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

Optimal. Leaf size=168 \[ \frac{16 b^3 \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^5}-\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^3}+\frac{2 b \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{429 c^3 x}-\frac{x \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]

[Out]

(16*b^3*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2))/(15015*c^5*x^5) - (8*b^2*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2
))/(3003*c^4*x^3) + (2*b*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2))/(429*c^3*x) - ((8*b*B - 13*A*c)*x*(b*x^2 + c*
x^4)^(5/2))/(143*c^2) + (B*x^3*(b*x^2 + c*x^4)^(5/2))/(13*c)

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Rubi [A]  time = 0.297179, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2039, 2016, 2002, 2014} \[ \frac{16 b^3 \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^5}-\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^3}+\frac{2 b \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{429 c^3 x}-\frac{x \left (b x^2+c x^4\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^3*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2))/(15015*c^5*x^5) - (8*b^2*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2
))/(3003*c^4*x^3) + (2*b*(8*b*B - 13*A*c)*(b*x^2 + c*x^4)^(5/2))/(429*c^3*x) - ((8*b*B - 13*A*c)*x*(b*x^2 + c*
x^4)^(5/2))/(143*c^2) + (B*x^3*(b*x^2 + c*x^4)^(5/2))/(13*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^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac{(8 b B-13 A c) \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx}{13 c}\\ &=-\frac{(8 b B-13 A c) x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}+\frac{(6 b (8 b B-13 A c)) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{143 c^2}\\ &=\frac{2 b (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{(8 b B-13 A c) x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac{\left (8 b^2 (8 b B-13 A c)\right ) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{429 c^3}\\ &=-\frac{8 b^2 (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{2 b (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{(8 b B-13 A c) x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}+\frac{\left (16 b^3 (8 b B-13 A c)\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{3003 c^4}\\ &=\frac{16 b^3 (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac{8 b^2 (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{2 b (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{(8 b B-13 A c) x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}\\ \end{align*}

Mathematica [A]  time = 0.0826835, size = 113, normalized size = 0.67 \[ \frac{x \left (b+c x^2\right )^3 \left (40 b^2 c^2 x^2 \left (13 A+14 B x^2\right )-16 b^3 c \left (13 A+20 B x^2\right )-70 b c^3 x^4 \left (13 A+12 B x^2\right )+105 c^4 x^6 \left (13 A+11 B x^2\right )+128 b^4 B\right )}{15015 c^5 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(128*b^4*B + 105*c^4*x^6*(13*A + 11*B*x^2) - 70*b*c^3*x^4*(13*A + 12*B*x^2) + 40*b^2*c^2*x^2*
(13*A + 14*B*x^2) - 16*b^3*c*(13*A + 20*B*x^2)))/(15015*c^5*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.005, size = 115, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -1155\,B{x}^{8}{c}^{4}-1365\,A{c}^{4}{x}^{6}+840\,Bb{c}^{3}{x}^{6}+910\,Ab{c}^{3}{x}^{4}-560\,B{b}^{2}{c}^{2}{x}^{4}-520\,A{b}^{2}{c}^{2}{x}^{2}+320\,B{b}^{3}c{x}^{2}+208\,A{b}^{3}c-128\,B{b}^{4} \right ) }{15015\,{c}^{5}{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^4*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/15015*(c*x^2+b)*(-1155*B*c^4*x^8-1365*A*c^4*x^6+840*B*b*c^3*x^6+910*A*b*c^3*x^4-560*B*b^2*c^2*x^4-520*A*b^2
*c^2*x^2+320*B*b^3*c*x^2+208*A*b^3*c-128*B*b^4)*(c*x^4+b*x^2)^(3/2)/c^5/x^3

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Maxima [A]  time = 1.24136, size = 203, normalized size = 1.21 \begin{align*} \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} A}{1155 \, c^{4}} + \frac{{\left (1155 \, c^{6} x^{12} + 1470 \, b c^{5} x^{10} + 35 \, b^{2} c^{4} x^{8} - 40 \, b^{3} c^{3} x^{6} + 48 \, b^{4} c^{2} x^{4} - 64 \, b^{5} c x^{2} + 128 \, b^{6}\right )} \sqrt{c x^{2} + b} B}{15015 \, c^{5}} \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)^(3/2),x, algorithm="maxima")

[Out]

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)*A
/c^4 + 1/15015*(1155*c^6*x^12 + 1470*b*c^5*x^10 + 35*b^2*c^4*x^8 - 40*b^3*c^3*x^6 + 48*b^4*c^2*x^4 - 64*b^5*c*
x^2 + 128*b^6)*sqrt(c*x^2 + b)*B/c^5

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Fricas [A]  time = 1.08841, size = 350, normalized size = 2.08 \begin{align*} \frac{{\left (1155 \, B c^{6} x^{12} + 105 \,{\left (14 \, B b c^{5} + 13 \, A c^{6}\right )} x^{10} + 35 \,{\left (B b^{2} c^{4} + 52 \, A b c^{5}\right )} x^{8} + 128 \, B b^{6} - 208 \, A b^{5} c - 5 \,{\left (8 \, B b^{3} c^{3} - 13 \, A b^{2} c^{4}\right )} x^{6} + 6 \,{\left (8 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{4} - 8 \,{\left (8 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15015 \, c^{5} 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)^(3/2),x, algorithm="fricas")

[Out]

1/15015*(1155*B*c^6*x^12 + 105*(14*B*b*c^5 + 13*A*c^6)*x^10 + 35*(B*b^2*c^4 + 52*A*b*c^5)*x^8 + 128*B*b^6 - 20
8*A*b^5*c - 5*(8*B*b^3*c^3 - 13*A*b^2*c^4)*x^6 + 6*(8*B*b^4*c^2 - 13*A*b^3*c^3)*x^4 - 8*(8*B*b^5*c - 13*A*b^4*
c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^5*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \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**4*(B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**4*(x**2*(b + c*x**2))**(3/2)*(A + B*x**2), x)

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Giac [B]  time = 1.24968, size = 440, normalized size = 2.62 \begin{align*} \frac{\frac{143 \,{\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 b \mathrm{sgn}\left (x\right )}{c^{3}} + \frac{13 \,{\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 b \mathrm{sgn}\left (x\right )}{c^{4}} + \frac{13 \,{\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 )} A \mathrm{sgn}\left (x\right )}{c^{3}} + \frac{5 \,{\left (693 \,{\left (c x^{2} + b\right )}^{\frac{13}{2}} - 4095 \,{\left (c x^{2} + b\right )}^{\frac{11}{2}} b + 10010 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} b^{2} - 12870 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} b^{3} + 9009 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b^{4} - 3003 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{5}\right )} B \mathrm{sgn}\left (x\right )}{c^{4}}}{45045 \, c} - \frac{16 \,{\left (8 \, B b^{\frac{13}{2}} - 13 \, A b^{\frac{11}{2}} c\right )} \mathrm{sgn}\left (x\right )}{15015 \, c^{5}} \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)^(3/2),x, algorithm="giac")

[Out]

1/45045*(143*(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*b*sgn(x)/c^3 + 13*(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*b*sgn(x)/c^4 + 13*(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)*A*sg
n(x)/c^3 + 5*(693*(c*x^2 + b)^(13/2) - 4095*(c*x^2 + b)^(11/2)*b + 10010*(c*x^2 + b)^(9/2)*b^2 - 12870*(c*x^2
+ b)^(7/2)*b^3 + 9009*(c*x^2 + b)^(5/2)*b^4 - 3003*(c*x^2 + b)^(3/2)*b^5)*B*sgn(x)/c^4)/c - 16/15015*(8*B*b^(1
3/2) - 13*A*b^(11/2)*c)*sgn(x)/c^5