3.3.46 \(\int \frac {(\frac {c}{a+b x^2})^{3/2}}{x} \, dx\) [246]

Optimal. Leaf size=71 \[ \frac {c \sqrt {\frac {c}{a+b x^2}}}{a}-\frac {c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}} \tanh ^{-1}\left (\sqrt {1+\frac {b x^2}{a}}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1973, 272, 53, 65, 214} \begin {gather*} \frac {c \sqrt {\frac {c}{a+b x^2}}}{a}-\frac {c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2)/x,x]

[Out]

(c*Sqrt[c/(a + b*x^2)])/a - (c*Sqrt[c/(a + b*x^2)]*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/a

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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

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

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Mathematica [A]
time = 0.04, size = 59, normalized size = 0.83 \begin {gather*} \frac {c \sqrt {\frac {c}{a+b x^2}} \left (\sqrt {a}-\sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x^2))^(3/2)/x,x]

[Out]

(c*Sqrt[c/(a + b*x^2)]*(Sqrt[a] - Sqrt[a + b*x^2]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))/a^(3/2)

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Maple [A]
time = 0.03, size = 64, normalized size = 0.90

method	result	size
default	\(-\frac {\left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \left (\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) a \sqrt {b \,x^{2}+a}-a^{\frac {3}{2}}\right )}{a^{\frac {5}{2}}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*a*(b*x^2+a)^(1/2)-a^(3/2))/a^(5/2)

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Maxima [A]
time = 0.48, size = 80, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, c {\left (\frac {c \log \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} - \sqrt {a c}}{a \sqrt {\frac {c}{b x^{2} + a}} + \sqrt {a c}}\right )}{\sqrt {a c} a} + \frac {2 \, \sqrt {\frac {c}{b x^{2} + a}}}{a}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 138, normalized size = 1.94 \begin {gather*} \left [\frac {c \sqrt {\frac {c}{a}} \log \left (-\frac {b c x^{2} + 2 \, a c - 2 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {\frac {c}{b x^{2} + a}} \sqrt {\frac {c}{a}}}{x^{2}}\right ) + 2 \, c \sqrt {\frac {c}{b x^{2} + a}}}{2 \, a}, \frac {c \sqrt {-\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} \sqrt {-\frac {c}{a}}}{c}\right ) + c \sqrt {\frac {c}{b x^{2} + a}}}{a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(c*sqrt(c/a)*log(-(b*c*x^2 + 2*a*c - 2*(a*b*x^2 + a^2)*sqrt(c/(b*x^2 + a))*sqrt(c/a))/x^2) + 2*c*sqrt(c/(
b*x^2 + a)))/a, (c*sqrt(-c/a)*arctan(a*sqrt(c/(b*x^2 + a))*sqrt(-c/a)/c) + c*sqrt(c/(b*x^2 + a)))/a]

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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}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.53, size = 59, normalized size = 0.83 \begin {gather*} c {\left (\frac {c \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a} + \frac {c}{\sqrt {b c x^{2} + a c} a}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {c}{b\,x^2+a}\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x^2))^(3/2)/x,x)

[Out]

int((c/(a + b*x^2))^(3/2)/x, x)

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