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

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

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 31

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

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 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 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])

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*}

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Mathematica [A]  time = 0.01, size = 44, normalized size = 0.59 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.22, size = 156, normalized size = 2.08 \begin {gather*} \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2}+\frac {a \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )}{4 b^3}+\frac {a \left (\sqrt {b^2}-b\right ) \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )}{4 b^3}-\frac {x^2}{4 \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*x^2/Sqrt[b^2] + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/(4*b^2) + (a*(b + Sqrt[b^2])*Log[-a - Sqrt[b^2]*x^2 + Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*b^3) + (a*(-b + Sqrt[b^2])*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^
2*x^4]])/(4*b^3)

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fricas [A]  time = 0.67, size = 22, normalized size = 0.29 \begin {gather*} \frac {b x^{2} - a \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end {gather*}

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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giac [A]  time = 0.17, size = 33, normalized size = 0.44 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 41, normalized size = 0.55 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (-b \,x^{2}+a \ln \left (b \,x^{2}+a \right )\right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2}} \end {gather*}

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 = 1.31, size = 23, normalized size = 0.31 \begin {gather*} \frac {x^{2}}{2 \, b} - \frac {a \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end {gather*}

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*x^2/b - 1/2*a*log(b*x^2 + a)/b^2

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mupad [B]  time = 4.52, size = 64, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,b^2}-\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )}{2\,{\left (b^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a + b*x^2)^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.18, size = 20, normalized size = 0.27 \begin {gather*} - \frac {a \log {\left (a + b x^{2} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} \end {gather*}

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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