\(\int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx\) [149]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 309 \[ \int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx=\frac {3}{5} x \sqrt [3]{a-b x} \sqrt [3]{a+b x}+\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a^4 \left (1-\frac {b^2 x^2}{a^2}\right )^{2/3} \left (1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right ) \sqrt {\frac {1+\sqrt [3]{1-\frac {b^2 x^2}{a^2}}+\left (1-\frac {b^2 x^2}{a^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}}} \] Output:

3/5*x*(-b*x+a)^(1/3)*(b*x+a)^(1/3)+2/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a 
^4*(1-b^2*x^2/a^2)^(2/3)*(1-(1-b^2*x^2/a^2)^(1/3))*((1+(1-b^2*x^2/a^2)^(1/ 
3)+(1-b^2*x^2/a^2)^(2/3))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3))^2)^(1/2)*Ellip 
ticF((1+3^(1/2)-(1-b^2*x^2/a^2)^(1/3))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3)),2 
*I-I*3^(1/2))/b^2/x/(-b*x+a)^(2/3)/(b*x+a)^(2/3)/(-(1-(1-b^2*x^2/a^2)^(1/3 
))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.22 \[ \int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx=-\frac {3 (a-b x)^{4/3} \sqrt [3]{a+b x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},\frac {a-b x}{2 a}\right )}{2\ 2^{2/3} b \sqrt [3]{\frac {a+b x}{a}}} \] Input:

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

Output:

(-3*(a - b*x)^(4/3)*(a + b*x)^(1/3)*Hypergeometric2F1[-1/3, 4/3, 7/3, (a - 
 b*x)/(2*a)])/(2*2^(2/3)*b*((a + b*x)/a)^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {46, 211, 234, 760}

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.

\(\displaystyle \int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx\)

\(\Big \downarrow \) 46

\(\displaystyle \frac {\sqrt [3]{a-b x} \sqrt [3]{a+b x} \int \sqrt [3]{a^2-b^2 x^2}dx}{\sqrt [3]{a^2-b^2 x^2}}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\sqrt [3]{a-b x} \sqrt [3]{a+b x} \left (\frac {2}{5} a^2 \int \frac {1}{\left (a^2-b^2 x^2\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a^2-b^2 x^2}\right )}{\sqrt [3]{a^2-b^2 x^2}}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {\sqrt [3]{a-b x} \sqrt [3]{a+b x} \left (\frac {3}{5} x \sqrt [3]{a^2-b^2 x^2}-\frac {3 a^2 \sqrt {-b^2 x^2} \int \frac {1}{\sqrt {-b^2 x^2}}d\sqrt [3]{a^2-b^2 x^2}}{5 b^2 x}\right )}{\sqrt [3]{a^2-b^2 x^2}}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {\sqrt [3]{a-b x} \sqrt [3]{a+b x} \left (\frac {3}{5} x \sqrt [3]{a^2-b^2 x^2}+\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right ) \sqrt {\frac {a^{4/3}+\left (a^2-b^2 x^2\right )^{2/3}+a^{2/3} \sqrt [3]{a^2-b^2 x^2}}{\left (\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}}{\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {a^{2/3} \left (a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )^2}}}\right )}{\sqrt [3]{a^2-b^2 x^2}}\)

Input:

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

Output:

((a - b*x)^(1/3)*(a + b*x)^(1/3)*((3*x*(a^2 - b^2*x^2)^(1/3))/5 + (2*3^(3/ 
4)*Sqrt[2 - Sqrt[3]]*a^2*(a^(2/3) - (a^2 - b^2*x^2)^(1/3))*Sqrt[(a^(4/3) + 
 a^(2/3)*(a^2 - b^2*x^2)^(1/3) + (a^2 - b^2*x^2)^(2/3))/((1 - Sqrt[3])*a^( 
2/3) - (a^2 - b^2*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(2/3) - 
 (a^2 - b^2*x^2)^(1/3))/((1 - Sqrt[3])*a^(2/3) - (a^2 - b^2*x^2)^(1/3))], 
-7 + 4*Sqrt[3]])/(5*b^2*x*Sqrt[-((a^(2/3)*(a^(2/3) - (a^2 - b^2*x^2)^(1/3) 
))/((1 - Sqrt[3])*a^(2/3) - (a^2 - b^2*x^2)^(1/3))^2)])))/(a^2 - b^2*x^2)^ 
(1/3)
 

Defintions of rubi rules used

rule 46
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(a 
 + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^FracPart[m])   I 
nt[(a*c + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 
 0] &&  !IntegerQ[2*m]
 

rule 211
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 234
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 760
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x + a)^(1/3)*(-b*x + a)^(1/3), x)
 

Sympy [F]

\[ \int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx=\int \sqrt [3]{a - b x} \sqrt [3]{a + b x}\, dx \] Input:

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

Output:

Integral((a - b*x)**(1/3)*(a + b*x)**(1/3), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{a-b x} \sqrt [3]{a+b x} \, dx=\frac {3 \left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {1}{3}} x}{5}+\frac {2 \left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {1}{3}}}{-b^{2} x^{2}+a^{2}}d x \right ) a^{2}}{5} \] Input:

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

Output:

(3*(a + b*x)**(1/3)*(a - b*x)**(1/3)*x + 2*int(((a + b*x)**(1/3)*(a - b*x) 
**(1/3))/(a**2 - b**2*x**2),x)*a**2)/5