\(\int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx\) [100]

Optimal result

Integrand size = 19, antiderivative size = 111 \[ \int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {d x \left (a+b x^3\right )^{2/3}}{3 b}+\frac {(3 b c-a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}-\frac {(3 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.47 \[ \int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {6 \sqrt [3]{b} d x \left (a+b x^3\right )^{2/3}+2 \sqrt {3} (3 b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+2 (-3 b c+a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+(3 b c-a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 b^{4/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {913, 769}

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.

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 769

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

Input:

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

Output:

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

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 913

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(224\) vs. \(2(88)=176\).

Time = 1.29 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.03

method	result	size
pseudoelliptic	\(\frac {6 d x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {1}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a d -6 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) b c +2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a d -6 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) b c -\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a d +3 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) b c}{18 b^{\frac {4}{3}}}\)	\(225\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 362, normalized size of antiderivative = 3.26 \[ \int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx=\left [\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b d x - 3 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c - a b d\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) - 2 \, {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{2}}, \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b d x - 2 \, {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \frac {6 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c - a b d\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}}}{18 \, b^{2}}\right ] \] Input:

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

Output:

[1/18*(6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2*c - a*b*d)*sqrt(-1/b 
^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/ 
3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/ 
b^(2/3)) + 2*a) - 2*(3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1 
/3))/x) + (3*b*c - a*d)*b^(2/3)*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^2, 1/18*(6*(b*x^3 + a)^(2/3)*b*d*x - 2*( 
3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c - a* 
d)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2 
/3))/x^2) - 6*sqrt(1/3)*(3*b^2*c - a*b*d)*arctan(sqrt(1/3)*(b^(1/3)*x + 2* 
(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3))/b^2]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx=\frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \] Input:

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

Output:

c*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**( 
1/3)*gamma(4/3)) + d*x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), b*x**3*exp_ 
polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (88) = 176\).

Time = 0.12 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.20 \[ \int \frac {c+d x^3}{\sqrt [3]{a+b x^3}} \, dx=-\frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}}\right )} c + \frac {1}{18} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{2}}\right )} d \] Input:

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

Output:

-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/ 
3))/b^(1/3) - log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3 
)/x^2)/b^(1/3) + 2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3))*c + 1/18*( 
2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/ 
b^(4/3) - a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/ 
x^2)/b^(4/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) - 6*(b*x^3 
+ a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2))*d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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