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

Optimal. Leaf size=58 \[ \frac {\sqrt {a x^2+b x^4}}{2 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^2+b x^4}}\right )}{2 b^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2018, 640, 620, 206} \begin {gather*} \frac {\sqrt {a x^2+b x^4}}{2 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^2+b x^4}}\right )}{2 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a*x^2 + b*x^4],x]

[Out]

Sqrt[a*x^2 + b*x^4]/(2*b) - (a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x^2 + b*x^4]])/(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 620

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

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 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 73, normalized size = 1.26 \begin {gather*} \frac {x \left (\sqrt {b} x \left (a+b x^2\right )-a \sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )}{2 b^{3/2} \sqrt {x^2 \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a*x^2 + b*x^4],x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.20, size = 68, normalized size = 1.17 \begin {gather*} \frac {a \log \left (-2 b^{3/2} \sqrt {a x^2+b x^4}+a b+2 b^2 x^2\right )}{4 b^{3/2}}+\frac {\sqrt {a x^2+b x^4}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 114, normalized size = 1.97 \begin {gather*} \left [\frac {a \sqrt {b} \log \left (-2 \, b x^{2} - a + 2 \, \sqrt {b x^{4} + a x^{2}} \sqrt {b}\right ) + 2 \, \sqrt {b x^{4} + a x^{2}} b}{4 \, b^{2}}, \frac {a \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{4} + a x^{2}} \sqrt {-b}}{b x^{2} + a}\right ) + \sqrt {b x^{4} + a x^{2}} b}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^4+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.20, size = 59, normalized size = 1.02 \begin {gather*} \frac {a \log \left ({\left | -2 \, {\left (\sqrt {b} x^{2} - \sqrt {b x^{4} + a x^{2}}\right )} \sqrt {b} - a \right |}\right )}{4 \, b^{\frac {3}{2}}} + \frac {\sqrt {b x^{4} + a x^{2}}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^4+a*x^2)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 64, normalized size = 1.10 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, \left (-a b \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\sqrt {b \,x^{2}+a}\, b^{\frac {3}{2}} x \right ) x}{2 \sqrt {b \,x^{4}+a \,x^{2}}\, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^4+a*x^2)^(1/2),x)

[Out]

1/2*x*(b*x^2+a)^(1/2)*(x*(b*x^2+a)^(1/2)*b^(3/2)-a*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*b)/(b*x^4+a*x^2)^(1/2)/b^(5/2
)

________________________________________________________________________________________

maxima [A]  time = 1.42, size = 52, normalized size = 0.90 \begin {gather*} -\frac {a \log \left (2 \, b x^{2} + a + 2 \, \sqrt {b x^{4} + a x^{2}} \sqrt {b}\right )}{4 \, b^{\frac {3}{2}}} + \frac {\sqrt {b x^{4} + a x^{2}}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^4+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.71, size = 53, normalized size = 0.91 \begin {gather*} \frac {\sqrt {b\,x^4+a\,x^2}}{2\,b}-\frac {a\,\ln \left (\frac {b\,x^2+\frac {a}{2}}{\sqrt {b}}+\sqrt {b\,x^4+a\,x^2}\right )}{4\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^4)^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {x^{2} \left (a + b x^{2}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**4+a*x**2)**(1/2),x)

[Out]

Integral(x**3/sqrt(x**2*(a + b*x**2)), x)

________________________________________________________________________________________