3.251 \(\int x^2 (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=80 \[ \frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac{4 b \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{\left (b x^2+c x^4\right )^{5/2}}{9 c x} \]

[Out]

(8*b^2*(b*x^2 + c*x^4)^(5/2))/(315*c^3*x^5) - (4*b*(b*x^2 + c*x^4)^(5/2))/(63*c^2*x^3) + (b*x^2 + c*x^4)^(5/2)
/(9*c*x)

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Rubi [A]  time = 0.111375, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \[ \frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac{4 b \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{\left (b x^2+c x^4\right )^{5/2}}{9 c x} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(8*b^2*(b*x^2 + c*x^4)^(5/2))/(315*c^3*x^5) - (4*b*(b*x^2 + c*x^4)^(5/2))/(63*c^2*x^3) + (b*x^2 + c*x^4)^(5/2)
/(9*c*x)

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 2002

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

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac{(4 b) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{9 c}\\ &=-\frac{4 b \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{\left (b x^2+c x^4\right )^{5/2}}{9 c x}+\frac{\left (8 b^2\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{63 c^2}\\ &=\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac{4 b \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{\left (b x^2+c x^4\right )^{5/2}}{9 c x}\\ \end{align*}

Mathematica [A]  time = 0.025244, size = 53, normalized size = 0.66 \[ \frac{x \left (b+c x^2\right )^3 \left (8 b^2-20 b c x^2+35 c^2 x^4\right )}{315 c^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(b + c*x^2)^3*(8*b^2 - 20*b*c*x^2 + 35*c^2*x^4))/(315*c^3*Sqrt[x^2*(b + c*x^2)])

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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 ( 35\,{c}^{2}{x}^{4}-20\,bc{x}^{2}+8\,{b}^{2} \right ) }{315\,{c}^{3}{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(c*x^2+b)*(35*c^2*x^4-20*b*c*x^2+8*b^2)*(c*x^4+b*x^2)^(3/2)/c^3/x^3

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Maxima [A]  time = 1.04395, size = 77, normalized size = 0.96 \begin{align*} \frac{{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt{c x^{2} + b}}{315 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*c^4*x^8 + 50*b*c^3*x^6 + 3*b^2*c^2*x^4 - 4*b^3*c*x^2 + 8*b^4)*sqrt(c*x^2 + b)/c^3

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Fricas [A]  time = 1.3391, size = 136, normalized size = 1.7 \begin{align*} \frac{{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*c^4*x^8 + 50*b*c^3*x^6 + 3*b^2*c^2*x^4 - 4*b^3*c*x^2 + 8*b^4)*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^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(x**2*(b + c*x**2))**(3/2), x)

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Giac [A]  time = 1.15817, size = 163, normalized size = 2.04 \begin{align*} -\frac{8 \, b^{\frac{9}{2}} \mathrm{sgn}\left (x\right )}{315 \, c^{3}} + \frac{\frac{3 \,{\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 )} b \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{{\left (35 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} - 135 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} b + 189 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{3}\right )} \mathrm{sgn}\left (x\right )}{c^{2}}}{315 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/315*b^(9/2)*sgn(x)/c^3 + 1/315*(3*(15*(c*x^2 + b)^(7/2) - 42*(c*x^2 + b)^(5/2)*b + 35*(c*x^2 + b)^(3/2)*b^2
)*b*sgn(x)/c^2 + (35*(c*x^2 + b)^(9/2) - 135*(c*x^2 + b)^(7/2)*b + 189*(c*x^2 + b)^(5/2)*b^2 - 105*(c*x^2 + b)
^(3/2)*b^3)*sgn(x)/c^2)/c