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

Optimal. Leaf size=94 \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{35 c^2 x}+\frac{2 b \left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^3}+\frac{B x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

[Out]

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

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Rubi [A]  time = 0.165806, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2039, 2016, 2000} \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{35 c^2 x}+\frac{2 b \left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^3}+\frac{B x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

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

Rule 2039

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(d*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(b*(m + n + p*(j + n) + 1)), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

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

Mathematica [A]  time = 0.041685, size = 64, normalized size = 0.68 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-2 b c \left (7 A+6 B x^2\right )+3 c^2 x^2 \left (7 A+5 B x^2\right )+8 b^2 B\right )}{105 c^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

((x^2*(b + c*x^2))^(3/2)*(8*b^2*B + 3*c^2*x^2*(7*A + 5*B*x^2) - 2*b*c*(7*A + 6*B*x^2)))/(105*c^3*x^3)

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Maple [A]  time = 0.006, size = 67, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -15\,B{c}^{2}{x}^{4}-21\,A{x}^{2}{c}^{2}+12\,B{x}^{2}bc+14\,Abc-8\,B{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^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/105*(c*x^2+b)*(-15*B*c^2*x^4-21*A*c^2*x^2+12*B*b*c*x^2+14*A*b*c-8*B*b^2)*(c*x^4+b*x^2)^(1/2)/c^3/x

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Maxima [A]  time = 1.10136, size = 112, normalized size = 1.19 \begin{align*} \frac{{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt{c x^{2} + b} A}{15 \, c^{2}} + \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} B}{105 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^2 + b)*A/c^2 + 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)*B/c^3

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Fricas [A]  time = 1.16213, size = 177, normalized size = 1.88 \begin{align*} \frac{{\left (15 \, B c^{3} x^{6} + 3 \,{\left (B b c^{2} + 7 \, A c^{3}\right )} x^{4} + 8 \, B b^{3} - 14 \, A b^{2} c -{\left (4 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\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^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*B*c^3*x^6 + 3*(B*b*c^2 + 7*A*c^3)*x^4 + 8*B*b^3 - 14*A*b^2*c - (4*B*b^2*c - 7*A*b*c^2)*x^2)*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} \sqrt{x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

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Giac [A]  time = 1.16053, size = 142, normalized size = 1.51 \begin{align*} \frac{\frac{7 \,{\left (3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b\right )} A \mathrm{sgn}\left (x\right )}{c} + \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 )} B \mathrm{sgn}\left (x\right )}{c^{2}}}{105 \, c} - \frac{2 \,{\left (4 \, B b^{\frac{7}{2}} - 7 \, A b^{\frac{5}{2}} c\right )} \mathrm{sgn}\left (x\right )}{105 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(7*(3*(c*x^2 + b)^(5/2) - 5*(c*x^2 + b)^(3/2)*b)*A*sgn(x)/c + (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)/c - 2/105*(4*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^3