3.153 \(\int \frac{x^8 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{x^3 \sqrt{b x^2+c x^4} (6 b B-5 A c)}{5 b c^2}-\frac{4 x \sqrt{b x^2+c x^4} (6 b B-5 A c)}{15 c^3}+\frac{8 b \sqrt{b x^2+c x^4} (6 b B-5 A c)}{15 c^4 x}-\frac{x^7 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

[Out]

-(((b*B - A*c)*x^7)/(b*c*Sqrt[b*x^2 + c*x^4])) + (8*b*(6*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(15*c^4*x) - (4*(6*
b*B - 5*A*c)*x*Sqrt[b*x^2 + c*x^4])/(15*c^3) + ((6*b*B - 5*A*c)*x^3*Sqrt[b*x^2 + c*x^4])/(5*b*c^2)

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Rubi [A]  time = 0.248238, antiderivative size = 139, 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 = {2037, 2016, 1588} \[ \frac{x^3 \sqrt{b x^2+c x^4} (6 b B-5 A c)}{5 b c^2}-\frac{4 x \sqrt{b x^2+c x^4} (6 b B-5 A c)}{15 c^3}+\frac{8 b \sqrt{b x^2+c x^4} (6 b B-5 A c)}{15 c^4 x}-\frac{x^7 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^7)/(b*c*Sqrt[b*x^2 + c*x^4])) + (8*b*(6*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(15*c^4*x) - (4*(6*
b*B - 5*A*c)*x*Sqrt[b*x^2 + c*x^4])/(15*c^3) + ((6*b*B - 5*A*c)*x^3*Sqrt[b*x^2 + c*x^4])/(5*b*c^2)

Rule 2037

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> -Si
mp[(e^(j - 1)*(b*c - a*d)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*b*n*(p + 1)), x] - Dist[(e^j*(a*
d*(m + j*p + 1) - b*c*(m + n + p*(j + n) + 1)))/(a*b*n*(p + 1)), Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p +
1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && Lt
Q[p, -1] && GtQ[j, 0] && LeQ[j, m] && (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 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{(b B-A c) x^7}{b c \sqrt{b x^2+c x^4}}+\frac{(6 b B-5 A c) \int \frac{x^6}{\sqrt{b x^2+c x^4}} \, dx}{b c}\\ &=-\frac{(b B-A c) x^7}{b c \sqrt{b x^2+c x^4}}+\frac{(6 b B-5 A c) x^3 \sqrt{b x^2+c x^4}}{5 b c^2}-\frac{(4 (6 b B-5 A c)) \int \frac{x^4}{\sqrt{b x^2+c x^4}} \, dx}{5 c^2}\\ &=-\frac{(b B-A c) x^7}{b c \sqrt{b x^2+c x^4}}-\frac{4 (6 b B-5 A c) x \sqrt{b x^2+c x^4}}{15 c^3}+\frac{(6 b B-5 A c) x^3 \sqrt{b x^2+c x^4}}{5 b c^2}+\frac{(8 b (6 b B-5 A c)) \int \frac{x^2}{\sqrt{b x^2+c x^4}} \, dx}{15 c^3}\\ &=-\frac{(b B-A c) x^7}{b c \sqrt{b x^2+c x^4}}+\frac{8 b (6 b B-5 A c) \sqrt{b x^2+c x^4}}{15 c^4 x}-\frac{4 (6 b B-5 A c) x \sqrt{b x^2+c x^4}}{15 c^3}+\frac{(6 b B-5 A c) x^3 \sqrt{b x^2+c x^4}}{5 b c^2}\\ \end{align*}

Mathematica [A]  time = 0.0508943, size = 82, normalized size = 0.59 \[ \frac{x \left (-8 b^2 c \left (5 A-3 B x^2\right )-2 b c^2 x^2 \left (10 A+3 B x^2\right )+c^3 x^4 \left (5 A+3 B x^2\right )+48 b^3 B\right )}{15 c^4 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(48*b^3*B - 8*b^2*c*(5*A - 3*B*x^2) + c^3*x^4*(5*A + 3*B*x^2) - 2*b*c^2*x^2*(10*A + 3*B*x^2)))/(15*c^4*Sqrt
[x^2*(b + c*x^2)])

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Maple [A]  time = 0.005, size = 91, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -3\,B{c}^{3}{x}^{6}-5\,A{x}^{4}{c}^{3}+6\,B{x}^{4}b{c}^{2}+20\,A{x}^{2}b{c}^{2}-24\,B{x}^{2}{b}^{2}c+40\,A{b}^{2}c-48\,B{b}^{3} \right ){x}^{3}}{15\,{c}^{4}} \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^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/15*(c*x^2+b)*(-3*B*c^3*x^6-5*A*c^3*x^4+6*B*b*c^2*x^4+20*A*b*c^2*x^2-24*B*b^2*c*x^2+40*A*b^2*c-48*B*b^3)*x^3
/c^4/(c*x^4+b*x^2)^(3/2)

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Maxima [A]  time = 1.19335, size = 111, normalized size = 0.8 \begin{align*} \frac{{\left (c^{2} x^{4} - 4 \, b c x^{2} - 8 \, b^{2}\right )} A}{3 \, \sqrt{c x^{2} + b} c^{3}} + \frac{{\left (c^{3} x^{6} - 2 \, b c^{2} x^{4} + 8 \, b^{2} c x^{2} + 16 \, b^{3}\right )} B}{5 \, \sqrt{c x^{2} + b} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.36381, size = 194, normalized size = 1.4 \begin{align*} \frac{{\left (3 \, B c^{3} x^{6} -{\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 48 \, B b^{3} - 40 \, A b^{2} c + 4 \,{\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15 \,{\left (c^{5} x^{3} + b c^{4} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*B*c^3*x^6 - (6*B*b*c^2 - 5*A*c^3)*x^4 + 48*B*b^3 - 40*A*b^2*c + 4*(6*B*b^2*c - 5*A*b*c^2)*x^2)*sqrt(c*
x^4 + b*x^2)/(c^5*x^3 + b*c^4*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8} \left (A + B x^{2}\right )}{\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**8*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**8*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{8}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^8/(c*x^4 + b*x^2)^(3/2), x)