\(\int \frac {1}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\) [141]

Optimal result

Integrand size = 15, antiderivative size = 296 \[ \int \frac {1}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=-\frac {3 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{\sqrt {a} b \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}+\frac {3 x^{2/3}}{b \sqrt {b \sqrt [3]{x}+a x}}+\frac {3 \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{a^{3/4} b^{3/4} \sqrt {b \sqrt [3]{x}+a x}}-\frac {3 \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

-3*(b+a*x^(2/3))*x^(1/3)/a^(1/2)/b/(b^(1/2)+a^(1/2)*x^(1/3))/(b*x^(1/3)+a* 
x)^(1/2)+3*x^(2/3)/b/(b*x^(1/3)+a*x)^(1/2)+3*(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)*EllipticE(sin(2*arc 
tan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))/a^(3/4)/b^(3/4)/(b*x^(1/3)+a*x) 
^(1/2)-3/2*(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))/a^(3/4)/b^(3/4)/(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 = 10.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=\frac {2 \sqrt {1+\frac {a x^{2/3}}{b}} x^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {a x^{2/3}}{b}\right )}{b \sqrt {b \sqrt [3]{x}+a x}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.66 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1912, 1924, 1938, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 1912

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

\(\Big \downarrow \) 1924

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

\(\Big \downarrow \) 1938

\(\displaystyle \frac {3 x^{2/3}}{b \sqrt {a x+b \sqrt [3]{x}}}-\frac {3 \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {\sqrt [6]{x}}{\sqrt {x^{2/3} a+b}}d\sqrt [3]{x}}{2 b \sqrt {a x+b \sqrt [3]{x}}}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {3 x^{2/3}}{b \sqrt {a x+b \sqrt [3]{x}}}-\frac {3 \sqrt [6]{x} \sqrt {a x^{2/3}+b} \left (\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a x^{4/3}+b}}-\frac {\int \frac {\sqrt {b}-\sqrt {a} x^{2/3}}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{\sqrt {a}}\right )}{b \sqrt {a x+b \sqrt [3]{x}}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {3 x^{2/3}}{b \sqrt {a x+b \sqrt [3]{x}}}-\frac {3 \sqrt [6]{x} \sqrt {a x^{2/3}+b} \left (\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{3/4} \sqrt {a x^{4/3}+b}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{a} \sqrt {a x^{4/3}+b}}-\frac {\sqrt [6]{x} \sqrt {a x^{4/3}+b}}{\sqrt {a} x^{2/3}+\sqrt {b}}}{\sqrt {a}}\right )}{b \sqrt {a x+b \sqrt [3]{x}}}\)

Input:

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

Output:

(3*x^(2/3))/(b*Sqrt[b*x^(1/3) + a*x]) - (3*Sqrt[b + a*x^(2/3)]*x^(1/6)*(-( 
(-((x^(1/6)*Sqrt[b + a*x^(4/3)])/(Sqrt[b] + Sqrt[a]*x^(2/3))) + (b^(1/4)*( 
Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^(2/3) 
)^2]*EllipticE[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(a^(1/4)*Sqrt[b 
+ a*x^(4/3)]))/Sqrt[a]) + (b^(1/4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*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])/(2*a^(3/4)*Sqrt[b + a*x^(4/3)])))/(b*Sqrt[b*x^(1/3) + 
a*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 834

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

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

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 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 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {3 x^{\frac {2}{3}}}{b \sqrt {\left (x^{\frac {2}{3}}+\frac {b}{a}\right ) x^{\frac {1}{3}} a}}-\frac {3 \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{2 b a \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(197\)
default	\(-\frac {3 \left (2 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b -\sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b -2 \sqrt {b \,x^{\frac {1}{3}}+a x}\, x^{\frac {2}{3}} a \right )}{2 a \,x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right ) b}\)	\(243\)

Input:

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

Output:

3*x^(2/3)/b/((x^(2/3)+b/a)*x^(1/3)*a)^(1/2)-3/2/b/a*(-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)* 
(-2/a*(-a*b)^(1/2)*EllipticE(((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^( 
1/2),1/2*2^(1/2))+1/a*(-a*b)^(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]

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

integrate(1/(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^5 + 2*a^3*b^3*x^3 + b^6*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.14 \[ \int \frac {1}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=\frac {2\,x\,{\left (\frac {a\,x^{2/3}}{b}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {3}{2};\ \frac {7}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{{\left (a\,x+b\,x^{1/3}\right )}^{3/2}} \] Input:

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

Output:

(2*x*((a*x^(2/3))/b + 1)^(3/2)*hypergeom([3/4, 3/2], 7/4, -(a*x^(2/3))/b)) 
/(a*x + b*x^(1/3))^(3/2)
 

Reduce [F]

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

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

Output:

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