3.1.30 \(\int \frac {(c+d x)^3}{\sqrt [3]{a+b x^3}} \, dx\) [30]

3.1.30.1 Optimal result

Integrand size = 19, antiderivative size = 255 \[ \int \frac {(c+d x)^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {3 c d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac {d^3 x \left (a+b x^3\right )^{2/3}}{3 b}+\frac {c^3 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {a d^3 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {3 c^2 d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac {c^3 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}+\frac {a d^3 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}} \]

output

3/2*c*d^2*(b*x^3+a)^(2/3)/b+1/3*d^3*x*(b*x^3+a)^(2/3)/b+3/2*c^2*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^3 
*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(1/3)+1/6*a*d^3*ln(-b^(1/3)*x+(b*x^3+a)^ 
(1/3))/b^(4/3)+1/3*c^3*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2)) 
/b^(1/3)*3^(1/2)-1/9*a*d^3*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1 
/2))/b^(4/3)*3^(1/2)
 

3.1.30.2 Mathematica [A] (verified)

Time = 10.07 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x)^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {1}{18} \left (\frac {27 c^2 d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac {27 \sqrt [3]{b} c d^2 \left (a+b x^3\right )^{2/3}+6 \sqrt [3]{b} d^3 x \left (a+b x^3\right )^{2/3}+2 \sqrt {3} \left (3 b c^3-a d^3\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )+\left (-6 b c^3+2 a d^3\right ) \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+3 b c^3 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )-a d^3 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{4/3}}\right ) \]

input

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

output

((27*c^2*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) + (27*b^(1/3)*c*d^2*(a + b*x^3)^(2/3) + 6*b^( 
1/3)*d^3*x*(a + b*x^3)^(2/3) + 2*Sqrt[3]*(3*b*c^3 - a*d^3)*ArcTan[(1 + (2* 
b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] + (-6*b*c^3 + 2*a*d^3)*Log[1 - (b^( 
1/3)*x)/(a + b*x^3)^(1/3)] + 3*b*c^3*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/ 
3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)] - a*d^3*Log[1 + (b^(2/3)*x^2)/(a + b*x 
^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/b^(4/3))/18
 

3.1.30.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.30.3.1 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])
 

3.1.30.4 Maple [F]

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

input

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

output

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

3.1.30.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.1.30.6 Sympy [A] (verification not implemented)

Time = 1.91 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x)^3}{\sqrt [3]{a+b x^3}} \, dx=3 c d^{2} \left (\begin {cases} \frac {x^{3}}{3 \sqrt [3]{a}} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c^{3} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {c^{2} d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{\sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} + \frac {d^{3} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \]

input

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

output

3*c*d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/(2* 
b), True)) + c**3*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)) + c**2*d*x**2*gamma(2/3)*hyper((1/3, 2/3) 
, (5/3,), b*x**3*exp_polar(I*pi)/a)/(a**(1/3)*gamma(5/3)) + d**3*x**4*gamm 
a(4/3)*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gam 
ma(7/3))
 

3.1.30.7 Maxima [F]

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

input

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

3.1.30.8 Giac [F]

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

input

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

output

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

3.1.30.9 Mupad [F(-1)]

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

input

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

output

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