\(\int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx\) [132]

Optimal result

Integrand size = 15, antiderivative size = 126 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{a^{5/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

2*(b*x^(1/3)+a*x)^(1/2)/a-b^(3/4)*(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))/a^(5/4)/(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 = 10.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x} \left (b+a x^{2/3}-b \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^{2/3}}{b}\right )\right )}{a \left (b+a x^{2/3}\right )} \] Input:

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

Output:

(2*Sqrt[b*x^(1/3) + a*x]*(b + a*x^(2/3) - b*Sqrt[1 + (a*x^(2/3))/b]*Hyperg 
eometric2F1[1/4, 1/2, 5/4, -((a*x^(2/3))/b)]))/(a*(b + a*x^(2/3)))
 

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1916, 1919, 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.

\(\displaystyle \int \frac {1}{\sqrt {a x+b \sqrt [3]{x}}} \, dx\)

\(\Big \downarrow \) 1916

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

\(\Big \downarrow \) 1919

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

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

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 1916

Int[1/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[-2*(Sqrt 
[a*x^j + b*x^n]/(b*(n - 2)*x^(n - 1))), x] - Simp[a*((2*n - j - 2)/(b*(n - 
2)))   Int[1/(x^(n - j)*Sqrt[a*x^j + b*x^n]), x], x] /; FreeQ[{a, b}, x] && 
 LtQ[2*(n - 1), j, n]
 

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 1919

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp 
[1/n   Subst[Int[(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]] 
 && EqQ[Simplify[m - n + 1], 0]
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {b \sqrt {-a b}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-2 x^{\frac {1}{3}} a b -2 a^{2} x}{\sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2}}\)	\(127\)
derivativedivides	\(\frac {2 \sqrt {b \,x^{\frac {1}{3}}+a x}}{a}-\frac {b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(135\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {1}{\sqrt {a x + b \sqrt [3]{x}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2\,x\,\sqrt {\frac {b}{a\,x^{2/3}}+1}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ -\frac {b}{a\,x^{2/3}}\right )}{\sqrt {a\,x+b\,x^{1/3}}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {6 x^{\frac {1}{6}} \sqrt {x^{\frac {2}{3}} a +b}-\left (\int \frac {\sqrt {x^{\frac {2}{3}} a +b}}{x^{\frac {5}{6}} b +\sqrt {x}\, a x}d x \right ) b}{3 a} \] Input:

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

Output:

(6*x**(1/6)*sqrt(x**(2/3)*a + b) - int(sqrt(x**(2/3)*a + b)/(x**(5/6)*b + 
sqrt(x)*a*x),x)*b)/(3*a)