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

Optimal. Leaf size=106 \[ -\frac{16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac{2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

[Out]

(-16*b^3*(b*x^2 + c*x^4)^(5/2))/(1155*c^4*x^5) + (8*b^2*(b*x^2 + c*x^4)^(5/2))/(231*c^3*x^3) - (2*b*(b*x^2 + c
*x^4)^(5/2))/(33*c^2*x) + (x*(b*x^2 + c*x^4)^(5/2))/(11*c)

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Rubi [A]  time = 0.195626, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \[ -\frac{16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac{2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-16*b^3*(b*x^2 + c*x^4)^(5/2))/(1155*c^4*x^5) + (8*b^2*(b*x^2 + c*x^4)^(5/2))/(231*c^3*x^3) - (2*b*(b*x^2 + c
*x^4)^(5/2))/(33*c^2*x) + (x*(b*x^2 + c*x^4)^(5/2))/(11*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^4 \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac{(6 b) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{11 c}\\ &=-\frac{2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c}+\frac{\left (8 b^2\right ) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{33 c^2}\\ &=\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac{2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac{\left (16 b^3\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{231 c^3}\\ &=-\frac{16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac{8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac{2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac{x \left (b x^2+c x^4\right )^{5/2}}{11 c}\\ \end{align*}

Mathematica [A]  time = 0.0320802, size = 64, normalized size = 0.6 \[ \frac{x \left (b+c x^2\right )^3 \left (40 b^2 c x^2-16 b^3-70 b c^2 x^4+105 c^3 x^6\right )}{1155 c^4 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(-16*b^3 + 40*b^2*c*x^2 - 70*b*c^2*x^4 + 105*c^3*x^6))/(1155*c^4*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.047, size = 61, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -105\,{c}^{3}{x}^{6}+70\,b{c}^{2}{x}^{4}-40\,{b}^{2}c{x}^{2}+16\,{b}^{3} \right ) }{1155\,{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^4*(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/1155*(c*x^2+b)*(-105*c^3*x^6+70*b*c^2*x^4-40*b^2*c*x^2+16*b^3)*(c*x^4+b*x^2)^(3/2)/c^4/x^3

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Maxima [A]  time = 1.04, size = 92, normalized size = 0.87 \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}}{1155 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(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)/c
^4

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Fricas [A]  time = 1.28476, size = 165, normalized size = 1.56 \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^{4} + b x^{2}}}{1155 \, c^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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^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} \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**4*(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.17888, size = 201, normalized size = 1.9 \begin{align*} \frac{16 \, b^{\frac{11}{2}} \mathrm{sgn}\left (x\right )}{1155 \, c^{4}} + \frac{\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 \mathrm{sgn}\left (x\right )}{c^{3}} + \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 )} \mathrm{sgn}\left (x\right )}{c^{3}}}{3465 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/1155*b^(11/2)*sgn(x)/c^4 + 1/3465*(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*sgn(x)/c^3 + (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)*sgn(x)/c^3)/c