3.625 \(\int \frac{x^3}{\sqrt{a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=75 \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac{a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0559249, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1111, 640, 608, 31} \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4}}{2 b^2}-\frac{a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0124745, size = 44, normalized size = 0.59 \[ \frac{\left (a+b x^2\right ) \left (b x^2-a \log \left (a+b x^2\right )\right )}{2 b^2 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.21, size = 41, normalized size = 0.6 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( -b{x}^{2}+a\ln \left ( b{x}^{2}+a \right ) \right ) }{2\,{b}^{2}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((b*x^2+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.999998, size = 63, normalized size = 0.84 \begin{align*} -\frac{a \sqrt{\frac{1}{b^{2}}} \log \left (x^{2} + \frac{a}{b}\right )}{2 \, b} + \frac{\sqrt{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*a*sqrt(b^(-2))*log(x^2 + a/b)/b + 1/2*sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)/b^2

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Fricas [A]  time = 1.29094, size = 49, normalized size = 0.65 \begin{align*} \frac{b x^{2} - a \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/2*(b*x^2 - a*log(b*x^2 + a))/b^2

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Sympy [A]  time = 0.318036, size = 20, normalized size = 0.27 \begin{align*} - \frac{a \log{\left (a + b x^{2} \right )}}{2 b^{2}} + \frac{x^{2}}{2 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/((b*x**2+a)**2)**(1/2),x)

[Out]

-a*log(a + b*x**2)/(2*b**2) + x**2/(2*b)

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Giac [A]  time = 1.12577, size = 45, normalized size = 0.6 \begin{align*} \frac{1}{2} \,{\left (\frac{x^{2}}{b} - \frac{a \log \left ({\left | b x^{2} + a \right |}\right )}{b^{2}}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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