\(\int \frac {(c+d x^3)^2}{\sqrt [3]{a+b x^3}} \, dx\) [99]

Optimal result

Integrand size = 21, antiderivative size = 172 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx=\frac {2 d (3 b c-a d) x \left (a+b x^3\right )^{2/3}}{9 b^2}+\frac {d^2 x^4 \left (a+b x^3\right )^{2/3}}{6 b}+\frac {\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{7/3}}-\frac {\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{7/3}} \] Output:

2/9*d*(-a*d+3*b*c)*x*(b*x^3+a)^(2/3)/b^2+1/6*d^2*x^4*(b*x^3+a)^(2/3)/b+1/2 
7*(2*a^2*d^2-6*a*b*c*d+9*b^2*c^2)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3 
))*3^(1/2))*3^(1/2)/b^(7/3)-1/18*(2*a^2*d^2-6*a*b*c*d+9*b^2*c^2)*ln(-b^(1/ 
3)*x+(b*x^3+a)^(1/3))/b^(7/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.30 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx=\frac {3 \sqrt [3]{b} d x \left (a+b x^3\right )^{2/3} \left (-4 a d+3 b \left (4 c+d x^3\right )\right )+2 \sqrt {3} \left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 \left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{7/3}} \] Input:

Integrate[(c + d*x^3)^2/(a + b*x^3)^(1/3),x]
 

Output:

(3*b^(1/3)*d*x*(a + b*x^3)^(2/3)*(-4*a*d + 3*b*(4*c + d*x^3)) + 2*Sqrt[3]* 
(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x 
+ 2*(a + b*x^3)^(1/3))] - 2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*Log[-(b^(1 
/3)*x) + (a + b*x^3)^(1/3)] + (9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*Log[b^(2 
/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*b^(7/3))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {933, 913, 769}

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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \frac {d (9 b c-4 a d) x^3+c (6 b c-a d)}{\sqrt [3]{b x^3+a}}dx}{6 b}+\frac {d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {2 \left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{3 b}+\frac {d x \left (a+b x^3\right )^{2/3} (9 b c-4 a d)}{3 b}}{6 b}+\frac {d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {\frac {2 \left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}+\frac {d x \left (a+b x^3\right )^{2/3} (9 b c-4 a d)}{3 b}}{6 b}+\frac {d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}\)

Input:

Int[(c + d*x^3)^2/(a + b*x^3)^(1/3),x]
 

Output:

(d*x*(a + b*x^3)^(2/3)*(c + d*x^3))/(6*b) + ((d*(9*b*c - 4*a*d)*x*(a + b*x 
^3)^(2/3))/(3*b) + (2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(ArcTan[(1 + (2* 
b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x 
) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))/(3*b))/(6*b)
 

Defintions of rubi rules used

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(-\frac {2 \left (-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {d \,x^{3}}{4}+c \right ) d x \,b^{\frac {4}{3}}+a \,d^{2} x \,b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {2}{3}}+\frac {\left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (a^{2} d^{2}-3 a b c d +\frac {9}{2} b^{2} c^{2}\right )}{3}\right )}{9 b^{\frac {7}{3}}}\)	\(166\)

Input:

int((d*x^3+c)^2/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
 

Output:

-2/9*(-3*(b*x^3+a)^(2/3)*(1/4*d*x^3+c)*d*x*b^(4/3)+a*d^2*x*b^(1/3)*(b*x^3+ 
a)^(2/3)+1/3*(3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^( 
1/3)/x)+ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/2*ln((b^(2/3)*x^2+b^(1/3)*(b* 
x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*(a^2*d^2-3*a*b*c*d+9/2*b^2*c^2))/b^( 
7/3)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 554, normalized size of antiderivative = 3.22 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx =\text {Too large to display} \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

[1/54*(3*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d + 2*a^2*b*d^2)*sqrt((-b)^(1/3) 
/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^( 
1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3)*x)*s 
qrt((-b)^(1/3)/b) + 2*a) - 2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-b)^(2/3 
)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + (9*b^2*c^2 - 6*a*b*c*d + 2*a 
^2*d^2)*(-b)^(2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + 
(b*x^3 + a)^(2/3))/x^2) + 3*(3*b^2*d^2*x^4 + 4*(3*b^2*c*d - a*b*d^2)*x)*(b 
*x^3 + a)^(2/3))/b^3, -1/54*(6*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d + 2*a^2* 
b*d^2)*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a) 
^(1/3))*sqrt(-(-b)^(1/3)/b)/x) + 2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-b 
)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (9*b^2*c^2 - 6*a*b*c*d 
 + 2*a^2*d^2)*(-b)^(2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3 
)*x + (b*x^3 + a)^(2/3))/x^2) - 3*(3*b^2*d^2*x^4 + 4*(3*b^2*c*d - a*b*d^2) 
*x)*(b*x^3 + a)^(2/3))/b^3]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx=\frac {c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {2 c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} + \frac {d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

c**2*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)) + 2*c*d*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**2*x**7*gamma(7/3)*hyper 
((1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(10/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (145) = 290\).

Time = 0.12 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx =\text {Too large to display} \] Input:

integrate((d*x^3+c)^2/(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^2 + 1/9* 
(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3)) 
/b^(4/3) - a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3) 
/x^2)/b^(4/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) - 6*(b*x^3 
 + a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2))*c*d - 1/54*(4*sqrt(3)*a^2*a 
rctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 2*a 
^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(7 
/3) + 4*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) - 3*(7*(b*x^3 + a) 
^(2/3)*a^2*b/x^2 - 4*(b*x^3 + a)^(5/3)*a^2/x^5)/(b^4 - 2*(b*x^3 + a)*b^3/x 
^3 + (b*x^3 + a)^2*b^2/x^6))*d^2
 

Giac [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(1/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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