\(\int \frac {(b \sqrt [3]{x}+a x)^{3/2}}{x^2} \, dx\) [123]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 144 \[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^2} \, dx=4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+\frac {4 a^{3/4} b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{\sqrt {b \sqrt [3]{x}+a x}} \] Output:

4*a*(b*x^(1/3)+a*x)^(1/2)-2*(b*x^(1/3)+a*x)^(3/2)/x+4*a^(3/4)*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))/(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 = 60, normalized size of antiderivative = 0.42 \[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^2} \, dx=-\frac {2 b \sqrt {b \sqrt [3]{x}+a x} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{4},\frac {1}{4},-\frac {a x^{2/3}}{b}\right )}{\sqrt {1+\frac {a x^{2/3}}{b}} x^{2/3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1924, 1926, 1927, 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.

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

\(\Big \downarrow \) 1924

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

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

Output:

3*((-2*(b*x^(1/3) + a*x)^(3/2))/(3*x) + 2*a*((2*Sqrt[b*x^(1/3) + a*x])/3 + 
 (2*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]*EllipticF[2*ArcTan[(a^(1/4)* 
x^(1/6))/b^(1/4)], 1/2])/(3*a^(1/4)*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 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 1927
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 + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(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] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 
Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90

method result size
default \(\frac {4 x^{\frac {1}{3}} \sqrt {-a b}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b +2 x^{\frac {4}{3}} a^{2}-2 b^{2}}{x^{\frac {1}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}}\) \(130\)
derivativedivides \(-\frac {2 b \sqrt {b \,x^{\frac {1}{3}}+a x}}{x^{\frac {2}{3}}}+2 a \sqrt {b \,x^{\frac {1}{3}}+a x}+\frac {4 b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b \,x^{\frac {1}{3}}+a x}}\) \(147\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^2} \, dx=\frac {2 x^{\frac {2}{3}} \sqrt {x^{\frac {2}{3}} a +b}\, a -10 \sqrt {x^{\frac {2}{3}} a +b}\, b -4 \sqrt {x}\, \left (\int \frac {\sqrt {x^{\frac {2}{3}} a +b}}{\sqrt {x}\, b x +x^{\frac {13}{6}} a}d x \right ) b^{2}}{\sqrt {x}} \] Input:

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

Output:

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