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

Optimal. Leaf size=101 \[ \frac {4 c \left (b+2 c x^2\right ) (5 b B-6 A c)}{15 b^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.22, antiderivative size = 101, 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, 613} \begin {gather*} \frac {4 c \left (b+2 c x^2\right ) (5 b B-6 A c)}{15 b^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(5*b*x^4*Sqrt[b*x^2 + c*x^4]) - (5*b*B - 6*A*c)/(15*b^2*x^2*Sqrt[b*x^2 + c*x^4]) + (4*c*(5*b*B - 6*A*c)*(b
+ 2*c*x^2))/(15*b^4*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 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^3 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}+\frac {\left (\frac {1}{2} (b B-2 A c)-2 (-b B+A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{5 b}\\ &=-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {(2 c (5 b B-6 A c)) \operatorname {Subst}\left (\int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{15 b^2}\\ &=-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}+\frac {4 c (5 b B-6 A c) \left (b+2 c x^2\right )}{15 b^4 \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 85, normalized size = 0.84 \begin {gather*} \frac {-3 A \left (b^3-2 b^2 c x^2+8 b c^2 x^4+16 c^3 x^6\right )-5 b B x^2 \left (b^2-4 b c x^2-8 c^2 x^4\right )}{15 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*B*x^2*(b^2 - 4*b*c*x^2 - 8*c^2*x^4) - 3*A*(b^3 - 2*b^2*c*x^2 + 8*b*c^2*x^4 + 16*c^3*x^6))/(15*b^4*x^4*Sq
rt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.42, size = 99, normalized size = 0.98 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 A b^3+6 A b^2 c x^2-24 A b c^2 x^4-48 A c^3 x^6-5 b^3 B x^2+20 b^2 B c x^4+40 b B c^2 x^6\right )}{15 b^4 x^6 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 98, normalized size = 0.97 \begin {gather*} \frac {{\left (8 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{6} + 4 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{4} - 3 \, A b^{3} - {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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maple [A]  time = 0.05, size = 94, normalized size = 0.93 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (48 A \,c^{3} x^{6}-40 B b \,c^{2} x^{6}+24 A b \,c^{2} x^{4}-20 B \,b^{2} c \,x^{4}-6 A \,b^{2} c \,x^{2}+5 B \,b^{3} x^{2}+3 A \,b^{3}\right )}{15 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{4} 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)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.50, size = 160, normalized size = 1.58 \begin {gather*} \frac {1}{3} \, B {\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 )} - \frac {1}{5} \, A {\left (\frac {16 \, c^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{4}} + \frac {8 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} - \frac {2 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{4}}\right )} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.42, size = 95, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (5\,B\,b^3\,x^2+3\,A\,b^3-20\,B\,b^2\,c\,x^4-6\,A\,b^2\,c\,x^2-40\,B\,b\,c^2\,x^6+24\,A\,b\,c^2\,x^4+48\,A\,c^3\,x^6\right )}{15\,b^4\,x^6\,\left (c\,x^2+b\right )} \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)^(3/2)),x)

[Out]

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

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

[Out]

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

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