3.3.47 \(\int \frac {(\frac {c}{a+b x^2})^{3/2}}{x^2} \, dx\) [247]

Optimal. Leaf size=48 \[ -\frac {c \sqrt {\frac {c}{a+b x^2}}}{a x}-\frac {2 b c x \sqrt {\frac {c}{a+b x^2}}}{a^2} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1973, 277, 197} \begin {gather*} -\frac {2 b c x \sqrt {\frac {c}{a+b x^2}}}{a^2}-\frac {c \sqrt {\frac {c}{a+b x^2}}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2)/x^2,x]

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

\begin {align*} \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x^2} \, dx &=\left (c \sqrt {\frac {c}{a+b x^2}} \sqrt {a+b x^2}\right ) \int \frac {1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx\\ &=-\frac {c \sqrt {\frac {c}{a+b x^2}}}{a x}-\frac {\left (2 b c \sqrt {\frac {c}{a+b x^2}} \sqrt {a+b x^2}\right ) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{a}\\ &=-\frac {c \sqrt {\frac {c}{a+b x^2}}}{a x}-\frac {2 b c x \sqrt {\frac {c}{a+b x^2}}}{a^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 32, normalized size = 0.67 \begin {gather*} -\frac {c \sqrt {\frac {c}{a+b x^2}} \left (a+2 b x^2\right )}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 37, normalized size = 0.77

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 b \,x^{2}+a \right ) \left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}}}{a^{2} x}\)	\(37\)
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 b \,x^{2}+a \right ) \left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}}}{a^{2} x}\)	\(37\)
trager	\(-\frac {\left (a c +b c \right ) \left (2 b \,x^{2}+a \right ) \sqrt {\frac {c}{b \,x^{2}+a}}}{a^{2} \left (a +b \right ) x}\)	\(42\)
risch	\(-\frac {\left (b \,x^{2}+a \right ) c \sqrt {\frac {c}{b \,x^{2}+a}}}{a^{2} x}-\frac {b c x \sqrt {\frac {c}{b \,x^{2}+a}}}{a^{2}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 46, normalized size = 0.96 \begin {gather*} -\frac {2 \, b^{2} c^{\frac {3}{2}} x^{4} + 3 \, a b c^{\frac {3}{2}} x^{2} + a^{2} c^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 32, normalized size = 0.67 \begin {gather*} -\frac {{\left (2 \, b c x^{2} + a c\right )} \sqrt {\frac {c}{b x^{2} + a}}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.94, size = 81, normalized size = 1.69 \begin {gather*} -{\left (\frac {b c x \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c x^{2} + a c} a^{2}} - \frac {2 \, \sqrt {b c} c \mathrm {sgn}\left (b x^{2} + a\right )}{{\left ({\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c}\right )}^{2} - a c\right )} a}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.83, size = 54, normalized size = 1.12 \begin {gather*} -\frac {\left (\frac {b\,c}{a}+\frac {2\,b^2\,c\,x^2}{a^2}\right )\,\sqrt {\frac {c}{b\,x^2+a}}\,\left (\frac {a}{b}+x^2\right )}{b\,x^3+a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x^2))^(3/2)/x^2,x)

[Out]

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

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