\(\int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx\) [530]

Optimal result

Integrand size = 19, antiderivative size = 172 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx=\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}+\frac {(b c-a d) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{\sqrt {3} b^{2/3} d^{4/3}}+\frac {(b c-a d) \log (a+b x)}{6 b^{2/3} d^{4/3}}+\frac {(b c-a d) \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{2 b^{2/3} d^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx=\frac {6 b^{2/3} \sqrt [3]{d} \sqrt [3]{a+b x} (c+d x)^{2/3}+2 \sqrt {3} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )+2 (b c-a d) \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )-(b c-a d) \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{6 b^{2/3} d^{4/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (133) = 266\).

Time = 0.09 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.59 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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