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

Optimal. Leaf size=131 \[ -\frac{8 b^2 \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^3}-\frac{x \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac{4 b \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{105 c^3 x}+\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c} \]

[Out]

(-8*b^2*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*c^4*x^3) + (4*b*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(10
5*c^3*x) - ((2*b*B - 3*A*c)*x*(b*x^2 + c*x^4)^(3/2))/(21*c^2) + (B*x^3*(b*x^2 + c*x^4)^(3/2))/(9*c)

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Rubi [A]  time = 0.219836, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2039, 2016, 2000} \[ -\frac{8 b^2 \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^3}-\frac{x \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac{4 b \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{105 c^3 x}+\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*c^4*x^3) + (4*b*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(10
5*c^3*x) - ((2*b*B - 3*A*c)*x*(b*x^2 + c*x^4)^(3/2))/(21*c^2) + (B*x^3*(b*x^2 + c*x^4)^(3/2))/(9*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^4 \left (A+B x^2\right ) \sqrt{b x^2+c x^4} \, dx &=\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac{(6 b B-9 A c) \int x^4 \sqrt{b x^2+c x^4} \, dx}{9 c}\\ &=-\frac{(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}+\frac{(4 b (2 b B-3 A c)) \int x^2 \sqrt{b x^2+c x^4} \, dx}{21 c^2}\\ &=\frac{4 b (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x}-\frac{(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac{\left (8 b^2 (2 b B-3 A c)\right ) \int \sqrt{b x^2+c x^4} \, dx}{105 c^3}\\ &=-\frac{8 b^2 (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{315 c^4 x^3}+\frac{4 b (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x}-\frac{(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}\\ \end{align*}

Mathematica [A]  time = 0.0596196, size = 82, normalized size = 0.63 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (24 b^2 c \left (A+B x^2\right )-6 b c^2 x^2 \left (6 A+5 B x^2\right )+5 c^3 x^4 \left (9 A+7 B x^2\right )-16 b^3 B\right )}{315 c^4 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + c*x^2))^(3/2)*(-16*b^3*B + 24*b^2*c*(A + B*x^2) - 6*b*c^2*x^2*(6*A + 5*B*x^2) + 5*c^3*x^4*(9*A + 7*
B*x^2)))/(315*c^4*x^3)

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Maple [A]  time = 0.006, size = 91, normalized size = 0.7 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ) \left ( 35\,B{c}^{3}{x}^{6}+45\,A{x}^{4}{c}^{3}-30\,B{x}^{4}b{c}^{2}-36\,A{x}^{2}b{c}^{2}+24\,B{x}^{2}{b}^{2}c+24\,A{b}^{2}c-16\,B{b}^{3} \right ) }{315\,{c}^{4}x}\sqrt{c{x}^{4}+b{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(c*x^2+b)*(35*B*c^3*x^6+45*A*c^3*x^4-30*B*b*c^2*x^4-36*A*b*c^2*x^2+24*B*b^2*c*x^2+24*A*b^2*c-16*B*b^3)*(
c*x^4+b*x^2)^(1/2)/c^4/x

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Maxima [A]  time = 1.20731, size = 143, normalized size = 1.09 \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} A}{105 \, c^{3}} + \frac{{\left (35 \, c^{4} x^{8} + 5 \, b c^{3} x^{6} - 6 \, b^{2} c^{2} x^{4} + 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} \sqrt{c x^{2} + b} B}{315 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)*(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)*A/c^3 + 1/315*(35*c^4*x^8 + 5*b*c^3*x^6
- 6*b^2*c^2*x^4 + 8*b^3*c*x^2 - 16*b^4)*sqrt(c*x^2 + b)*B/c^4

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Fricas [A]  time = 1.11853, size = 230, normalized size = 1.76 \begin{align*} \frac{{\left (35 \, B c^{4} x^{8} + 5 \,{\left (B b c^{3} + 9 \, A c^{4}\right )} x^{6} - 16 \, B b^{4} + 24 \, A b^{3} c - 3 \,{\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{4} + 4 \,{\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, c^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*B*c^4*x^8 + 5*(B*b*c^3 + 9*A*c^4)*x^6 - 16*B*b^4 + 24*A*b^3*c - 3*(2*B*b^2*c^2 - 3*A*b*c^3)*x^4 + 4*
(2*B*b^3*c - 3*A*b^2*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*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 )} \left (A + B x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16876, size = 180, normalized size = 1.37 \begin{align*} \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 )} A \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 )} B \mathrm{sgn}\left (x\right )}{c^{3}}}{315 \, c} + \frac{8 \,{\left (2 \, B b^{\frac{9}{2}} - 3 \, A b^{\frac{7}{2}} c\right )} \mathrm{sgn}\left (x\right )}{315 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*A*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)*B*sgn(x)/c^3)/c
+ 8/315*(2*B*b^(9/2) - 3*A*b^(7/2)*c)*sgn(x)/c^4