3.249 \(\int x^6 (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=134 \[ \frac{128 b^4 \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac{64 b^3 \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]

[Out]

(128*b^4*(b*x^2 + c*x^4)^(5/2))/(15015*c^5*x^5) - (64*b^3*(b*x^2 + c*x^4)^(5/2))/(3003*c^4*x^3) + (16*b^2*(b*x
^2 + c*x^4)^(5/2))/(429*c^3*x) - (8*b*x*(b*x^2 + c*x^4)^(5/2))/(143*c^2) + (x^3*(b*x^2 + c*x^4)^(5/2))/(13*c)

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Rubi [A]  time = 0.252684, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \[ \frac{128 b^4 \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac{64 b^3 \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]

Antiderivative was successfully verified.

[In]

Int[x^6*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(128*b^4*(b*x^2 + c*x^4)^(5/2))/(15015*c^5*x^5) - (64*b^3*(b*x^2 + c*x^4)^(5/2))/(3003*c^4*x^3) + (16*b^2*(b*x
^2 + c*x^4)^(5/2))/(429*c^3*x) - (8*b*x*(b*x^2 + c*x^4)^(5/2))/(143*c^2) + (x^3*(b*x^2 + c*x^4)^(5/2))/(13*c)

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^6 \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac{(8 b) \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx}{13 c}\\ &=-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}+\frac{\left (48 b^2\right ) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{143 c^2}\\ &=\frac{16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac{\left (64 b^3\right ) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{429 c^3}\\ &=-\frac{64 b^3 \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}+\frac{\left (128 b^4\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{3003 c^4}\\ &=\frac{128 b^4 \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac{64 b^3 \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac{16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac{8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac{x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}\\ \end{align*}

Mathematica [A]  time = 0.0382531, size = 75, normalized size = 0.56 \[ \frac{x \left (b+c x^2\right )^3 \left (560 b^2 c^2 x^4-320 b^3 c x^2+128 b^4-840 b c^3 x^6+1155 c^4 x^8\right )}{15015 c^5 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(b + c*x^2)^3*(128*b^4 - 320*b^3*c*x^2 + 560*b^2*c^2*x^4 - 840*b*c^3*x^6 + 1155*c^4*x^8))/(15015*c^5*Sqrt[x
^2*(b + c*x^2)])

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Maple [A]  time = 0.046, size = 72, normalized size = 0.5 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ) \left ( 1155\,{x}^{8}{c}^{4}-840\,b{x}^{6}{c}^{3}+560\,{b}^{2}{x}^{4}{c}^{2}-320\,{b}^{3}{x}^{2}c+128\,{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^6*(c*x^4+b*x^2)^(3/2),x)

[Out]

1/15015*(c*x^2+b)*(1155*c^4*x^8-840*b*c^3*x^6+560*b^2*c^2*x^4-320*b^3*c*x^2+128*b^4)*(c*x^4+b*x^2)^(3/2)/c^5/x
^3

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Maxima [A]  time = 1.04213, size = 107, normalized size = 0.8 \begin{align*} \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}}{15015 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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 + 1
28*b^6)*sqrt(c*x^2 + b)/c^5

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Fricas [A]  time = 1.38608, size = 198, normalized size = 1.48 \begin{align*} \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^{4} + b x^{2}}}{15015 \, c^{5} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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 + 1
28*b^6)*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^{6} \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**6*(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**6*(x**2*(b + c*x**2))**(3/2), x)

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Giac [A]  time = 1.22254, size = 240, normalized size = 1.79 \begin{align*} -\frac{128 \, b^{\frac{13}{2}} \mathrm{sgn}\left (x\right )}{15015 \, c^{5}} + \frac{\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 \mathrm{sgn}\left (x\right )}{c^{4}} + \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 )} \mathrm{sgn}\left (x\right )}{c^{4}}}{45045 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

-128/15015*b^(13/2)*sgn(x)/c^5 + 1/45045*(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*sgn(x)/c^4 + 5*(693*(c*x^2 + b)^(1
3/2) - 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)*sgn(x)/c^4)/c