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

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

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Rubi [A]  time = 0.25, antiderivative size = 139, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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 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])

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*}

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Mathematica [A]  time = 0.08, size = 82, normalized size = 0.59 \begin {gather*} \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 )}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.03, size = 96, normalized size = 0.69 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-40 A b^2 c-20 A b c^2 x^2+5 A c^3 x^4+48 b^3 B+24 b^2 B c x^2-6 b B c^2 x^4+3 B c^3 x^6\right )}{15 c^4 x \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 93, normalized size = 0.67 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{2} + A\right )} x^{8}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.05, size = 91, normalized size = 0.65 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-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}\right ) x^{3}}{15 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{4}} \end {gather*}

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 = 1.58, size = 82, normalized size = 0.59 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.40, size = 92, normalized size = 0.66 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (48\,B\,b^3+24\,B\,b^2\,c\,x^2-40\,A\,b^2\,c-6\,B\,b\,c^2\,x^4-20\,A\,b\,c^2\,x^2+3\,B\,c^3\,x^6+5\,A\,c^3\,x^4\right )}{15\,c^4\,x\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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