\(\int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx\) [274]

Optimal result

Integrand size = 21, antiderivative size = 56 \[ \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}}+\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 x^{3/2}} \]

[Out]

-2/3*(b*x^3+a*x^2)^(1/2)/a/x^(5/2)+4/3*b*(b*x^3+a*x^2)^(1/2)/a^2/x^(3/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2041, 2039} \[ \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx=\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 x^{3/2}}-\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}} \]

[In]

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

[Out]

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

Rule 2039

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 2041

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] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(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])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}}-\frac {(2 b) \int \frac {1}{\sqrt {x} \sqrt {a x^2+b x^3}} \, dx}{3 a} \\ & = -\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}}+\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 (a-2 b x) \sqrt {x^2 (a+b x)}}{3 a^2 x^{5/2}} \]

[In]

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

[Out]

(-2*(a - 2*b*x)*Sqrt[x^2*(a + b*x)])/(3*a^2*x^(5/2))

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.55

method	result	size
risch	\(-\frac {2 \left (b x +a \right ) \left (-2 b x +a \right )}{3 \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {x}\, a^{2}}\)	\(31\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-2 b x +a \right )}{3 \sqrt {x}\, a^{2} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(33\)
default	\(-\frac {2 \left (b x +a \right ) \left (-2 b x +a \right )}{3 \sqrt {x}\, a^{2} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx=\frac {2 \, \sqrt {b x^{3} + a x^{2}} {\left (2 \, b x - a\right )}}{3 \, a^{2} x^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{3}}{a^{2}} - \frac {3 \, b^{3}}{a}\right )} \sqrt {b x + a} b}{3 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}} {\left | b \right |} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

2/3*(2*(b*x + a)*b^3/a^2 - 3*b^3/a)*sqrt(b*x + a)*b/(((b*x + a)*b - a*b)^(3/2)*abs(b)*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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