\(\int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx\) [316]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}+\frac {8 b \sqrt {a x^3+b x^4}}{15 a^2 x^3}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{15 a^3 x^2} \]

[Out]

-2/5*(b*x^4+a*x^3)^(1/2)/a/x^4+8/15*b*(b*x^4+a*x^3)^(1/2)/a^2/x^3-16/15*b^2*(b*x^4+a*x^3)^(1/2)/a^3/x^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2025} \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=-\frac {16 b^2 \sqrt {a x^3+b x^4}}{15 a^3 x^2}+\frac {8 b \sqrt {a x^3+b x^4}}{15 a^2 x^3}-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4} \]

[In]

Int[1/(x^2*Sqrt[a*x^3 + b*x^4]),x]

[Out]

(-2*Sqrt[a*x^3 + b*x^4])/(5*a*x^4) + (8*b*Sqrt[a*x^3 + b*x^4])/(15*a^2*x^3) - (16*b^2*Sqrt[a*x^3 + b*x^4])/(15
*a^3*x^2)

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 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^3+b x^4}}{5 a x^4}-\frac {(4 b) \int \frac {1}{x \sqrt {a x^3+b x^4}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}+\frac {8 b \sqrt {a x^3+b x^4}}{15 a^2 x^3}+\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {a x^3+b x^4}} \, dx}{15 a^2} \\ & = -\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4}+\frac {8 b \sqrt {a x^3+b x^4}}{15 a^2 x^3}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{15 a^3 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=-\frac {2 \sqrt {x^3 (a+b x)} \left (3 a^2-4 a b x+8 b^2 x^2\right )}{15 a^3 x^4} \]

[In]

Integrate[1/(x^2*Sqrt[a*x^3 + b*x^4]),x]

[Out]

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

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.49

method	result	size
pseudoelliptic	\(-\frac {2 \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right ) \sqrt {x^{3} \left (b x +a \right )}}{15 a^{3} x^{4}}\)	\(39\)
trager	\(-\frac {2 \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right ) \sqrt {b \,x^{4}+a \,x^{3}}}{15 a^{3} x^{4}}\)	\(41\)
risch	\(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x \sqrt {x^{3} \left (b x +a \right )}\, a^{3}}\)	\(44\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x \,a^{3} \sqrt {b \,x^{4}+a \,x^{3}}}\)	\(46\)
default	\(-\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b \,x^{2}+a x}\, \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 x^{2} \sqrt {b \,x^{4}+a \,x^{3}}\, a^{3}}\)	\(61\)

[In]

int(1/x^2/(b*x^4+a*x^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=-\frac {2 \, \sqrt {b x^{4} + a x^{3}} {\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{4}} \]

[In]

integrate(1/x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=\int \frac {1}{x^{2} \sqrt {x^{3} \left (a + b x\right )}}\, dx \]

[In]

integrate(1/x**2/(b*x**4+a*x**3)**(1/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=\int { \frac {1}{\sqrt {b x^{4} + a x^{3}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^4 + a*x^3)*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=\frac {2 \, {\left (20 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} b + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a \sqrt {b} + 3 \, a^{2}\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="giac")

[Out]

2/15*(20*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*b + 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a*sqrt(b) + 3*a^2)/((sqrt(b)
*x - sqrt(b*x^2 + a*x))^5*sgn(x))

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx=-\frac {2\,\sqrt {b\,x^4+a\,x^3}\,\left (3\,a^2-4\,a\,b\,x+8\,b^2\,x^2\right )}{15\,a^3\,x^4} \]

[In]

int(1/(x^2*(a*x^3 + b*x^4)^(1/2)),x)

[Out]

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