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

Optimal. Leaf size=78 \[ \frac{8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac{4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

[Out]

(8*b^2*(b*x^2 + c*x^4)^(3/2))/(105*c^3*x^3) - (4*b*(b*x^2 + c*x^4)^(3/2))/(35*c^2*x) + (x*(b*x^2 + c*x^4)^(3/2
))/(7*c)

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Rubi [A]  time = 0.0942747, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2000} \[ \frac{8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac{4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[b*x^2 + c*x^4],x]

[Out]

(8*b^2*(b*x^2 + c*x^4)^(3/2))/(105*c^3*x^3) - (4*b*(b*x^2 + c*x^4)^(3/2))/(35*c^2*x) + (x*(b*x^2 + c*x^4)^(3/2
))/(7*c)

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])

Rule 2000

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rubi steps

\begin{align*} \int x^4 \sqrt{b x^2+c x^4} \, dx &=\frac{x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac{(4 b) \int x^2 \sqrt{b x^2+c x^4} \, dx}{7 c}\\ &=-\frac{4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{3/2}}{7 c}+\frac{\left (8 b^2\right ) \int \sqrt{b x^2+c x^4} \, dx}{35 c^2}\\ &=\frac{8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac{4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{3/2}}{7 c}\\ \end{align*}

Mathematica [A]  time = 0.0226631, size = 46, normalized size = 0.59 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (8 b^2-12 b c x^2+15 c^2 x^4\right )}{105 c^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[b*x^2 + c*x^4],x]

[Out]

((x^2*(b + c*x^2))^(3/2)*(8*b^2 - 12*b*c*x^2 + 15*c^2*x^4))/(105*c^3*x^3)

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Maple [A]  time = 0.046, size = 50, normalized size = 0.6 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ) \left ( 15\,{c}^{2}{x}^{4}-12\,bc{x}^{2}+8\,{b}^{2} \right ) }{105\,{c}^{3}x}\sqrt{c{x}^{4}+b{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(c*x^2+b)*(15*c^2*x^4-12*b*c*x^2+8*b^2)*(c*x^4+b*x^2)^(1/2)/c^3/x

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Maxima [A]  time = 1.00055, size = 62, normalized size = 0.79 \begin{align*} \frac{{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt{c x^{2} + b}}{105 \, c^{3}} \end{align*}

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]

1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b^3)*sqrt(c*x^2 + b)/c^3

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Fricas [A]  time = 1.59245, size = 113, normalized size = 1.45 \begin{align*} \frac{{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt{c x^{4} + b x^{2}}}{105 \, c^{3} x} \end{align*}

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]

1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2)/(c^3*x)

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

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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Giac [A]  time = 1.2429, size = 76, normalized size = 0.97 \begin{align*} -\frac{8 \, b^{\frac{7}{2}} \mathrm{sgn}\left (x\right )}{105 \, c^{3}} + \frac{{\left (15 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{2}\right )} \mathrm{sgn}\left (x\right )}{105 \, c^{3}} \end{align*}

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]

-8/105*b^(7/2)*sgn(x)/c^3 + 1/105*(15*(c*x^2 + b)^(7/2) - 42*(c*x^2 + b)^(5/2)*b + 35*(c*x^2 + b)^(3/2)*b^2)*s
gn(x)/c^3