)*c*d**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)) + a**(1/3)*d**4*x**5*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), b*x**3*e 
xp_polar(I*pi)/a)/(3*gamma(8/3)) + 6*c**2*d**2*Piecewise((a**(1/3)*x**3/3, 
 Eq(b, 0)), ((a + b*x**3)**(4/3)/(4*b), True))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^4 \sqrt [3]{a+b x^3} \, dx=\frac {135 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,c^{2} d^{2}+36 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a c \,d^{3} x +5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,d^{4} x^{2}+45 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{4} x +120 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{3} d \,x^{2}+135 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,c^{2} d^{2} x^{3}+72 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,d^{3} x^{4}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,d^{4} x^{5}-36 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} c \,d^{3}+45 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a b \,c^{4}-10 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} d^{4}+120 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a b \,c^{3} d}{90 b} \] Input:

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

Output:

(135*(a + b*x**3)**(1/3)*a*c**2*d**2 + 36*(a + b*x**3)**(1/3)*a*c*d**3*x + 
 5*(a + b*x**3)**(1/3)*a*d**4*x**2 + 45*(a + b*x**3)**(1/3)*b*c**4*x + 120 
*(a + b*x**3)**(1/3)*b*c**3*d*x**2 + 135*(a + b*x**3)**(1/3)*b*c**2*d**2*x 
**3 + 72*(a + b*x**3)**(1/3)*b*c*d**3*x**4 + 15*(a + b*x**3)**(1/3)*b*d**4 
*x**5 - 36*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**2*c*d**3 + 45*int((a 
 + b*x**3)**(1/3)/(a + b*x**3),x)*a*b*c**4 - 10*int(((a + b*x**3)**(1/3)*x 
)/(a + b*x**3),x)*a**2*d**4 + 120*int(((a + b*x**3)**(1/3)*x)/(a + b*x**3) 
,x)*a*b*c**3*d)/(90*b)