\(\int \frac {(c+d x)^4}{(a+b x^3)^{2/3}} \, dx\) [155]

Optimal result

Integrand size = 19, antiderivative size = 231 \[ \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {6 c^2 d^2 \sqrt [3]{a+b x^3}}{b}+\frac {2 c d^3 x \sqrt [3]{a+b x^3}}{b}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{3 b}-\frac {2 d \left (6 b c^3-a d^3\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3}}+\frac {c \left (b c^3-2 a d^3\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{b \left (a+b x^3\right )^{2/3}}-\frac {d \left (6 b c^3-a d^3\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3}} \] Output:

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

Mathematica [A] (verified)

Time = 10.16 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.72 \[ \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {3 b c^4 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )+d \left (\left (6 b c^3-a d^3\right ) x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {b x^3}{a+b x^3}\right )+d \left (\left (18 c^2+d^2 x^2\right ) \left (a+b x^3\right )+3 b c d x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )\right )\right )}{3 b \left (a+b x^3\right )^{2/3}} \] Input:

Integrate[(c + d*x)^4/(a + b*x^3)^(2/3),x]
 

Output:

(3*b*c^4*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3 
)/a)] + d*((6*b*c^3 - a*d^3)*x^2*Hypergeometric2F1[2/3, 1, 5/3, (b*x^3)/(a 
 + b*x^3)] + d*((18*c^2 + d^2*x^2)*(a + b*x^3) + 3*b*c*d*x^4*(1 + (b*x^3)/ 
a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3)/a)])))/(3*b*(a + b*x^3 
)^(2/3))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.32, 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.

\(\displaystyle \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

Input:

Int[(c + d*x)^4/(a + b*x^3)^(2/3),x]
 

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)^4/(b*x^3+a)^(2/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [A] (verification not implemented)

Time = 2.76 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx=6 c^{2} d^{2} \left (\begin {cases} \frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {\sqrt [3]{a + b x^{3}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {c^{4} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {4 c^{3} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} + \frac {4 c d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{4} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {8}{3}\right )} \] Input:

integrate((d*x+c)**4/(b*x**3+a)**(2/3),x)
 

Output:

6*c**2*d**2*Piecewise((x**3/(3*a**(2/3)), Eq(b, 0)), ((a + b*x**3)**(1/3)/ 
b, True)) + c**4*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)) + 4*c**3*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)) + 4*c*d**3*x* 
*4*gamma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2 
/3)*gamma(7/3)) + d**4*x**5*gamma(5/3)*hyper((2/3, 5/3), (8/3,), b*x**3*ex 
p_polar(I*pi)/a)/(3*a**(2/3)*gamma(8/3))
 

Maxima [F]

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

integrate((d*x+c)^4/(b*x^3+a)^(2/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x+c)^4/(b*x^3+a)^(2/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^4}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {{\left (c+d\,x\right )}^4}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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