\(\int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx\) [51]

Optimal result

Integrand size = 19, antiderivative size = 142 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}-\frac {64 a^2 \sqrt {b \sqrt {x}+a x}}{105 b^3 x^{3/2}}+\frac {256 a^3 \sqrt {b \sqrt {x}+a x}}{315 b^4 x}-\frac {512 a^4 \sqrt {b \sqrt {x}+a x}}{315 b^5 \sqrt {x}} \] Output:

-4/9*(b*x^(1/2)+a*x)^(1/2)/b/x^(5/2)+32/63*a*(b*x^(1/2)+a*x)^(1/2)/b^2/x^2 
-64/105*a^2*(b*x^(1/2)+a*x)^(1/2)/b^3/x^(3/2)+256/315*a^3*(b*x^(1/2)+a*x)^ 
(1/2)/b^4/x-512/315*a^4*(b*x^(1/2)+a*x)^(1/2)/b^5/x^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x} \left (35 b^4-40 a b^3 \sqrt {x}+48 a^2 b^2 x-64 a^3 b x^{3/2}+128 a^4 x^2\right )}{315 b^5 x^{5/2}} \] Input:

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

Output:

(-4*Sqrt[b*Sqrt[x] + a*x]*(35*b^4 - 40*a*b^3*Sqrt[x] + 48*a^2*b^2*x - 64*a 
^3*b*x^(3/2) + 128*a^4*x^2))/(315*b^5*x^(5/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1922, 1922, 1922, 1922, 1920}

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.

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

\(\Big \downarrow \) 1922

\(\displaystyle -\frac {8 a \int \frac {1}{x^{5/2} \sqrt {\sqrt {x} b+a x}}dx}{9 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle -\frac {8 a \left (-\frac {6 a \int \frac {1}{x^2 \sqrt {\sqrt {x} b+a x}}dx}{7 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{7 b x^2}\right )}{9 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle -\frac {8 a \left (-\frac {6 a \left (-\frac {4 a \int \frac {1}{x^{3/2} \sqrt {\sqrt {x} b+a x}}dx}{5 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{5 b x^{3/2}}\right )}{7 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{7 b x^2}\right )}{9 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle -\frac {8 a \left (-\frac {6 a \left (-\frac {4 a \left (-\frac {2 a \int \frac {1}{x \sqrt {\sqrt {x} b+a x}}dx}{3 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{3 b x}\right )}{5 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{5 b x^{3/2}}\right )}{7 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{7 b x^2}\right )}{9 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}}\)

\(\Big \downarrow \) 1920

\(\displaystyle -\frac {8 a \left (-\frac {6 a \left (-\frac {4 a \left (\frac {8 a \sqrt {a x+b \sqrt {x}}}{3 b^2 \sqrt {x}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{3 b x}\right )}{5 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{5 b x^{3/2}}\right )}{7 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{7 b x^2}\right )}{9 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}}\)

Input:

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

Output:

(-4*Sqrt[b*Sqrt[x] + a*x])/(9*b*x^(5/2)) - (8*a*((-4*Sqrt[b*Sqrt[x] + a*x] 
)/(7*b*x^2) - (6*a*((-4*Sqrt[b*Sqrt[x] + a*x])/(5*b*x^(3/2)) - (4*a*((-4*S 
qrt[b*Sqrt[x] + a*x])/(3*b*x) + (8*a*Sqrt[b*Sqrt[x] + a*x])/(3*b^2*Sqrt[x] 
)))/(5*b)))/(7*b)))/(9*b)
 

Defintions of rubi rules used

rule 1920

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] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[ 
n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1922

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, j, m, n, 
p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) 
/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {4 \sqrt {b \sqrt {x}+x a}}{9 b \,x^{\frac {5}{2}}}-\frac {16 a \left (-\frac {2 \sqrt {b \sqrt {x}+x a}}{7 b \,x^{2}}-\frac {6 a \left (-\frac {2 \sqrt {b \sqrt {x}+x a}}{5 b \,x^{\frac {3}{2}}}-\frac {4 a \left (-\frac {2 \sqrt {b \sqrt {x}+x a}}{3 b x}+\frac {4 a \sqrt {b \sqrt {x}+x a}}{3 b^{2} \sqrt {x}}\right )}{5 b}\right )}{7 b}\right )}{9 b}\)	\(119\)
default	\(-\frac {\sqrt {b \sqrt {x}+x a}\, \left (1260 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{\frac {9}{2}} a^{\frac {9}{2}}-630 \sqrt {b \sqrt {x}+x a}\, x^{\frac {11}{2}} a^{\frac {11}{2}}-315 x^{\frac {11}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {b \sqrt {x}+x a}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{5} b -630 x^{\frac {11}{2}} a^{\frac {11}{2}} \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}+315 x^{\frac {11}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{5} b +492 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{\frac {7}{2}} a^{\frac {5}{2}} b^{2}+140 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{\frac {5}{2}} \sqrt {a}\, b^{4}-748 a^{\frac {7}{2}} \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} b \,x^{4}-300 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{3} x^{3}\right )}{315 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, b^{6} x^{\frac {11}{2}} \sqrt {a}}\)	\(262\)

Input:

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

Output:

-4/9*(b*x^(1/2)+x*a)^(1/2)/b/x^(5/2)-16/9*a/b*(-2/7*(b*x^(1/2)+x*a)^(1/2)/ 
b/x^2-6/7*a/b*(-2/5*(b*x^(1/2)+x*a)^(1/2)/b/x^(3/2)-4/5*a/b*(-2/3*(b*x^(1/ 
2)+x*a)^(1/2)/b/x+4/3*a*(b*x^(1/2)+x*a)^(1/2)/b^2/x^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (64 \, a^{3} b x^{2} + 40 \, a b^{3} x - {\left (128 \, a^{4} x^{2} + 48 \, a^{2} b^{2} x + 35 \, b^{4}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{315 \, b^{5} x^{3}} \] Input:

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

Output:

4/315*(64*a^3*b*x^2 + 40*a*b^3*x - (128*a^4*x^2 + 48*a^2*b^2*x + 35*b^4)*s 
qrt(x))*sqrt(a*x + b*sqrt(x))/(b^5*x^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (1008 \, a^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{4} + 1680 \, a^{\frac {3}{2}} b {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{3} + 1080 \, a b^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{2} + 315 \, \sqrt {a} b^{3} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + 35 \, b^{4}\right )}}{315 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{9}} \] Input:

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

Output:

4/315*(1008*a^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^4 + 1680*a^(3/2) 
*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^3 + 1080*a*b^2*(sqrt(a)*sqrt( 
x) - sqrt(a*x + b*sqrt(x)))^2 + 315*sqrt(a)*b^3*(sqrt(a)*sqrt(x) - sqrt(a* 
x + b*sqrt(x))) + 35*b^4)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^9
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\frac {\frac {256 x^{\frac {7}{4}} \sqrt {\sqrt {x}\, a +b}\, a^{3} b}{315}+\frac {32 x^{\frac {3}{4}} \sqrt {\sqrt {x}\, a +b}\, a \,b^{3}}{63}-\frac {512 x^{\frac {9}{4}} \sqrt {\sqrt {x}\, a +b}\, a^{4}}{315}-\frac {64 x^{\frac {5}{4}} \sqrt {\sqrt {x}\, a +b}\, a^{2} b^{2}}{105}-\frac {4 x^{\frac {1}{4}} \sqrt {\sqrt {x}\, a +b}\, b^{4}}{9}+\frac {512 \sqrt {x}\, \sqrt {a}\, a^{4} x^{2}}{315}}{\sqrt {x}\, b^{5} x^{2}} \] Input:

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

Output:

(4*(64*x**(3/4)*sqrt(sqrt(x)*a + b)*a**3*b*x + 40*x**(3/4)*sqrt(sqrt(x)*a 
+ b)*a*b**3 - 128*x**(1/4)*sqrt(sqrt(x)*a + b)*a**4*x**2 - 48*x**(1/4)*sqr 
t(sqrt(x)*a + b)*a**2*b**2*x - 35*x**(1/4)*sqrt(sqrt(x)*a + b)*b**4 + 128* 
sqrt(x)*sqrt(a)*a**4*x**2))/(315*sqrt(x)*b**5*x**2)