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

Optimal. Leaf size=66 \[ -\frac {\left (b+2 c x^2\right ) (3 b B-4 A c)}{3 b^3 \sqrt {b x^2+c x^4}}-\frac {A}{3 b x^2 \sqrt {b x^2+c x^4}} \]

[Out]

-1/3*A/b/x^2/(c*x^4+b*x^2)^(1/2)-1/3*(-4*A*c+3*B*b)*(2*c*x^2+b)/b^3/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2034, 792, 613} \[ -\frac {\left (b+2 c x^2\right ) (3 b B-4 A c)}{3 b^3 \sqrt {b x^2+c x^4}}-\frac {A}{3 b x^2 \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {A}{3 b x^2 \sqrt {b x^2+c x^4}}+\frac {\left (b B-A c+\frac {1}{2} (b B-2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{3 b}\\ &=-\frac {A}{3 b x^2 \sqrt {b x^2+c x^4}}-\frac {(3 b B-4 A c) \left (b+2 c x^2\right )}{3 b^3 \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.97 \[ \frac {A \left (-b^2+4 b c x^2+8 c^2 x^4\right )-3 b B x^2 \left (b+2 c x^2\right )}{3 b^3 x^2 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.91, size = 72, normalized size = 1.09 \[ -\frac {{\left (2 \, {\left (3 \, B b c - 4 \, A c^{2}\right )} x^{4} + A b^{2} + {\left (3 \, B b^{2} - 4 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 66, normalized size = 1.00 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-8 A \,c^{2} x^{4}+6 B b c \,x^{4}-4 A b c \,x^{2}+3 B \,b^{2} x^{2}+b^{2} A \right )}{3 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.39, size = 112, normalized size = 1.70 \[ -B {\left (\frac {2 \, c x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b}\right )} + \frac {1}{3} \, A {\left (\frac {8 \, c^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 70, normalized size = 1.06 \[ -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (3\,B\,b^2\,x^2+A\,b^2+6\,B\,b\,c\,x^4-4\,A\,b\,c\,x^2-8\,A\,c^2\,x^4\right )}{3\,b^3\,x^4\,\left (c\,x^2+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x^{2}}{x \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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