3.488 \(\int \frac {x^2}{\sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=49 \[ \frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 49, 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 = {321, 217, 206} \[ \frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b*x^2],x]

[Out]

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

Rule 206

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

Rule 217

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.00 \[ \frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b*x^2],x]

[Out]

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

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fricas [A]  time = 0.94, size = 93, normalized size = 1.90 \[ \left [\frac {2 \, \sqrt {b x^{2} + a} b x + a \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{4 \, b^{2}}, \frac {\sqrt {b x^{2} + a} b x + a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{2 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.18, size = 40, normalized size = 0.82 \[ \frac {\sqrt {b x^{2} + a} x}{2 \, b} + \frac {a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 39, normalized size = 0.80 \[ -\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, x}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.26, size = 31, normalized size = 0.63 \[ \frac {\sqrt {b x^{2} + a} x}{2 \, b} - \frac {a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.82, size = 56, normalized size = 1.14 \[ \left \{\begin {array}{cl} \frac {x^3}{3\,\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {x\,\sqrt {b\,x^2+a}}{2\,b}-\frac {a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^2)^(1/2),x)

[Out]

piecewise(b == 0, x^3/(3*a^(1/2)), b ~= 0, (x*(a + b*x^2)^(1/2))/(2*b) - (a*log(2*b^(1/2)*x + 2*(a + b*x^2)^(1
/2)))/(2*b^(3/2)))

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sympy [A]  time = 2.25, size = 42, normalized size = 0.86 \[ \frac {\sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2 b} - \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**2+a)**(1/2),x)

[Out]

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

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