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

Optimal. Leaf size=131 \[ -\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^5}-\frac{\left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{99 c^2 x}+\frac{4 b \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^3}+\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

[Out]

(-8*b^2*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*c^4*x^5) + (4*b*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/
(693*c^3*x^3) - ((6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*c^2*x) + (B*x*(b*x^2 + c*x^4)^(5/2))/(11*c)

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Rubi [A]  time = 0.241394, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2039, 2016, 2002, 2014} \[ -\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^5}-\frac{\left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{99 c^2 x}+\frac{4 b \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^3}+\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-8*b^2*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*c^4*x^5) + (4*b*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/
(693*c^3*x^3) - ((6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*c^2*x) + (B*x*(b*x^2 + c*x^4)^(5/2))/(11*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 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 (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac{(6 b B-11 A c) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{11 c}\\ &=-\frac{(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c}+\frac{(4 b (6 b B-11 A c)) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{99 c^2}\\ &=\frac{4 b (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{693 c^3 x^3}-\frac{(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac{\left (8 b^2 (6 b B-11 A c)\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{693 c^3}\\ &=-\frac{8 b^2 (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 c^4 x^5}+\frac{4 b (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{693 c^3 x^3}-\frac{(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac{B x \left (b x^2+c x^4\right )^{5/2}}{11 c}\\ \end{align*}

Mathematica [A]  time = 0.067365, size = 92, normalized size = 0.7 \[ \frac{x \left (b+c x^2\right )^3 \left (8 b^2 c \left (11 A+15 B x^2\right )-10 b c^2 x^2 \left (22 A+21 B x^2\right )+35 c^3 x^4 \left (11 A+9 B x^2\right )-48 b^3 B\right )}{3465 c^4 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(-48*b^3*B + 35*c^3*x^4*(11*A + 9*B*x^2) + 8*b^2*c*(11*A + 15*B*x^2) - 10*b*c^2*x^2*(22*A + 2
1*B*x^2)))/(3465*c^4*Sqrt[x^2*(b + c*x^2)])

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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 ( 315\,B{c}^{3}{x}^{6}+385\,A{x}^{4}{c}^{3}-210\,B{x}^{4}b{c}^{2}-220\,A{x}^{2}b{c}^{2}+120\,B{x}^{2}{b}^{2}c+88\,A{b}^{2}c-48\,B{b}^{3} \right ) }{3465\,{c}^{4}{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*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)

[Out]

1/3465*(c*x^2+b)*(315*B*c^3*x^6+385*A*c^3*x^4-210*B*b*c^2*x^4-220*A*b*c^2*x^2+120*B*b^2*c*x^2+88*A*b^2*c-48*B*
b^3)*(c*x^4+b*x^2)^(3/2)/c^4/x^3

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Maxima [A]  time = 1.28856, size = 173, normalized size = 1.32 \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} A}{315 \, c^{3}} + \frac{{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt{c x^{2} + b} B}{1155 \, c^{4}} \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)^(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)*A/c^3 + 1/1155*(105*c^
5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)*B/c^4

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Fricas [A]  time = 1.20251, size = 294, normalized size = 2.24 \begin{align*} \frac{{\left (315 \, B c^{5} x^{10} + 35 \,{\left (12 \, B b c^{4} + 11 \, A c^{5}\right )} x^{8} + 5 \,{\left (3 \, B b^{2} c^{3} + 110 \, A b c^{4}\right )} x^{6} - 48 \, B b^{5} + 88 \, A b^{4} c - 3 \,{\left (6 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4} + 4 \,{\left (6 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{3465 \, c^{4} 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)^(3/2),x, algorithm="fricas")

[Out]

1/3465*(315*B*c^5*x^10 + 35*(12*B*b*c^4 + 11*A*c^5)*x^8 + 5*(3*B*b^2*c^3 + 110*A*b*c^4)*x^6 - 48*B*b^5 + 88*A*
b^4*c - 3*(6*B*b^3*c^2 - 11*A*b^2*c^3)*x^4 + 4*(6*B*b^4*c - 11*A*b^3*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^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \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)**(3/2),x)

[Out]

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

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Giac [B]  time = 1.22388, size = 363, normalized size = 2.77 \begin{align*} \frac{\frac{33 \,{\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 b \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{11 \,{\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 b \mathrm{sgn}\left (x\right )}{c^{3}} + \frac{11 \,{\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 )} A \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{{\left (315 \,{\left (c x^{2} + b\right )}^{\frac{11}{2}} - 1540 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} b + 2970 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} b^{2} - 2772 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b^{3} + 1155 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{4}\right )} B \mathrm{sgn}\left (x\right )}{c^{3}}}{3465 \, c} + \frac{8 \,{\left (6 \, B b^{\frac{11}{2}} - 11 \, A b^{\frac{9}{2}} c\right )} \mathrm{sgn}\left (x\right )}{3465 \, c^{4}} \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)^(3/2),x, algorithm="giac")

[Out]

1/3465*(33*(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*b*sgn(x)/c^2 + 11*(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*b*sgn(x
)/c^3 + 11*(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)*A*sgn(x)/c^2 + (315*(c*x^2 + b)^(11/2) - 1540*(c*x^2 + b)^(9/2)*b + 2970*(c*x^2 + b)^(7/2)*b^2 - 2772*(c
*x^2 + b)^(5/2)*b^3 + 1155*(c*x^2 + b)^(3/2)*b^4)*B*sgn(x)/c^3)/c + 8/3465*(6*B*b^(11/2) - 11*A*b^(9/2)*c)*sgn
(x)/c^4