3.5.99 \(\int \frac {x^2}{(a+b x^2)^{3/2}} \, dx\) [499]

Optimal. Leaf size=43 \[ -\frac {x}{b \sqrt {a+b x^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {294, 223, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {x}{b \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^2)^(3/2),x]

[Out]

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

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 294

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 46, normalized size = 1.07 \begin {gather*} -\frac {x}{b \sqrt {a+b x^2}}-\frac {\log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.03, size = 37, normalized size = 0.86

method	result	size
default	\(-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 29, normalized size = 0.67 \begin {gather*} -\frac {x}{\sqrt {b x^{2} + a} b} + \frac {\operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.47, size = 130, normalized size = 3.02 \begin {gather*} \left [-\frac {2 \, \sqrt {b x^{2} + a} b x - {\left (b x^{2} + a\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {\sqrt {b x^{2} + a} b x + {\left (b x^{2} + a\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{b^{3} x^{2} + a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.77, size = 37, normalized size = 0.86 \begin {gather*} \frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.52, size = 39, normalized size = 0.91 \begin {gather*} -\frac {x}{\sqrt {b x^{2} + a} b} - \frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.09, size = 36, normalized size = 0.84 \begin {gather*} \frac {\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{b^{3/2}}-\frac {x}{b\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^2)^(3/2),x)

[Out]

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

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