\(\int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx\) [116]

Optimal result

Integrand size = 19, antiderivative size = 188 \[ \int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{11 x^2}-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{77 b x^{4/3}}+\frac {20 a^2 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}+\frac {10 a^{11/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 b^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

-6/11*(b*x^(1/3)+a*x)^(1/2)/x^2-12/77*a*(b*x^(1/3)+a*x)^(1/2)/b/x^(4/3)+20 
/77*a^2*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(2/3)+10/77*a^(11/4)*(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)*Inverse 
JacobiAM(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)),1/2*2^(1/2))/b^(9/4)/(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 = 59, normalized size of antiderivative = 0.31 \[ \int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx=-\frac {6 \sqrt {b \sqrt [3]{x}+a x} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{2},-\frac {7}{4},-\frac {a x^{2/3}}{b}\right )}{11 \sqrt {1+\frac {a x^{2/3}}{b}} x^2} \] Input:

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

Output:

(-6*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-11/4, -1/2, -7/4, -((a*x^(2/3 
))/b)])/(11*Sqrt[1 + (a*x^(2/3))/b]*x^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.61 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1924, 1926, 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.

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

\(\Big \downarrow \) 1924

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

3*((-2*Sqrt[b*x^(1/3) + a*x])/(11*x^2) + (2*a*((-2*Sqrt[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)
 

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 1926

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

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 = 0.74 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {6 \sqrt {b \,x^{\frac {1}{3}}+a x}}{11 x^{2}}-\frac {12 a \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b \,x^{\frac {4}{3}}}+\frac {20 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b^{2} x^{\frac {2}{3}}}+\frac {10 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(179\)
default	\(-\frac {6 \sqrt {b \,x^{\frac {1}{3}}+a x}}{11 x^{2}}-\frac {12 a \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b \,x^{\frac {4}{3}}}+\frac {20 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b^{2} x^{\frac {2}{3}}}+\frac {10 a^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(179\)

Input:

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

Output:

-6/11*(b*x^(1/3)+a*x)^(1/2)/x^2-12/77*a*(b*x^(1/3)+a*x)^(1/2)/b/x^(4/3)+20 
/77*a^2*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(2/3)+10/77*a^2/b^2*(-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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {b \sqrt [3]{x}+a x}}{x^3} \, dx=\frac {-\frac {2 \sqrt {x^{\frac {2}{3}} a +b}}{3}-\frac {2 x^{\frac {11}{6}} \left (\int \frac {\sqrt {x^{\frac {2}{3}} a +b}}{x^{\frac {17}{6}} b +\sqrt {x}\, a \,x^{3}}d x \right ) b}{9}}{x^{\frac {11}{6}}} \] Input:

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

Output:

(2*( - 3*sqrt(x**(2/3)*a + b) - x**(5/6)*int(sqrt(x**(2/3)*a + b)/(x**(5/6 
)*b*x**2 + sqrt(x)*a*x**3),x)*b*x))/(9*x**(5/6)*x)