Integrand size = 19, antiderivative size = 61 \[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\frac {16 a \left (b+2 a \sqrt {x}\right )}{3 b^3 \sqrt {b \sqrt {x}+a x}}-\frac {4}{3 b \sqrt {x} \sqrt {b \sqrt {x}+a x}} \] Output:
16/3*a*(b+2*a*x^(1/2))/b^3/(b*x^(1/2)+a*x)^(1/2)-4/3/b/x^(1/2)/(b*x^(1/2)+ a*x)^(1/2)
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x} \left (b^2-4 a b \sqrt {x}-8 a^2 x\right )}{3 b^3 \left (b+a \sqrt {x}\right ) x} \] Input:
Integrate[1/(x*(b*Sqrt[x] + a*x)^(3/2)),x]
Output:
(-4*Sqrt[b*Sqrt[x] + a*x]*(b^2 - 4*a*b*Sqrt[x] - 8*a^2*x))/(3*b^3*(b + a*S qrt[x])*x)
Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1921, 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.
\(\displaystyle \int \frac {1}{x \left (a x+b \sqrt {x}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {4 \int \frac {1}{x^{3/2} \sqrt {\sqrt {x} b+a x}}dx}{b}+\frac {4}{b \sqrt {x} \sqrt {a x+b \sqrt {x}}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {4 \left (-\frac {2 a \int \frac {1}{x \sqrt {\sqrt {x} b+a x}}dx}{3 b}-\frac {4 \sqrt {a x+b \sqrt {x}}}{3 b x}\right )}{b}+\frac {4}{b \sqrt {x} \sqrt {a x+b \sqrt {x}}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {4 \left (\frac {8 a \sqrt {a x+b \sqrt {x}}}{3 b^2 \sqrt {x}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{3 b x}\right )}{b}+\frac {4}{b \sqrt {x} \sqrt {a x+b \sqrt {x}}}\) |
Input:
Int[1/(x*(b*Sqrt[x] + a*x)^(3/2)),x]
Output:
4/(b*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x]) + (4*((-4*Sqrt[b*Sqrt[x] + a*x])/(3*b* x) + (8*a*Sqrt[b*Sqrt[x] + a*x])/(3*b^2*Sqrt[x])))/b
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])
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, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
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])
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {16 a \left (b +2 \sqrt {x}\, a \right )}{3 b^{3} \sqrt {b \sqrt {x}+x a}}-\frac {4}{3 b \sqrt {x}\, \sqrt {b \sqrt {x}+x a}}\) | \(46\) |
default | \(\frac {\sqrt {b \sqrt {x}+x a}\, \left (24 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{\frac {5}{2}} a^{\frac {7}{2}}-6 \sqrt {b \sqrt {x}+x a}\, x^{\frac {7}{2}} a^{\frac {9}{2}}+3 x^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{4} b -6 x^{\frac {7}{2}} a^{\frac {9}{2}} \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}-3 x^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {b \sqrt {x}+x a}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{4} b +44 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{2} a^{\frac {5}{2}} b -12 \sqrt {b \sqrt {x}+x a}\, x^{3} a^{\frac {7}{2}} b +6 x^{3} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{2}-12 x^{3} a^{\frac {7}{2}} \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, b -6 x^{3} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {b \sqrt {x}+x a}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{2}-12 x^{\frac {5}{2}} a^{\frac {7}{2}} \left (\sqrt {x}\, \left (\sqrt {x}\, a +b \right )\right )^{\frac {3}{2}}+16 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} x^{\frac {3}{2}} a^{\frac {3}{2}} b^{2}-6 \sqrt {b \sqrt {x}+x a}\, x^{\frac {5}{2}} a^{\frac {5}{2}} b^{2}+3 x^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{3}-6 x^{\frac {5}{2}} a^{\frac {5}{2}} \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, b^{2}-3 x^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {x}\, a +2 \sqrt {b \sqrt {x}+x a}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{3}-4 \left (b \sqrt {x}+x a \right )^{\frac {3}{2}} \sqrt {a}\, b^{3} x \right )}{3 \sqrt {\sqrt {x}\, \left (\sqrt {x}\, a +b \right )}\, b^{4} x^{\frac {5}{2}} \sqrt {a}\, \left (\sqrt {x}\, a +b \right )^{2}}\) | \(524\) |
Input:
int(1/x/(b*x^(1/2)+x*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
16/3*a*(b+2*x^(1/2)*a)/b^3/(b*x^(1/2)+x*a)^(1/2)-4/3/b/x^(1/2)/(b*x^(1/2)+ x*a)^(1/2)
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=-\frac {4 \, {\left (4 \, a^{2} b x - b^{3} - {\left (8 \, a^{3} x - 5 \, a b^{2}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{3 \, {\left (a^{2} b^{3} x^{2} - b^{5} x\right )}} \] Input:
integrate(1/x/(b*x^(1/2)+a*x)^(3/2),x, algorithm="fricas")
Output:
-4/3*(4*a^2*b*x - b^3 - (8*a^3*x - 5*a*b^2)*sqrt(x))*sqrt(a*x + b*sqrt(x)) /(a^2*b^3*x^2 - b^5*x)
\[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\int \frac {1}{x \left (a x + b \sqrt {x}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x/(b*x**(1/2)+a*x)**(3/2),x)
Output:
Integral(1/(x*(a*x + b*sqrt(x))**(3/2)), x)
\[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x} \,d x } \] Input:
integrate(1/x/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((a*x + b*sqrt(x))^(3/2)*x), x)
\[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x} \,d x } \] Input:
integrate(1/x/(b*x^(1/2)+a*x)^(3/2),x, algorithm="giac")
Output:
integrate(1/((a*x + b*sqrt(x))^(3/2)*x), x)
Timed out. \[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a\,x+b\,\sqrt {x}\right )}^{3/2}} \,d x \] Input:
int(1/(x*(a*x + b*x^(1/2))^(3/2)),x)
Output:
int(1/(x*(a*x + b*x^(1/2))^(3/2)), x)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x \left (b \sqrt {x}+a x\right )^{3/2}} \, dx=\frac {\frac {16 \sqrt {x}\, \sqrt {\sqrt {x}\, a +b}\, a b}{3}+\frac {32 \sqrt {\sqrt {x}\, a +b}\, a^{2} x}{3}-\frac {4 \sqrt {\sqrt {x}\, a +b}\, b^{2}}{3}-\frac {32 x^{\frac {3}{4}} \sqrt {a}\, a b}{3}-\frac {32 x^{\frac {5}{4}} \sqrt {a}\, a^{2}}{3}}{x^{\frac {1}{4}} b^{3} \left (\sqrt {x}\, b +a x \right )} \] Input:
int(1/x/(b*x^(1/2)+a*x)^(3/2),x)
Output:
(4*(4*sqrt(x)*sqrt(sqrt(x)*a + b)*a*b + 8*sqrt(sqrt(x)*a + b)*a**2*x - sqr t(sqrt(x)*a + b)*b**2 - 8*x**(3/4)*sqrt(a)*a*b - 8*x**(1/4)*sqrt(a)*a**2*x ))/(3*x**(1/4)*b**3*(sqrt(x)*b + a*x))