\(\int \sqrt [3]{a-b x} (a+b x)^{2/3} \, dx\) [141]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 152 \[ \int \sqrt [3]{a-b x} (a+b x)^{2/3} \, dx=\frac {2 a \sqrt [3]{a-b x} (a+b x)^{2/3}}{3 b}-\frac {(a-b x)^{4/3} (a+b x)^{2/3}}{2 b}-\frac {4 a^2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a-b x}}\right )}{3 \sqrt {3} b}-\frac {2 a^2 \log (a-b x)}{9 b}-\frac {2 a^2 \log \left (1+\frac {\sqrt [3]{a+b x}}{\sqrt [3]{a-b x}}\right )}{3 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05 \[ \int \sqrt [3]{a-b x} (a+b x)^{2/3} \, dx=\frac {3 \sqrt [3]{a-b x} (a+b x)^{2/3} (a+3 b x)-8 \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{a+b x}}{-2 \sqrt [3]{a-b x}+\sqrt [3]{a+b x}}\right )-8 a^2 \log \left (b \left (\sqrt [3]{a-b x}+\sqrt [3]{a+b x}\right )\right )+4 a^2 \log \left ((a-b x)^{2/3}-\sqrt [3]{a-b x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{18 b} \] Input:

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

Output:

(3*(a - b*x)^(1/3)*(a + b*x)^(2/3)*(a + 3*b*x) - 8*Sqrt[3]*a^2*ArcTan[(Sqr 
t[3]*(a + b*x)^(1/3))/(-2*(a - b*x)^(1/3) + (a + b*x)^(1/3))] - 8*a^2*Log[ 
b*((a - b*x)^(1/3) + (a + b*x)^(1/3))] + 4*a^2*Log[(a - b*x)^(2/3) - (a - 
b*x)^(1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)])/(18*b)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {60, 60, 72}

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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {2}{3} a \int \frac {\sqrt [3]{a-b x}}{\sqrt [3]{a+b x}}dx-\frac {(a-b x)^{4/3} (a+b x)^{2/3}}{2 b}\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 72

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 72
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> 
 With[{q = Rt[-d/b, 3]}, Simp[Sqrt[3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a + b* 
x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08 \[ \int \sqrt [3]{a-b x} (a+b x)^{2/3} \, dx=-\frac {8 \, \sqrt {3} a^{2} \arctan \left (-\frac {\sqrt {3} {\left (b x + a\right )} - 2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}}{3 \, {\left (b x + a\right )}}\right ) + 8 \, a^{2} \log \left (\frac {b x + {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}} + a}{b x + a}\right ) - 4 \, a^{2} \log \left (\frac {b x - {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (-b x + a\right )}^{\frac {2}{3}} + a}{b x + a}\right ) - 3 \, {\left (3 \, b x + a\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}}{18 \, b} \] Input:

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

Output:

-1/18*(8*sqrt(3)*a^2*arctan(-1/3*(sqrt(3)*(b*x + a) - 2*sqrt(3)*(b*x + a)^ 
(2/3)*(-b*x + a)^(1/3))/(b*x + a)) + 8*a^2*log((b*x + (b*x + a)^(2/3)*(-b* 
x + a)^(1/3) + a)/(b*x + a)) - 4*a^2*log((b*x - (b*x + a)^(2/3)*(-b*x + a) 
^(1/3) + (b*x + a)^(1/3)*(-b*x + a)^(2/3) + a)/(b*x + a)) - 3*(3*b*x + a)* 
(b*x + a)^(2/3)*(-b*x + a)^(1/3))/b
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((-b*x+a)^(1/3)*(b*x+a)^(2/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [3]{a-b x} (a+b x)^{2/3} \, dx=\frac {9 \left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} a +3 \left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} b x +8 \left (\int \frac {\left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} x}{-b^{2} x^{2}+a^{2}}d x \right ) a \,b^{2}}{6 b} \] Input:

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

Output:

(9*(a + b*x)**(2/3)*(a - b*x)**(1/3)*a + 3*(a + b*x)**(2/3)*(a - b*x)**(1/ 
3)*b*x + 8*int(((a + b*x)**(2/3)*(a - b*x)**(1/3)*x)/(a**2 - b**2*x**2),x) 
*a*b**2)/(6*b)