\(\int \frac {x \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [682]

Optimal result

Integrand size = 22, antiderivative size = 234 \[ \int \frac {x \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {\sqrt [3]{b} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d}+\frac {\sqrt [3]{b c-a d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} d}-\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 \sqrt [3]{c} d}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d}+\frac {\sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.90 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.81 \[ \int \frac {x \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {-4 \sqrt {3} \sqrt [3]{b} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-\frac {2 \sqrt {-6-6 i \sqrt {3}} \sqrt [3]{b c-a d} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{c}}-4 \sqrt [3]{b} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\frac {2 i \left (i+\sqrt {3}\right ) \sqrt [3]{b c-a d} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{c}}+2 \sqrt [3]{b} \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\frac {\left (1-i \sqrt {3}\right ) \sqrt [3]{b c-a d} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{c}}}{12 d} \] Input:

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

Output:

(-4*Sqrt[3]*b^(1/3)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^ 
(1/3))] - (2*Sqrt[-6 - (6*I)*Sqrt[3]]*(b*c - a*d)^(1/3)*ArcTan[(3*(b*c - a 
*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b 
*x^3)^(1/3))])/c^(1/3) - 4*b^(1/3)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + 
 ((2*I)*(I + Sqrt[3])*(b*c - a*d)^(1/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I 
*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/c^(1/3) + 2*b^(1/3)*Log[b^(2/3)*x^2 
+ b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)] + ((1 - I*Sqrt[3])*(b*c 
 - a*d)^(1/3)*Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c 
- a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/ 
3)])/c^(1/3))/(12*d)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {984, 853, 992}

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.

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

\(\Big \downarrow \) 984

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

\(\Big \downarrow \) 853

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

\(\Big \downarrow \) 992

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
 

rule 984

Int[((x_)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol 
] :> Simp[b/d   Int[x*(a + b*x^n)^(p - 1), x], x] - Simp[(b*c - a*d)/d   In 
t[x*((a + b*x^n)^(p - 1)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[p, 0] && IntBinomialQ[a, b, c, d, 1, 
 1, n, p, -1, x]
 

rule 992

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> 
With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3 
))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c* 
q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && 
 NeQ[b*c - a*d, 0]
 

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.39

method	result	size
pseudoelliptic	\(\frac {\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) \left (a d -b c \right )}{2}+\frac {b^{\frac {1}{3}} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}}{2}+\left (-a d +b c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )+b^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}-b^{\frac {1}{3}} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}+\left (a d -b c \right ) \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} d c}\)	\(325\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.41 \[ \int \frac {x \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {2 \, \sqrt {3} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) - 2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}}}{3 \, b x}\right ) + 2 \, \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 2 \, \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (-\frac {x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (\frac {x^{2} \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{6 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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