3.4.73 \(\int \frac {(a+b x^2)^{3/2}}{x} \, dx\) [373]

Optimal. Leaf size=54 \[ a \sqrt {a+b x^2}+\frac {1}{3} \left (a+b x^2\right )^{3/2}-a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 52, 65, 214} \begin {gather*} a^{3/2} \left (-\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+a \sqrt {a+b x^2}+\frac {1}{3} \left (a+b x^2\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x,x]

[Out]

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

Rule 52

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 50, normalized size = 0.93 \begin {gather*} \frac {1}{3} \sqrt {a+b x^2} \left (4 a+b x^2\right )-a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x,x]

[Out]

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

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Maple [A]
time = 0.03, size = 53, normalized size = 0.98

method	result	size
default	\(\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 40, normalized size = 0.74 \begin {gather*} -a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} + \sqrt {b x^{2} + a} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a)^(3/2) + sqrt(b*x^2 + a)*a

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Fricas [A]
time = 1.10, size = 100, normalized size = 1.85 \begin {gather*} \left [\frac {1}{2} \, a^{\frac {3}{2}} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + \frac {1}{3} \, {\left (b x^{2} + 4 \, a\right )} \sqrt {b x^{2} + a}, \sqrt {-a} a \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + \frac {1}{3} \, {\left (b x^{2} + 4 \, a\right )} \sqrt {b x^{2} + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*a^(3/2)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 1/3*(b*x^2 + 4*a)*sqrt(b*x^2 + a), sqrt(-a)
*a*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + 1/3*(b*x^2 + 4*a)*sqrt(b*x^2 + a)]

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Sympy [A]
time = 1.09, size = 78, normalized size = 1.44 \begin {gather*} \frac {4 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{3} + \frac {a^{\frac {3}{2}} \log {\left (\frac {b x^{2}}{a} \right )}}{2} - a^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )} + \frac {\sqrt {a} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**(3/2)*sqrt(1 + b*x**2/a)/3 + a**(3/2)*log(b*x**2/a)/2 - a**(3/2)*log(sqrt(1 + b*x**2/a) + 1) + sqrt(a)*b*
x**2*sqrt(1 + b*x**2/a)/3

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Giac [A]
time = 1.03, size = 48, normalized size = 0.89 \begin {gather*} \frac {a^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} + \sqrt {b x^{2} + a} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.70, size = 42, normalized size = 0.78 \begin {gather*} a\,\sqrt {b\,x^2+a}-a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+\frac {{\left (b\,x^2+a\right )}^{3/2}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x,x)

[Out]

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

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