3.4.75 \(\int \frac {(a+b x^2)^{3/2}}{x^5} \, dx\) [375]

Optimal. Leaf size=68 \[ -\frac {3 b \sqrt {a+b x^2}}{8 x^2}-\frac {\left (a+b x^2\right )^{3/2}}{4 x^4}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]

[Out]

-1/4*(b*x^2+a)^(3/2)/x^4-3/8*b^2*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)-3/8*b*(b*x^2+a)^(1/2)/x^2

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Rubi [A]
time = 0.03, antiderivative size = 68, 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, 43, 65, 214} \begin {gather*} -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {3 b \sqrt {a+b x^2}}{8 x^2}-\frac {\left (a+b x^2\right )^{3/2}}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^5,x]

[Out]

(-3*b*Sqrt[a + b*x^2])/(8*x^2) - (a + b*x^2)^(3/2)/(4*x^4) - (3*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*Sqrt[
a])

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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

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Mathematica [A]
time = 0.10, size = 59, normalized size = 0.87 \begin {gather*} \frac {\left (-2 a-5 b x^2\right ) \sqrt {a+b x^2}}{8 x^4}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^5,x]

[Out]

((-2*a - 5*b*x^2)*Sqrt[a + b*x^2])/(8*x^4) - (3*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*Sqrt[a])

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Maple [A]
time = 0.04, size = 101, normalized size = 1.49

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (5 b \,x^{2}+2 a \right )}{8 x^{4}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 \sqrt {a}}\)	\(57\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4/a/x^4*(b*x^2+a)^(5/2)+1/4*b/a*(-1/2/a/x^2*(b*x^2+a)^(5/2)+3/2*b/a*(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 = 0.37, size = 90, normalized size = 1.32 \begin {gather*} -\frac {3 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {b x^{2} + a} b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="maxima")

[Out]

-3/8*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/8*(b*x^2 + a)^(3/2)*b^2/a^2 + 3/8*sqrt(b*x^2 + a)*b^2/a - 1
/8*(b*x^2 + a)^(5/2)*b/(a^2*x^2) - 1/4*(b*x^2 + a)^(5/2)/(a*x^4)

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Fricas [A]
time = 1.73, size = 136, normalized size = 2.00 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a x^{4}}, \frac {3 \, \sqrt {-a} b^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/16*(3*sqrt(a)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(5*a*b*x^2 + 2*a^2)*sqrt(b*x^
2 + a))/(a*x^4), 1/8*(3*sqrt(-a)*b^2*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (5*a*b*x^2 + 2*a^2)*sqrt(b*x^2 + a
))/(a*x^4)]

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Sympy [A]
time = 1.75, size = 71, normalized size = 1.04 \begin {gather*} - \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{4 x^{3}} - \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{8 x} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**5,x)

[Out]

-a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(4*x**3) - 5*b**(3/2)*sqrt(a/(b*x**2) + 1)/(8*x) - 3*b**2*asinh(sqrt(a)/(sqrt(
b)*x))/(8*sqrt(a))

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Giac [A]
time = 0.87, size = 70, normalized size = 1.03 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x^{2} + a} a b^{3}}{b^{2} x^{4}}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.91, size = 52, normalized size = 0.76 \begin {gather*} \frac {3\,a\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {5\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x^5,x)

[Out]

(3*a*(a + b*x^2)^(1/2))/(8*x^4) - (3*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(1/2)) - (5*(a + b*x^2)^(3/2))
/(8*x^4)

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