\(\int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx\) [517]

Optimal result

Integrand size = 19, antiderivative size = 220 \[ \int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx=\frac {(b c-a d) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b d}+\frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 b}+\frac {(b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} b^{4/3} d^{5/3}}+\frac {(b c-a d)^2 \log (c+d x)}{18 b^{4/3} d^{5/3}}+\frac {(b c-a d)^2 \log \left (1-\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 b^{4/3} d^{5/3}} \] Output:

1/6*(-a*d+b*c)*(b*x+a)^(2/3)*(d*x+c)^(1/3)/b/d+1/2*(b*x+a)^(5/3)*(d*x+c)^( 
1/3)/b+1/9*(-a*d+b*c)^2*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)*3^(1/ 
2)/b^(1/3)/(d*x+c)^(1/3))*3^(1/2)/b^(4/3)/d^(5/3)+1/18*(-a*d+b*c)^2*ln(d*x 
+c)/b^(4/3)/d^(5/3)+1/6*(-a*d+b*c)^2*ln(1-d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d 
*x+c)^(1/3))/b^(4/3)/d^(5/3)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.13 \[ \int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx=\frac {3 \sqrt [3]{b} d^{2/3} (a+b x)^{2/3} \sqrt [3]{c+d x} (2 a d+b (c+3 d x))-2 \sqrt {3} (b c-a d)^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )+2 (b c-a d)^2 \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )-(b c-a d)^2 \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{18 b^{4/3} d^{5/3}} \] Input:

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

Output:

(3*b^(1/3)*d^(2/3)*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(2*a*d + b*(c + 3*d*x)) 
 - 2*Sqrt[3]*(b*c - a*d)^2*ArcTan[(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))/(2*d^( 
1/3)*(a + b*x)^(1/3) + b^(1/3)*(c + d*x)^(1/3))] + 2*(b*c - a*d)^2*Log[d^( 
1/3)*(a + b*x)^(1/3) - b^(1/3)*(c + d*x)^(1/3)] - (b*c - a*d)^2*Log[d^(2/3 
)*(a + b*x)^(2/3) + b^(1/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(1/3) + b^(2 
/3)*(c + d*x)^(2/3)])/(18*b^(4/3)*d^(5/3))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {60, 60, 71}

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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \int \frac {(a+b x)^{2/3}}{(c+d x)^{2/3}}dx}{6 b}+\frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \left (\frac {(a+b x)^{2/3} \sqrt [3]{c+d x}}{d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}}dx}{3 d}\right )}{6 b}+\frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 b}\)

\(\Big \downarrow \) 71

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

Input:

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

Output:

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

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 71

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[2*q*((a + b*x)^(1/3)/( 
Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[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])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[d/b]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 717, normalized size of antiderivative = 3.26 \[ \int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/18*(3*sqrt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*sqrt(-(b*d^2)^( 
1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 + 3*(b*d^2)^(1/3)*(b*x + a)^(2/3) 
*(d*x + c)^(1/3)*d + 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - 
(b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (b*d^2)^(1/3)*(b*d*x + a*d 
))*sqrt(-(b*d^2)^(1/3)/b)) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/ 
3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b* 
x + a)) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/3)*log(((b*x + a)^(1/ 
3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + ( 
b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*x + b^2*c*d^2 + 2*a* 
b*d^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^2*d^3), -1/18*(6*sqrt(1/3)*(b^3 
*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*sqrt((b*d^2)^(1/3)/b)*arctan(sqrt(1/3) 
*(2*(b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + 
 a*d))*sqrt((b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2)) - 2*(b^2*c^2 - 2*a*b*c*d + 
 a^2*d^2)*(b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2) 
^(2/3)*(b*x + a))/(b*x + a)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/ 
3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x + a)^(2/3 
)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) - 3*(3*b^2*d^3 
*x + b^2*c*d^2 + 2*a*b*d^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^2*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx=\frac {9 \left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a c +3 \left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a d x +6 \left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} b c x +2 \left (\int \frac {\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a^{3} d^{3}-6 \left (\int \frac {\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) a \,b^{2} c^{2} d +4 \left (\int \frac {\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} x}{a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+a^{2} d^{2} x +3 a b c d x +2 b^{2} c^{2} x +a^{2} c d +2 a b \,c^{2}}d x \right ) b^{3} c^{3}}{6 a d +12 b c} \] Input:

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

Output:

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