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

Optimal. Leaf size=80 \[ -\frac {16 b^2 \sqrt {a x^3+b x^4}}{15 a^3 x^2}+\frac {8 b \sqrt {a x^3+b x^4}}{15 a^2 x^3}-\frac {2 \sqrt {a x^3+b x^4}}{5 a x^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 2000} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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 2000

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 2016

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*} \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 {(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]  time = 0.02, size = 42, normalized size = 0.52 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.18, size = 44, normalized size = 0.55 \begin {gather*} -\frac {2 \left (3 a^2-4 a b x+8 b^2 x^2\right ) \sqrt {a x^3+b x^4}}{15 a^3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.39, size = 40, normalized size = 0.50 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.24, size = 43, normalized size = 0.54 \begin {gather*} -\frac {2 \, {\left (3 \, {\left (b + \frac {a}{x}\right )}^{\frac {5}{2}} - 10 \, {\left (b + \frac {a}{x}\right )}^{\frac {3}{2}} b + 15 \, \sqrt {b + \frac {a}{x}} b^{2}\right )}}{15 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*(b + a/x)^(5/2) - 10*(b + a/x)^(3/2)*b + 15*sqrt(b + a/x)*b^2)/a^3

________________________________________________________________________________________

maple [A]  time = 0.04, size = 46, normalized size = 0.58 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}-4 a b x +3 a^{2}\right )}{15 \sqrt {b \,x^{4}+a \,x^{3}}\, a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x^{4} + a x^{3}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 5.14, size = 40, normalized size = 0.50 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \sqrt {x^{3} \left (a + b x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________