3.2.36 \(\int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 c \sqrt {b x^2+c x^4} (5 b B-4 A c)}{15 b^3 x^2}-\frac {\sqrt {b x^2+c x^4} (5 b B-4 A c)}{15 b^2 x^4}-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6} \]

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Rubi [A]  time = 0.21, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \begin {gather*} \frac {2 c \sqrt {b x^2+c x^4} (5 b B-4 A c)}{15 b^3 x^2}-\frac {\sqrt {b x^2+c x^4} (5 b B-4 A c)}{15 b^2 x^4}-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(5*b*x^6) - ((5*b*B - 4*A*c)*Sqrt[b*x^2 + c*x^4])/(15*b^2*x^4) + (2*c*(5*b*B - 4*A*c)
*Sqrt[b*x^2 + c*x^4])/(15*b^3*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 658

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))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], 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] && ILtQ[Simplify[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^5 \sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^3 \sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6}+\frac {\left (-3 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{5 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6}-\frac {(5 b B-4 A c) \sqrt {b x^2+c x^4}}{15 b^2 x^4}-\frac {(c (5 b B-4 A c)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{15 b^2}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6}-\frac {(5 b B-4 A c) \sqrt {b x^2+c x^4}}{15 b^2 x^4}+\frac {2 c (5 b B-4 A c) \sqrt {b x^2+c x^4}}{15 b^3 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 64, normalized size = 0.67 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (A \left (-3 b^2+4 b c x^2-8 c^2 x^4\right )-5 b B x^2 \left (b-2 c x^2\right )\right )}{15 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.34, size = 66, normalized size = 0.69 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 A b^2+4 A b c x^2-8 A c^2 x^4-5 b^2 B x^2+10 b B c x^4\right )}{15 b^3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^5*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*A*b^2 - 5*b^2*B*x^2 + 4*A*b*c*x^2 + 10*b*B*c*x^4 - 8*A*c^2*x^4))/(15*b^3*x^6)

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fricas [A]  time = 0.41, size = 62, normalized size = 0.65 \begin {gather*} \frac {{\left (2 \, {\left (5 \, B b c - 4 \, A c^{2}\right )} x^{4} - 3 \, A b^{2} - {\left (5 \, B b^{2} - 4 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 153, normalized size = 1.59 \begin {gather*} \frac {15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3} B \sqrt {c} + 5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} B b + 20 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} A c + 15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} A b \sqrt {c} + 3 \, A b^{2}}{15 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/15*(15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^3*B*sqrt(c) + 5*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*B*b + 20*(s
qrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*A*c + 15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*A*b*sqrt(c) + 3*A*b^2)/(sqrt(
c)*x^2 - sqrt(c*x^4 + b*x^2))^5

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maple [A]  time = 0.05, size = 70, normalized size = 0.73 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (8 A \,c^{2} x^{4}-10 B b c \,x^{4}-4 A b c \,x^{2}+5 B \,b^{2} x^{2}+3 b^{2} A \right )}{15 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/15*(c*x^2+b)*(8*A*c^2*x^4-10*B*b*c*x^4-4*A*b*c*x^2+5*B*b^2*x^2+3*A*b^2)/x^4/b^3/(c*x^4+b*x^2)^(1/2)

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maxima [A]  time = 1.49, size = 119, normalized size = 1.24 \begin {gather*} \frac {1}{3} \, B {\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 {1}{15} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{b x^{6}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*B*(2*sqrt(c*x^4 + b*x^2)*c/(b^2*x^2) - sqrt(c*x^4 + b*x^2)/(b*x^4)) - 1/15*A*(8*sqrt(c*x^4 + b*x^2)*c^2/(b
^3*x^2) - 4*sqrt(c*x^4 + b*x^2)*c/(b^2*x^4) + 3*sqrt(c*x^4 + b*x^2)/(b*x^6))

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mupad [B]  time = 0.25, size = 62, normalized size = 0.65 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (5\,B\,b^2\,x^2+3\,A\,b^2-10\,B\,b\,c\,x^4-4\,A\,b\,c\,x^2+8\,A\,c^2\,x^4\right )}{15\,b^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

-((b*x^2 + c*x^4)^(1/2)*(3*A*b^2 + 5*B*b^2*x^2 + 8*A*c^2*x^4 - 4*A*b*c*x^2 - 10*B*b*c*x^4))/(15*b^3*x^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x^{2}}{x^{5} \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**5/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral((A + B*x**2)/(x**5*sqrt(x**2*(b + c*x**2))), x)

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