\(\int \frac {1}{x^3 (b \sqrt [3]{x}+a x)^{3/2}} \, dx\) [144]

Optimal result

Integrand size = 19, antiderivative size = 246 \[ \int \frac {1}{x^3 \left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=\frac {3}{b x^{7/3} \sqrt {b \sqrt [3]{x}+a x}}-\frac {17 \sqrt {b \sqrt [3]{x}+a x}}{5 b^2 x^{8/3}}+\frac {221 a \sqrt {b \sqrt [3]{x}+a x}}{55 b^3 x^2}-\frac {1989 a^2 \sqrt {b \sqrt [3]{x}+a x}}{385 b^4 x^{4/3}}+\frac {663 a^3 \sqrt {b \sqrt [3]{x}+a x}}{77 b^5 x^{2/3}}+\frac {663 a^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{154 b^{21/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

3/b/x^(7/3)/(b*x^(1/3)+a*x)^(1/2)-17/5*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(8/3)+2 
21/55*a*(b*x^(1/3)+a*x)^(1/2)/b^3/x^2-1989/385*a^2*(b*x^(1/3)+a*x)^(1/2)/b 
^4/x^(4/3)+663/77*a^3*(b*x^(1/3)+a*x)^(1/2)/b^5/x^(2/3)+663/154*a^(15/4)*( 
b^(1/2)+a^(1/2)*x^(1/3))*((b+a*x^(2/3))/(b^(1/2)+a^(1/2)*x^(1/3))^2)^(1/2) 
*x^(1/6)*InverseJacobiAM(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)),1/2*2^(1/2))/b^ 
(21/4)/(b*x^(1/3)+a*x)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^3 \left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=-\frac {2 \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},\frac {3}{2},-\frac {11}{4},-\frac {a x^{2/3}}{b}\right )}{5 b x^{7/3} \sqrt {b \sqrt [3]{x}+a x}} \] Input:

Integrate[1/(x^3*(b*x^(1/3) + a*x)^(3/2)),x]
 

Output:

(-2*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[-15/4, 3/2, -11/4, -((a*x^(2 
/3))/b)])/(5*b*x^(7/3)*Sqrt[b*x^(1/3) + a*x])
 

Rubi [A] (warning: unable to verify)

Time = 0.84 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1924, 1929, 1931, 1931, 1931, 1931, 1917, 266, 761}

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.

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

\(\Big \downarrow \) 1924

\(\displaystyle 3 \int \frac {1}{x^{7/3} \left (\sqrt [3]{x} b+a x\right )^{3/2}}d\sqrt [3]{x}\)

\(\Big \downarrow \) 1929

\(\displaystyle 3 \left (\frac {17 \int \frac {1}{x^{8/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 1931

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \int \frac {1}{x^2 \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 1931

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \int \frac {1}{x^{4/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 1931

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \left (-\frac {5 a \int \frac {1}{x^{2/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 1931

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \left (-\frac {5 a \left (-\frac {a \int \frac {1}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{3 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 1917

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \left (-\frac {5 a \left (-\frac {a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {x^{2/3} a+b} \sqrt [6]{x}}d\sqrt [3]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 266

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \left (-\frac {5 a \left (-\frac {2 a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle 3 \left (\frac {17 \left (-\frac {13 a \left (-\frac {9 a \left (-\frac {5 a \left (-\frac {a^{3/4} \sqrt [6]{x} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {a x^{2/3}+b} \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b \sqrt [3]{x}} \sqrt {a x^{4/3}+b}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )}{11 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 b x^2}\right )}{15 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{15 b x^{8/3}}\right )}{2 b}+\frac {1}{b x^{7/3} \sqrt {a x+b \sqrt [3]{x}}}\right )\)

Input:

Int[1/(x^3*(b*x^(1/3) + a*x)^(3/2)),x]
 

Output:

3*(1/(b*x^(7/3)*Sqrt[b*x^(1/3) + a*x]) + (17*((-2*Sqrt[b*x^(1/3) + a*x])/( 
15*b*x^(8/3)) - (13*a*((-2*Sqrt[b*x^(1/3) + a*x])/(11*b*x^2) - (9*a*((-2*S 
qrt[b*x^(1/3) + a*x])/(7*b*x^(4/3)) - (5*a*((-2*Sqrt[b*x^(1/3) + a*x])/(3* 
b*x^(2/3)) - (a^(3/4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[b + a*x^(2/3)]*x^(1 
/6)*Sqrt[(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^(2/3))^2]*EllipticF[2*ArcTan 
[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[b*x^(1/3) + a*x]*Sqrt[b 
 + a*x^(4/3)])))/(7*b)))/(11*b)))/(15*b)))/(2*b))
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 1924

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp 
[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x 
], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j 
] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1 
]
 

rule 1929

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*(n - j 
)*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))   In 
t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] & 
&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, 
 -1]
 

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 [A] (verified)

Time = 4.04 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {3 x^{\frac {1}{3}} a^{4}}{b^{5} \sqrt {\left (x^{\frac {2}{3}}+\frac {b}{a}\right ) x^{\frac {1}{3}} a}}-\frac {2 \sqrt {b \,x^{\frac {1}{3}}+a x}}{5 b^{2} x^{\frac {8}{3}}}+\frac {56 a \sqrt {b \,x^{\frac {1}{3}}+a x}}{55 b^{3} x^{2}}-\frac {834 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 b^{4} x^{\frac {4}{3}}}+\frac {432 a^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b^{5} x^{\frac {2}{3}}}+\frac {663 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{154 b^{5} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(231\)
default	\(\frac {3315 x^{\frac {14}{3}} \sqrt {-a b}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{3}-884 x^{\frac {11}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2} b^{2}+2652 x^{\frac {13}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{3} b +476 x^{3} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a \,b^{3}+2310 \sqrt {b \,x^{\frac {1}{3}}+a x}\, x^{5} a^{4}+4320 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{4} x^{5}-308 x^{\frac {7}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, b^{4}}{770 b^{5} x^{5} \left (b +a \,x^{\frac {2}{3}}\right )}\)	\(261\)

Input:

int(1/x^3/(b*x^(1/3)+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

3*x^(1/3)*a^4/b^5/((x^(2/3)+b/a)*x^(1/3)*a)^(1/2)-2/5*(b*x^(1/3)+a*x)^(1/2 
)/b^2/x^(8/3)+56/55*a*(b*x^(1/3)+a*x)^(1/2)/b^3/x^2-834/385*a^2*(b*x^(1/3) 
+a*x)^(1/2)/b^4/x^(4/3)+432/77*a^3*(b*x^(1/3)+a*x)^(1/2)/b^5/x^(2/3)+663/1 
54*a^3/b^5*(-a*b)^(1/2)*((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)* 
(-2*(x^(1/3)-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x^(1/3)/(-a*b)^(1/2 
)*a)^(1/2)/(b*x^(1/3)+a*x)^(1/2)*EllipticF(((x^(1/3)+1/a*(-a*b)^(1/2))*a/( 
-a*b)^(1/2))^(1/2),1/2*2^(1/2))
 

Fricas [F]

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

integrate(1/x^3/(b*x^(1/3)+a*x)^(3/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

integral((a^4*x^3 + 3*a^2*b^2*x^(5/3) - 2*a*b^3*x - (2*a^3*b*x^2 - b^4)*x^ 
(1/3))*sqrt(a*x + b*x^(1/3))/(a^6*x^8 + 2*a^3*b^3*x^6 + b^6*x^4), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x^3/(b*x^(1/3)+a*x)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/x^3/(b*x^(1/3)+a*x)^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^3 \left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=\int \frac {1}{x^{\frac {10}{3}} \sqrt {x^{\frac {1}{3}} b +a x}\, b +\sqrt {x^{\frac {1}{3}} b +a x}\, a \,x^{4}}d x \] Input:

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

Output:

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