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

Optimal. Leaf size=70 \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {a x \sqrt {a+b x^2}}{8 b}+\frac {1}{4} x^3 \sqrt {a+b x^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {279, 321, 217, 206} \[ -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {1}{4} x^3 \sqrt {a+b x^2}+\frac {a x \sqrt {a+b x^2}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a + b*x^2],x]

[Out]

(a*x*Sqrt[a + b*x^2])/(8*b) + (x^3*Sqrt[a + b*x^2])/4 - (a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*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 279

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

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

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Mathematica [A]  time = 0.02, size = 64, normalized size = 0.91 \[ \sqrt {a+b x^2} \left (\frac {a x}{8 b}+\frac {x^3}{4}\right )-\frac {a^2 \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{8 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a + b*x^2],x]

[Out]

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

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fricas [A]  time = 0.91, size = 119, normalized size = 1.70 \[ \left [\frac {a^{2} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} x^{3} + a b x\right )} \sqrt {b x^{2} + a}}{16 \, b^{2}}, \frac {a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, b^{2} x^{3} + a b x\right )} \sqrt {b x^{2} + a}}{8 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.71, size = 50, normalized size = 0.71 \[ \frac {1}{8} \, \sqrt {b x^{2} + a} {\left (2 \, x^{2} + \frac {a}{b}\right )} x + \frac {a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 57, normalized size = 0.81 \[ -\frac {a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, a x}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} x}{4 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 49, normalized size = 0.70 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {\sqrt {b x^{2} + a} a x}{8 \, b} - \frac {a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\sqrt {b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^(1/2),x)

[Out]

int(x^2*(a + b*x^2)^(1/2), x)

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sympy [A]  time = 3.53, size = 92, normalized size = 1.31 \[ \frac {a^{\frac {3}{2}} x}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 \sqrt {a} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {b x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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