3.19.100 \(\int (a+\frac {b}{x^2})^{3/2} x^2 \, dx\) [1900]

Optimal. Leaf size=61 \[ b \sqrt {a+\frac {b}{x^2}} x+\frac {1}{3} \left (a+\frac {b}{x^2}\right )^{3/2} x^3-b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 61, 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 = {342, 283, 223, 212} \begin {gather*} -b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )+b x \sqrt {a+\frac {b}{x^2}}+\frac {1}{3} x^3 \left (a+\frac {b}{x^2}\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)*x^2,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 74, normalized size = 1.21 \begin {gather*} \frac {\sqrt {a+\frac {b}{x^2}} x \left (\sqrt {b+a x^2} \left (4 b+a x^2\right )-3 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )\right )}{3 \sqrt {b+a x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)*x^2,x]

[Out]

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

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Maple [A]
time = 0.03, size = 78, normalized size = 1.28

method	result	size
default	\(-\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right )-\left (a \,x^{2}+b \right )^{\frac {3}{2}}-3 \sqrt {a \,x^{2}+b}\, b \right )}{3 \left (a \,x^{2}+b \right )^{\frac {3}{2}}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 68, normalized size = 1.11 \begin {gather*} \frac {1}{3} \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} x^{3} + \sqrt {a + \frac {b}{x^{2}}} b x + \frac {1}{2} \, b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 129, normalized size = 2.11 \begin {gather*} \left [\frac {1}{2} \, b^{\frac {3}{2}} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + \frac {1}{3} \, {\left (a x^{3} + 4 \, b x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}, \sqrt {-b} b \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + \frac {1}{3} \, {\left (a x^{3} + 4 \, b x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.25, size = 78, normalized size = 1.28 \begin {gather*} \frac {a \sqrt {b} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3} + \frac {4 b^{\frac {3}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3} + \frac {b^{\frac {3}{2}} \log {\left (\frac {a x^{2}}{b} \right )}}{2} - b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.85, size = 89, normalized size = 1.46 \begin {gather*} \frac {b^{2} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + \frac {1}{3} \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + \sqrt {a x^{2} + b} b \mathrm {sgn}\left (x\right ) - \frac {{\left (3 \, b^{2} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 4 \, \sqrt {-b} b^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,{\left (a+\frac {b}{x^2}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/x^2)^(3/2),x)

[Out]

int(x^2*(a + b/x^2)^(3/2), x)

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