\(\int \frac {A+B x^3}{x (a+b x^3)^{2/3}} \, dx\) [346]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=\frac {B \sqrt [3]{a+b x^3}}{b}-\frac {A \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}-\frac {A \log (x)}{2 a^{2/3}}+\frac {A \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}} \] Output:

B*(b*x^3+a)^(1/3)/b-1/3*A*arctan(1/3*(a^(1/3)+2*(b*x^3+a)^(1/3))*3^(1/2)/a 
^(1/3))*3^(1/2)/a^(2/3)-1/2*A*ln(x)/a^(2/3)+1/2*A*ln(a^(1/3)-(b*x^3+a)^(1/ 
3))/a^(2/3)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=\frac {6 a^{2/3} B \sqrt [3]{a+b x^3}-2 \sqrt {3} A b \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 A b \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )-A b \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 a^{2/3} b} \] Input:

Integrate[(A + B*x^3)/(x*(a + b*x^3)^(2/3)),x]
 

Output:

(6*a^(2/3)*B*(a + b*x^3)^(1/3) - 2*Sqrt[3]*A*b*ArcTan[(1 + (2*(a + b*x^3)^ 
(1/3))/a^(1/3))/Sqrt[3]] + 2*A*b*Log[-a^(1/3) + (a + b*x^3)^(1/3)] - A*b*L 
og[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(6*a^(2/3)*b)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 90, 69, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {B x^3+A}{x^3 \left (b x^3+a\right )^{2/3}}dx^3\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(A + B*x^3)/(x*(a + b*x^3)^(2/3)),x]
 

Output:

((3*B*(a + b*x^3)^(1/3))/b + A*(-((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3 
))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x^3]/(2*a^(2/3)) + (3*Log[a^(1/3) - ( 
a + b*x^3)^(1/3)])/(2*a^(2/3))))/3
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 69

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

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 948

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(\frac {-A \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 a^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) b +A \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) b -\frac {A \ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) b}{2}+3 B \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}}{3 a^{\frac {2}{3}} b}\)	\(106\)

Input:

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

Output:

1/3*(-A*3^(1/2)*arctan(2/3*3^(1/2)/a^(1/3)*(b*x^3+a)^(1/3)+1/3*3^(1/2))*b+ 
A*ln((b*x^3+a)^(1/3)-a^(1/3))*b-1/2*A*ln((b*x^3+a)^(2/3)+a^(1/3)*(b*x^3+a) 
^(1/3)+a^(2/3))*b+3*B*(b*x^3+a)^(1/3)*a^(2/3))/a^(2/3)/b
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.42 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {6 \, \sqrt {\frac {1}{3}} A {\left (a^{2}\right )}^{\frac {1}{6}} a b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (a^{2}\right )}^{\frac {1}{6}} {\left ({\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{a^{2}}\right ) + A {\left (a^{2}\right )}^{\frac {2}{3}} b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, A {\left (a^{2}\right )}^{\frac {2}{3}} b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} B a^{2}}{6 \, a^{2} b} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 6.79 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=- \frac {A \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{2} \Gamma \left (\frac {5}{3}\right )} + B \left (\begin {cases} \frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {\sqrt [3]{a + b x^{3}}}{b} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((B*x**3+A)/x/(b*x**3+a)**(2/3),x)
 

Output:

-A*gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**3))/(3*b** 
(2/3)*x**2*gamma(5/3)) + B*Piecewise((x**3/(3*a**(2/3)), Eq(b, 0)), ((a + 
b*x**3)**(1/3)/b, True))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {1}{6} \, A {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {2 \, \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{a^{\frac {2}{3}}}\right )} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} B}{b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt {3} A \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {A \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}}} + \frac {A \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {2}{3}}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} B}{b} \] Input:

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

Output:

-1/3*sqrt(3)*A*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3)) 
/a^(2/3) - 1/6*A*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/ 
3))/a^(2/3) + 1/3*A*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/a^(2/3) + (b*x^3 
 + a)^(1/3)*B/b
 

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x^3}{x \left (a+b x^3\right )^{2/3}} \, dx=\frac {B\,{\left (b\,x^3+a\right )}^{1/3}}{b}-\frac {\ln \left (3\,A\,{\left (b\,x^3+a\right )}^{1/3}+\frac {3\,a^{1/3}\,\left (A-\sqrt {3}\,A\,1{}\mathrm {i}\right )}{2}\right )\,\left (A-\sqrt {3}\,A\,1{}\mathrm {i}\right )}{6\,a^{2/3}}-\frac {\ln \left (3\,A\,{\left (b\,x^3+a\right )}^{1/3}+\frac {3\,a^{1/3}\,\left (A+\sqrt {3}\,A\,1{}\mathrm {i}\right )}{2}\right )\,\left (A+\sqrt {3}\,A\,1{}\mathrm {i}\right )}{6\,a^{2/3}}+\frac {A\,\ln \left (3\,A\,a^{1/3}-3\,A\,{\left (b\,x^3+a\right )}^{1/3}\right )}{3\,a^{2/3}} \] Input:

int((A + B*x^3)/(x*(a + b*x^3)^(2/3)),x)
 

Output:

(B*(a + b*x^3)^(1/3))/b - (log(3*A*(a + b*x^3)^(1/3) + (3*a^(1/3)*(A - 3^( 
1/2)*A*1i))/2)*(A - 3^(1/2)*A*1i))/(6*a^(2/3)) - (log(3*A*(a + b*x^3)^(1/3 
) + (3*a^(1/3)*(A + 3^(1/2)*A*1i))/2)*(A + 3^(1/2)*A*1i))/(6*a^(2/3)) + (A 
*log(3*A*a^(1/3) - 3*A*(a + b*x^3)^(1/3)))/(3*a^(2/3))
 

Reduce [F]

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

int((B*x^3+A)/x/(b*x^3+a)^(2/3),x)
 

Output:

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