3.2.50 \(\int \frac {x^4}{\sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=50 \[ \frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x} \]

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Rubi [A]  time = 0.08, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 1588} \begin {gather*} \frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2016

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

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {b x^2+c x^4}} \, dx &=\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {(2 b) \int \frac {x^2}{\sqrt {b x^2+c x^4}} \, dx}{3 c}\\ &=-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}+\frac {x \sqrt {b x^2+c x^4}}{3 c}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.68 \begin {gather*} \frac {\left (c x^2-2 b\right ) \sqrt {x^2 \left (b+c x^2\right )}}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 34, normalized size = 0.68 \begin {gather*} \frac {\left (c x^2-2 b\right ) \sqrt {b x^2+c x^4}}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.59, size = 30, normalized size = 0.60 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{3 \, c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/sqrt(c*x^4 + b*x^2), x)

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maple [A]  time = 0.00, size = 37, normalized size = 0.74 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-c \,x^{2}+2 b \right ) x}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.50, size = 34, normalized size = 0.68 \begin {gather*} \frac {c^{2} x^{4} - b c x^{2} - 2 \, b^{2}}{3 \, \sqrt {c x^{2} + b} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.25, size = 33, normalized size = 0.66 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {2\,b}{3\,c^2}-\frac {x^2}{3\,c}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2 + c*x^4)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/sqrt(x**2*(b + c*x**2)), x)

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