\(\int \frac {1}{x (b x^{2/3}+a x)^{3/2}} \, dx\) [178]

Optimal result

Integrand size = 19, antiderivative size = 146 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {105 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{9/2}} \] Output:

6/b/x^(2/3)/(b*x^(2/3)+a*x)^(1/2)-7*(b*x^(2/3)+a*x)^(1/2)/b^2/x^(4/3)+35/4 
*a*(b*x^(2/3)+a*x)^(1/2)/b^3/x-105/8*a^2*(b*x^(2/3)+a*x)^(1/2)/b^4/x^(2/3) 
+105/8*a^3*arctanh(b^(1/2)*x^(1/3)/(b*x^(2/3)+a*x)^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 4.73 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {-\sqrt {b} \left (8 b^3-14 a b^2 \sqrt [3]{x}+35 a^2 b x^{2/3}+105 a^3 x\right )+105 a^3 \sqrt {b+a \sqrt [3]{x}} x \text {arctanh}\left (\frac {\sqrt {b+a \sqrt [3]{x}}}{\sqrt {b}}\right )}{8 b^{9/2} x^{2/3} \sqrt {b x^{2/3}+a x}} \] Input:

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

Output:

(-(Sqrt[b]*(8*b^3 - 14*a*b^2*x^(1/3) + 35*a^2*b*x^(2/3) + 105*a^3*x)) + 10 
5*a^3*Sqrt[b + a*x^(1/3)]*x*ArcTanh[Sqrt[b + a*x^(1/3)]/Sqrt[b]])/(8*b^(9/ 
2)*x^(2/3)*Sqrt[b*x^(2/3) + a*x])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1929, 1931, 1931, 1931, 1935, 219}

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.

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

\(\Big \downarrow \) 1929

\(\displaystyle \frac {7 \int \frac {1}{x^{5/3} \sqrt {x^{2/3} b+a x}}dx}{b}+\frac {6}{b x^{2/3} \sqrt {a x+b x^{2/3}}}\)

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1935

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

6/(b*x^(2/3)*Sqrt[b*x^(2/3) + a*x]) + (7*(-(Sqrt[b*x^(2/3) + a*x]/(b*x^(4/ 
3))) - (5*a*((-3*Sqrt[b*x^(2/3) + a*x])/(2*b*x) - (3*a*((-3*Sqrt[b*x^(2/3) 
 + a*x])/(b*x^(2/3)) + (3*a*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x 
]])/b^(3/2)))/(4*b)))/(6*b)))/b
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1929

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*(n - j 
)*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))   In 
t[(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] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, 
 -1]
 

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]
 

rule 1935

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60

method	result	size
derivativedivides	\(\frac {\left (x^{\frac {1}{3}} a +b \right ) \left (105 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}}{\sqrt {b}}\right ) \sqrt {x^{\frac {1}{3}} a +b}\, a^{3} x +14 b^{\frac {5}{2}} a \,x^{\frac {1}{3}}-35 b^{\frac {3}{2}} a^{2} x^{\frac {2}{3}}-105 x \,a^{3} \sqrt {b}-8 b^{\frac {7}{2}}\right )}{8 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} b^{\frac {9}{2}}}\)	\(88\)
default	\(\frac {\left (x^{\frac {1}{3}} a +b \right ) \left (105 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}}{\sqrt {b}}\right ) \sqrt {x^{\frac {1}{3}} a +b}\, a^{3} x +14 b^{\frac {5}{2}} a \,x^{\frac {1}{3}}-35 b^{\frac {3}{2}} a^{2} x^{\frac {2}{3}}-105 x \,a^{3} \sqrt {b}-8 b^{\frac {7}{2}}\right )}{8 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} b^{\frac {9}{2}}}\)	\(88\)

Input:

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

Output:

1/8*(x^(1/3)*a+b)*(105*arctanh((x^(1/3)*a+b)^(1/2)/b^(1/2))*(x^(1/3)*a+b)^ 
(1/2)*a^3*x+14*b^(5/2)*a*x^(1/3)-35*b^(3/2)*a^2*x^(2/3)-105*x*a^3*b^(1/2)- 
8*b^(7/2))/(b*x^(2/3)+a*x)^(3/2)/b^(9/2)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=-\frac {105 \, a^{3} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} - \frac {6 \, a^{3}}{\sqrt {a x^{\frac {1}{3}} + b} b^{4}} - \frac {57 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{3} - 136 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{3} b + 87 \, \sqrt {a x^{\frac {1}{3}} + b} a^{3} b^{2}}{8 \, a^{3} b^{4} x} \] Input:

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

Output:

-105/8*a^3*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) - 6*a^3/(sq 
rt(a*x^(1/3) + b)*b^4) - 1/8*(57*(a*x^(1/3) + b)^(5/2)*a^3 - 136*(a*x^(1/3 
) + b)^(3/2)*a^3*b + 87*sqrt(a*x^(1/3) + b)*a^3*b^2)/(a^3*b^4*x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {-105 \sqrt {b}\, \sqrt {x^{\frac {1}{3}} a +b}\, \mathrm {log}\left (\sqrt {x^{\frac {1}{3}} a +b}-\sqrt {b}\right ) a^{3} x +105 \sqrt {b}\, \sqrt {x^{\frac {1}{3}} a +b}\, \mathrm {log}\left (\sqrt {x^{\frac {1}{3}} a +b}+\sqrt {b}\right ) a^{3} x -70 x^{\frac {2}{3}} a^{2} b^{2}+28 x^{\frac {1}{3}} a \,b^{3}-210 a^{3} b x -16 b^{4}}{16 \sqrt {x^{\frac {1}{3}} a +b}\, b^{5} x} \] Input:

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

Output:

( - 105*sqrt(b)*sqrt(x**(1/3)*a + b)*log(sqrt(x**(1/3)*a + b) - sqrt(b))*a 
**3*x + 105*sqrt(b)*sqrt(x**(1/3)*a + b)*log(sqrt(x**(1/3)*a + b) + sqrt(b 
))*a**3*x - 70*x**(2/3)*a**2*b**2 + 28*x**(1/3)*a*b**3 - 210*a**3*b*x - 16 
*b**4)/(16*sqrt(x**(1/3)*a + b)*b**5*x)