3.279 \(\int \frac {1}{x^3 (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {1}{b x^4 \sqrt {b x^2+c x^4}} \]

[Out]

1/b/x^4/(c*x^4+b*x^2)^(1/2)-6/5*(c*x^4+b*x^2)^(1/2)/b^2/x^6+8/5*c*(c*x^4+b*x^2)^(1/2)/b^3/x^4-16/5*c^2*(c*x^4+
b*x^2)^(1/2)/b^4/x^2

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Rubi [A]  time = 0.18, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ -\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {1}{b x^4 \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^4*Sqrt[b*x^2 + c*x^4]) - (6*Sqrt[b*x^2 + c*x^4])/(5*b^2*x^6) + (8*c*Sqrt[b*x^2 + c*x^4])/(5*b^3*x^4) -
(16*c^2*Sqrt[b*x^2 + c*x^4])/(5*b^4*x^2)

Rule 2014

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] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2015

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

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^3 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{b x^4 \sqrt {b x^2+c x^4}}+\frac {6 \int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx}{b}\\ &=\frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}-\frac {(24 c) \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{5 b^2}\\ &=\frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}+\frac {\left (16 c^2\right ) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{5 b^3}\\ &=\frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 57, normalized size = 0.56 \[ \frac {-b^3+2 b^2 c x^2-8 b c^2 x^4-16 c^3 x^6}{5 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-b^3 + 2*b^2*c*x^2 - 8*b*c^2*x^4 - 16*c^3*x^6)/(5*b^4*x^4*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.65, size = 63, normalized size = 0.62 \[ -\frac {{\left (16 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} - 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{5 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(16*c^3*x^6 + 8*b*c^2*x^4 - 2*b^2*c*x^2 + b^3)*sqrt(c*x^4 + b*x^2)/(b^4*c*x^8 + b^5*x^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 59, normalized size = 0.58 \[ -\frac {\left (c \,x^{2}+b \right ) \left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right )}{5 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(c*x^2+b)*(16*c^3*x^6+8*b*c^2*x^4-2*b^2*c*x^2+b^3)/x^2/b^4/(c*x^4+b*x^2)^(3/2)

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maxima [A]  time = 1.51, size = 89, normalized size = 0.87 \[ -\frac {16 \, c^{3} x^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {8 \, c^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {2 \, c}{5 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} - \frac {1}{5 \, \sqrt {c x^{4} + b x^{2}} b x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 60, normalized size = 0.59 \[ -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (b^3-2\,b^2\,c\,x^2+8\,b\,c^2\,x^4+16\,c^3\,x^6\right )}{5\,b^4\,x^6\,\left (c\,x^2+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

-((b*x^2 + c*x^4)^(1/2)*(b^3 + 16*c^3*x^6 - 2*b^2*c*x^2 + 8*b*c^2*x^4))/(5*b^4*x^6*(b + c*x^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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