\(\int \frac {1}{(a x^3+b x^6)^{4/3}} \, dx\) [7]

Optimal result

Integrand size = 15, antiderivative size = 217 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}-\frac {4 \left (a x^3+b x^6\right )^{2/3}}{3 a^2 x^5}-\frac {4 b x \sqrt [3]{a+b x^3} \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} \sqrt [3]{a x^3+b x^6}}+\frac {2 b x \sqrt [3]{a+b x^3} \log (x)}{3 a^{7/3} \sqrt [3]{a x^3+b x^6}}-\frac {2 b x \sqrt [3]{a+b x^3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{3 a^{7/3} \sqrt [3]{a x^3+b x^6}} \] Output:

1/a/x^2/(b*x^6+a*x^3)^(1/3)-4/3*(b*x^6+a*x^3)^(2/3)/a^2/x^5-4/9*b*x*(b*x^3 
+a)^(1/3)*arctan(1/3*(a^(1/3)+2*(b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/ 
a^(7/3)/(b*x^6+a*x^3)^(1/3)+2/3*b*x*(b*x^3+a)^(1/3)*ln(x)/a^(7/3)/(b*x^6+a 
*x^3)^(1/3)-2/3*b*x*(b*x^3+a)^(1/3)*ln(a^(1/3)-(b*x^3+a)^(1/3))/a^(7/3)/(b 
*x^6+a*x^3)^(1/3)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=-\frac {3 a^{4/3}+12 \sqrt [3]{a} b x^3+4 \sqrt {3} b x^3 \sqrt [3]{a+b x^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 b x^3 \sqrt [3]{a+b x^3} \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )-2 b x^3 \sqrt [3]{a+b x^3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{9 a^{7/3} x^2 \sqrt [3]{x^3 \left (a+b x^3\right )}} \] Input:

Integrate[(a*x^3 + b*x^6)^(-4/3),x]
 

Output:

-1/9*(3*a^(4/3) + 12*a^(1/3)*b*x^3 + 4*Sqrt[3]*b*x^3*(a + b*x^3)^(1/3)*Arc 
Tan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 4*b*x^3*(a + b*x^3)^(1/ 
3)*Log[-a^(1/3) + (a + b*x^3)^(1/3)] - 2*b*x^3*(a + b*x^3)^(1/3)*Log[a^(2/ 
3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(a^(7/3)*x^2*(x^3*(a 
+ b*x^3))^(1/3))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.78, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1912, 1931, 1917, 798, 67, 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.

\(\displaystyle \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx\)

\(\Big \downarrow \) 1912

\(\displaystyle \frac {4 \int \frac {1}{x^3 \sqrt [3]{b x^6+a x^3}}dx}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {4 \left (-\frac {b \int \frac {1}{\sqrt [3]{b x^6+a x^3}}dx}{3 a}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \int \frac {1}{x \sqrt [3]{b x^3+a}}dx}{3 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \int \frac {1}{x^3 \sqrt [3]{b x^3+a}}dx^3}{9 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \left (\frac {3}{2} \int \frac {1}{x^6+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 \sqrt [3]{a}}\right )}{9 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \left (\frac {3}{2} \int \frac {1}{x^6+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 \sqrt [3]{a}}\right )}{9 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \left (-\frac {3 \int \frac {1}{-x^6-3}d\left (\frac {2 \sqrt [3]{b x^3+a}}{\sqrt [3]{a}}+1\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 \sqrt [3]{a}}\right )}{9 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {4 \left (-\frac {b x \sqrt [3]{a+b x^3} \left (\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 \sqrt [3]{a}}\right )}{9 a \sqrt [3]{a x^3+b x^6}}-\frac {\left (a x^3+b x^6\right )^{2/3}}{3 a x^5}\right )}{a}+\frac {1}{a x^2 \sqrt [3]{a x^3+b x^6}}\)

Input:

Int[(a*x^3 + b*x^6)^(-4/3),x]
 

Output:

1/(a*x^2*(a*x^3 + b*x^6)^(1/3)) + (4*(-1/3*(a*x^3 + b*x^6)^(2/3)/(a*x^5) - 
 (b*x*(a + b*x^3)^(1/3)*((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3 
))/Sqrt[3]])/a^(1/3) - Log[x^3]/(2*a^(1/3)) + (3*Log[a^(1/3) - (a + b*x^3) 
^(1/3)])/(2*a^(1/3))))/(9*a*(a*x^3 + b*x^6)^(1/3))))/a
 

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 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   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 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 798

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

rule 1912

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + 
 b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ 
(a*(n - j)*(p + 1))   Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, 
b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ 
m + j*p + 1, 0]
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{6}+a \,x^{3}\right )^{\frac {4}{3}}}d x\]

Input:

int(1/(b*x^6+a*x^3)^(4/3),x)
 

Output:

int(1/(b*x^6+a*x^3)^(4/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx =\text {Too large to display} \] Input:

integrate(1/(b*x^6+a*x^3)^(4/3),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[1/9*(6*sqrt(1/3)*(a*b^2*x^8 + a^2*b*x^5)*sqrt((-a)^(1/3)/a)*log((2*b*x^5 
+ 3*a*x^2 - 3*(b*x^6 + a*x^3)^(1/3)*(-a)^(2/3)*x - 3*sqrt(1/3)*((-a)^(1/3) 
*a*x^2 - (b*x^6 + a*x^3)^(1/3)*a*x + 2*(b*x^6 + a*x^3)^(2/3)*(-a)^(2/3))*s 
qrt((-a)^(1/3)/a))/x^5) - 4*(b^2*x^8 + a*b*x^5)*(-a)^(2/3)*log(((-a)^(1/3) 
*x + (b*x^6 + a*x^3)^(1/3))/x) + 2*(b^2*x^8 + a*b*x^5)*(-a)^(2/3)*log(((-a 
)^(2/3)*x^2 - (b*x^6 + a*x^3)^(1/3)*(-a)^(1/3)*x + (b*x^6 + a*x^3)^(2/3))/ 
x^2) - 3*(b*x^6 + a*x^3)^(2/3)*(4*a*b*x^3 + a^2))/(a^3*b*x^8 + a^4*x^5), - 
1/9*(12*sqrt(1/3)*(a*b^2*x^8 + a^2*b*x^5)*sqrt(-(-a)^(1/3)/a)*arctan(-sqrt 
(1/3)*((-a)^(1/3)*x - 2*(b*x^6 + a*x^3)^(1/3))*sqrt(-(-a)^(1/3)/a)/x) + 4* 
(b^2*x^8 + a*b*x^5)*(-a)^(2/3)*log(((-a)^(1/3)*x + (b*x^6 + a*x^3)^(1/3))/ 
x) - 2*(b^2*x^8 + a*b*x^5)*(-a)^(2/3)*log(((-a)^(2/3)*x^2 - (b*x^6 + a*x^3 
)^(1/3)*(-a)^(1/3)*x + (b*x^6 + a*x^3)^(2/3))/x^2) + 3*(b*x^6 + a*x^3)^(2/ 
3)*(4*a*b*x^3 + a^2))/(a^3*b*x^8 + a^4*x^5)]
 

Sympy [F]

\[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=\int \frac {1}{\left (a x^{3} + b x^{6}\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(b*x**6+a*x**3)**(4/3),x)
 

Output:

Integral((a*x**3 + b*x**6)**(-4/3), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=\int { \frac {1}{{\left (b x^{6} + a x^{3}\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(b*x^6+a*x^3)^(4/3),x, algorithm="maxima")
 

Output:

integrate((b*x^6 + a*x^3)^(-4/3), x)
 

Giac [F]

\[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=\int { \frac {1}{{\left (b x^{6} + a x^{3}\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(b*x^6+a*x^3)^(4/3),x, algorithm="giac")
 

Output:

integrate((b*x^6 + a*x^3)^(-4/3), x)
 

Mupad [B] (verification not implemented)

Time = 18.90 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.19 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=-\frac {x\,{\left (\frac {a}{b\,x^3}+1\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {4}{3},\frac {7}{3};\ \frac {10}{3};\ -\frac {a}{b\,x^3}\right )}{7\,{\left (b\,x^6+a\,x^3\right )}^{4/3}} \] Input:

int(1/(a*x^3 + b*x^6)^(4/3),x)
 

Output:

-(x*(a/(b*x^3) + 1)^(4/3)*hypergeom([4/3, 7/3], 10/3, -a/(b*x^3)))/(7*(a*x 
^3 + b*x^6)^(4/3))
 

Reduce [F]

\[ \int \frac {1}{\left (a x^3+b x^6\right )^{4/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,x^{4}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{7}}d x \] Input:

int(1/(b*x^6+a*x^3)^(4/3),x)
 

Output:

int(1/((a + b*x**3)**(1/3)*a*x**4 + (a + b*x**3)**(1/3)*b*x**7),x)