\(\int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx\) [275]

Optimal result

Integrand size = 21, antiderivative size = 86 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}} \]

[Out]

-2/5*(b*x^3+a*x^2)^(1/2)/a/x^(7/2)+8/15*b*(b*x^3+a*x^2)^(1/2)/a^2/x^(5/2)-16/15*b^2*(b*x^3+a*x^2)^(1/2)/a^3/x^
(3/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2041, 2039} \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}} \]

[In]

Int[1/(x^(5/2)*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(-2*Sqrt[a*x^2 + b*x^3])/(5*a*x^(7/2)) + (8*b*Sqrt[a*x^2 + b*x^3])/(15*a^2*x^(5/2)) - (16*b^2*Sqrt[a*x^2 + b*x
^3])/(15*a^3*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}}{5 a x^{7/2}}-\frac {(4 b) \int \frac {1}{x^{3/2} \sqrt {a x^2+b x^3}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}+\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a x^2+b x^3}} \, dx}{15 a^2} \\ & = -\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}+\frac {8 b \sqrt {a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{15 a^3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {x^2 (a+b x)} \left (3 a^2-4 a b x+8 b^2 x^2\right )}{15 a^3 x^{7/2}} \]

[In]

Integrate[1/(x^(5/2)*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(-2*Sqrt[x^2*(a + b*x)]*(3*a^2 - 4*a*b*x + 8*b^2*x^2))/(15*a^3*x^(7/2))

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.51

method	result	size
risch	\(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 \sqrt {x^{2} \left (b x +a \right )}\, x^{\frac {3}{2}} a^{3}}\)	\(44\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x^{\frac {3}{2}} a^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(46\)
default	\(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x^{\frac {3}{2}} a^{3} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(46\)

[In]

int(1/x^(5/2)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15/(x^2*(b*x+a))^(1/2)/x^(3/2)*(b*x+a)*(8*b^2*x^2-4*a*b*x+3*a^2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, a^{3} x^{\frac {7}{2}}} \]

[In]

integrate(1/x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(8*b^2*x^2 - 4*a*b*x + 3*a^2)*sqrt(b*x^3 + a*x^2)/(a^3*x^(7/2))

Sympy [F]

\[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{\frac {5}{2}} \sqrt {x^{2} \left (a + b x\right )}}\, dx \]

[In]

integrate(1/x**(5/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(1/(x**(5/2)*sqrt(x**2*(a + b*x))), x)

Maxima [F]

\[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^3 + a*x^2)*x^(5/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\frac {32 \, {\left (10 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 5 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{2}\right )} b^{\frac {5}{2}}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

32/15*(10*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 5*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^2)*b^(5/2)/(((sqrt
(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{5/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{5/2}\,\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

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

[Out]

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