3.2.61 \(\int \frac {x}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\) [161]

3.2.61.1 Optimal result

Integrand size = 17, antiderivative size = 149 \[ \int \frac {x}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {5 b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \]

output

-3*x/a/(b*x^(1/3)+a*x)^(1/2)+5*(b*x^(1/3)+a*x)^(1/2)/a^2-5/2*b^(3/4)*x^(1/ 
6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/cos(2*arctan(a^(1/4)*x 
^(1/6)/b^(1/4)))*EllipticF(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1 
/2))*(x^(1/3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1/2))^2) 
^(1/2)/a^(9/4)/(b*x^(1/3)+a*x)^(1/2)
 

3.2.61.2 Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {x}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx=\frac {\sqrt {b \sqrt [3]{x}+a x} \left (5 b+2 a x^{2/3}-5 b \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^{2/3}}{b}\right )\right )}{a^2 \left (b+a x^{2/3}\right )} \]

input

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

output

(Sqrt[b*x^(1/3) + a*x]*(5*b + 2*a*x^(2/3) - 5*b*Sqrt[1 + (a*x^(2/3))/b]*Hy 
pergeometric2F1[1/4, 1/2, 5/4, -((a*x^(2/3))/b)]))/(a^2*(b + a*x^(2/3)))
 

3.2.61.3 Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1924, 1928, 1930, 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.

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

\(\Big \downarrow \) 1924

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

\(\Big \downarrow \) 1928

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

input

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

output

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

3.2.61.3.1 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 1928

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

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

3.2.61.4 Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {3 x^{\frac {1}{3}} b}{a^{2} \sqrt {\left (x^{\frac {2}{3}}+\frac {b}{a}\right ) x^{\frac {1}{3}} a}}+\frac {2 \sqrt {b \,x^{\frac {1}{3}}+a x}}{a^{2}}-\frac {5 b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(160\)
default	\(-\frac {5 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, b -6 \sqrt {b \,x^{\frac {1}{3}}+a x}\, x^{\frac {1}{3}} a b -4 x^{\frac {1}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a b -4 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2} x}{2 x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right ) a^{3}}\)	\(184\)

input

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

output

3*x^(1/3)/a^2*b/((x^(2/3)+b/a)*x^(1/3)*a)^(1/2)+2*(b*x^(1/3)+a*x)^(1/2)/a^ 
2-5/2*b/a^3*(-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/(-a*b)^( 
1/2))^(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))
 

3.2.61.5 Fricas [F]

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

input

integrate(x/(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^4 + 2*a^3*b^3*x^2 + b^6), x)
 

3.2.61.6 Sympy [F]

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

input

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

output

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

3.2.61.7 Maxima [F]

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

input

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

output

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

3.2.61.8 Giac [F]

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

input

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

output

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

3.2.61.9 Mupad [F(-1)]

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

input

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

output

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