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

Optimal. Leaf size=61 \[ -\frac {\sqrt {b x^2+c x^4} (3 b B-2 A c)}{3 b^2 x^2}-\frac {A \sqrt {b x^2+c x^4}}{3 b x^4} \]

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Rubi [A]  time = 0.17, antiderivative size = 61, 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, 650} \begin {gather*} -\frac {\sqrt {b x^2+c x^4} (3 b B-2 A c)}{3 b^2 x^2}-\frac {A \sqrt {b x^2+c x^4}}{3 b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 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^3 \sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^2 \sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {A \sqrt {b x^2+c x^4}}{3 b x^4}+\frac {\left (-2 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{3 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{3 b x^4}-\frac {(3 b B-2 A c) \sqrt {b x^2+c x^4}}{3 b^2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (A \left (b-2 c x^2\right )+3 b B x^2\right )}{3 b^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 44, normalized size = 0.72 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-A b+2 A c x^2-3 b B x^2\right )}{3 b^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 38, normalized size = 0.62 \begin {gather*} -\frac {\sqrt {c x^{4} + b x^{2}} {\left ({\left (3 \, B b - 2 \, A c\right )} x^{2} + A b\right )}}{3 \, b^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 88, normalized size = 1.44 \begin {gather*} \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} B + 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} A \sqrt {c} + A b}{3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 47, normalized size = 0.77 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+3 B b \,x^{2}+A b \right )}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.47, size = 70, normalized size = 1.15 \begin {gather*} \frac {1}{3} \, A {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}}}{b x^{4}}\right )} - \frac {\sqrt {c x^{4} + b x^{2}} B}{b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.20, size = 39, normalized size = 0.64 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (A\,b-2\,A\,c\,x^2+3\,B\,b\,x^2\right )}{3\,b^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**3/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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