3.19.99 \(\int (a+\frac {b}{x^2})^{3/2} x^3 \, dx\) [1899]

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

[Out]

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

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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+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {3}{8} b x^2 \sqrt {a+\frac {b}{x^2}}+\frac {1}{4} x^4 \left (a+\frac {b}{x^2}\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)*x^3,x]

[Out]

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

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Mathematica [A]
time = 0.08, size = 69, normalized size = 1.01 \begin {gather*} \frac {1}{8} \sqrt {a+\frac {b}{x^2}} x \left (5 b x+2 a x^3-\frac {3 b^2 \log \left (-\sqrt {a} x+\sqrt {b+a x^2}\right )}{\sqrt {a} \sqrt {b+a x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)*x^3,x]

[Out]

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

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Maple [A]
time = 0.03, size = 84, normalized size = 1.24

method	result	size
risch	\(\frac {x^{2} \left (2 a \,x^{2}+5 b \right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{8}+\frac {3 b^{2} \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+b}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{8 \sqrt {a}\, \sqrt {a \,x^{2}+b}}\)	\(77\)
default	\(\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (2 x \left (a \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {a}+3 \sqrt {a}\, \sqrt {a \,x^{2}+b}\, b x +3 \ln \left (x \sqrt {a}+\sqrt {a \,x^{2}+b}\right ) b^{2}\right )}{8 \left (a \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {a}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x^2+b)/x^2)^(3/2)*x^3*(2*x*(a*x^2+b)^(3/2)*a^(1/2)+3*a^(1/2)*(a*x^2+b)^(1/2)*b*x+3*ln(x*a^(1/2)+(a*x^2
+b)^(1/2))*b^2)/(a*x^2+b)^(3/2)/a^(1/2)

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Maxima [A]
time = 0.53, size = 98, normalized size = 1.44 \begin {gather*} -\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{16 \, \sqrt {a}} + \frac {5 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {a + \frac {b}{x^{2}}} a b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} a + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*b^2*log((sqrt(a + b/x^2) - sqrt(a))/(sqrt(a + b/x^2) + sqrt(a)))/sqrt(a) + 1/8*(5*(a + b/x^2)^(3/2)*b^2
- 3*sqrt(a + b/x^2)*a*b^2)/((a + b/x^2)^2 - 2*(a + b/x^2)*a + a^2)

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Fricas [A]
time = 0.39, size = 157, normalized size = 2.31 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (2 \, a^{2} x^{4} + 5 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, a}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - {\left (2 \, a^{2} x^{4} + 5 \, a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.73, size = 70, normalized size = 1.03 \begin {gather*} \frac {a \sqrt {b} x^{3} \sqrt {\frac {a x^{2}}{b} + 1}}{4} + \frac {5 b^{\frac {3}{2}} x \sqrt {\frac {a x^{2}}{b} + 1}}{8} + \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{8 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.67, size = 68, normalized size = 1.00 \begin {gather*} -\frac {3 \, b^{2} \log \left ({\left | -\sqrt {a} x + \sqrt {a x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {a}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, \sqrt {a}} + \frac {1}{8} \, {\left (2 \, a x^{2} \mathrm {sgn}\left (x\right ) + 5 \, b \mathrm {sgn}\left (x\right )\right )} \sqrt {a x^{2} + b} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*b^2*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))*sgn(x)/sqrt(a) + 3/16*b^2*log(abs(b))*sgn(x)/sqrt(a) + 1/8*(2*
a*x^2*sgn(x) + 5*b*sgn(x))*sqrt(a*x^2 + b)*x

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Mupad [B]
time = 1.65, size = 52, normalized size = 0.76 \begin {gather*} \frac {5\,x^4\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}{8}+\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {3\,a\,x^4\,\sqrt {a+\frac {b}{x^2}}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/x^2)^(3/2),x)

[Out]

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

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