\(\int \frac {(a+b x^3)^{2/3}}{c+d x^3} \, dx\) [140]

Optimal result

Integrand size = 21, antiderivative size = 233 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d}-\frac {(b c-a d)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} d}-\frac {(b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 c^{2/3} d}+\frac {(b c-a d)^{2/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} d}-\frac {b^{2/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.91 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.82 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {4 \sqrt {3} b^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+\frac {2 \sqrt {-6+6 i \sqrt {3}} (b c-a d)^{2/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{2/3}}-4 b^{2/3} \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-\frac {2 i \left (-i+\sqrt {3}\right ) (b c-a d)^{2/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{2/3}}+2 b^{2/3} \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\frac {\left (1+i \sqrt {3}\right ) (b c-a d)^{2/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{2/3}}}{12 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {916, 769, 901}

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.

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

\(\Big \downarrow \) 916

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 769

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

rule 901

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

rule 916

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

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.40

method	result	size
pseudoelliptic	\(\frac {\frac {\left (-a d +b c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\frac {b^{\frac {2}{3}} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{2}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) \left (a d -b c \right )-b^{\frac {2}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}-b^{\frac {2}{3}} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (a d -b c \right ) \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} d c}\)	\(326\)

Input:

int((b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (186) = 372\).

Time = 0.13 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {2 \, \sqrt {3} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 2 \, \sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}}}{3 \, b x}\right ) - 2 \, \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}{x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {{\left (b c - a d\right )} x^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )}}{x^{2}}\right )}{6 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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