3.374 \(\int \frac {(a+b x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=63 \[ -\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} b \sqrt {a+b x^2}-\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ -\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} b \sqrt {a+b x^2}-\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^3,x]

[Out]

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

Rule 47

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

Rule 50

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 63

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 208

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

Rule 266

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^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} (3 b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{2} b \sqrt {a+b x^2}-\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} (3 a b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {3}{2} b \sqrt {a+b x^2}-\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {1}{2} (3 a) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=\frac {3}{2} b \sqrt {a+b x^2}-\frac {\left (a+b x^2\right )^{3/2}}{2 x^2}-\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.59 \[ \frac {b \left (a+b x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b x^2}{a}+1\right )}{5 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^3,x]

[Out]

(b*(a + b*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x^2)/a])/(5*a^2)

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fricas [A]  time = 1.01, size = 119, normalized size = 1.89 \[ \left [\frac {3 \, \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, b x^{2} - a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, \frac {3 \, \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, b x^{2} - a\right )} \sqrt {b x^{2} + a}}{2 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.68, size = 63, normalized size = 1.00 \[ \frac {\frac {3 \, a b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x^{2} + a} b^{2} - \frac {\sqrt {b x^{2} + a} a b}{x^{2}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 75, normalized size = 1.19 \[ -\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {3 \sqrt {b \,x^{2}+a}\, b}{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} b}{2 a}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 63, normalized size = 1.00 \[ -\frac {3}{2} \, \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b x^{2} + a} b + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b}{2 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*sqrt(a)*b*arcsinh(a/(sqrt(a*b)*abs(x))) + 3/2*sqrt(b*x^2 + a)*b + 1/2*(b*x^2 + a)^(3/2)*b/a - 1/2*(b*x^2
+ a)^(5/2)/(a*x^2)

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mupad [B]  time = 4.82, size = 47, normalized size = 0.75 \[ b\,\sqrt {b\,x^2+a}-\frac {a\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {3\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x^3,x)

[Out]

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

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sympy [A]  time = 2.33, size = 88, normalized size = 1.40 \[ - \frac {3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {a^{2}}{2 \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {a \sqrt {b}}{2 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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