\(\int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx\) [1628]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.13 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx=\frac {3 b^{2/3} \sqrt [3]{d} \sqrt [3]{a+b x} (c+d x)^{2/3} (2 b c+a d+3 b d x)+2 \sqrt {3} (b c-a d)^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )+2 (b c-a d)^2 \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )-(b c-a d)^2 \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{18 b^{5/3} d^{4/3}} \] Input:

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

Output:

(3*b^(2/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(2*b*c + a*d + 3*b*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^(5/3)*d^(4/3))
 

Rubi [A] (verified)

Time = 0.22 (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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 716, normalized size of antiderivative = 3.25 \[ \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx =\text {Too large to display} \] Input:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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