3.152 \(\int \frac{A+B x^2}{x^5 (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{8 c^2 \left (b+2 c x^2\right ) (7 b B-8 A c)}{35 b^5 \sqrt{b x^2+c x^4}}+\frac{2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt{b x^2+c x^4}}-\frac{7 b B-8 A c}{35 b^2 x^4 \sqrt{b x^2+c x^4}}-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}} \]

[Out]

-A/(7*b*x^6*Sqrt[b*x^2 + c*x^4]) - (7*b*B - 8*A*c)/(35*b^2*x^4*Sqrt[b*x^2 + c*x^4]) + (2*c*(7*b*B - 8*A*c))/(3
5*b^3*x^2*Sqrt[b*x^2 + c*x^4]) - (8*c^2*(7*b*B - 8*A*c)*(b + 2*c*x^2))/(35*b^5*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.266861, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 613} \[ -\frac{8 c^2 \left (b+2 c x^2\right ) (7 b B-8 A c)}{35 b^5 \sqrt{b x^2+c x^4}}+\frac{2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt{b x^2+c x^4}}-\frac{7 b B-8 A c}{35 b^2 x^4 \sqrt{b x^2+c x^4}}-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

-A/(7*b*x^6*Sqrt[b*x^2 + c*x^4]) - (7*b*B - 8*A*c)/(35*b^2*x^4*Sqrt[b*x^2 + c*x^4]) + (2*c*(7*b*B - 8*A*c))/(3
5*b^3*x^2*Sqrt[b*x^2 + c*x^4]) - (8*c^2*(7*b*B - 8*A*c)*(b + 2*c*x^2))/(35*b^5*Sqrt[b*x^2 + c*x^4])

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}}+\frac{\left (\frac{1}{2} (b B-2 A c)-3 (-b B+A c)\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{7 b}\\ &=-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}}-\frac{7 b B-8 A c}{35 b^2 x^4 \sqrt{b x^2+c x^4}}-\frac{(3 c (7 b B-8 A c)) \operatorname{Subst}\left (\int \frac{1}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{35 b^2}\\ &=-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}}-\frac{7 b B-8 A c}{35 b^2 x^4 \sqrt{b x^2+c x^4}}+\frac{2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt{b x^2+c x^4}}+\frac{\left (4 c^2 (7 b B-8 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{35 b^3}\\ &=-\frac{A}{7 b x^6 \sqrt{b x^2+c x^4}}-\frac{7 b B-8 A c}{35 b^2 x^4 \sqrt{b x^2+c x^4}}+\frac{2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt{b x^2+c x^4}}-\frac{8 c^2 (7 b B-8 A c) \left (b+2 c x^2\right )}{35 b^5 \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0340571, size = 75, normalized size = 0.54 \[ \frac{x^2 \left (-2 b^2 c x^2+b^3+8 b c^2 x^4+16 c^3 x^6\right ) (8 A c-7 b B)-5 A b^4}{35 b^5 x^6 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-5*A*b^4 + (-7*b*B + 8*A*c)*x^2*(b^3 - 2*b^2*c*x^2 + 8*b*c^2*x^4 + 16*c^3*x^6))/(35*b^5*x^6*Sqrt[x^2*(b + c*x
^2)])

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Maple [A]  time = 0.006, size = 118, normalized size = 0.9 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -128\,A{c}^{4}{x}^{8}+112\,Bb{c}^{3}{x}^{8}-64\,Ab{c}^{3}{x}^{6}+56\,B{b}^{2}{c}^{2}{x}^{6}+16\,A{b}^{2}{c}^{2}{x}^{4}-14\,B{b}^{3}c{x}^{4}-8\,A{b}^{3}c{x}^{2}+7\,B{b}^{4}{x}^{2}+5\,A{b}^{4} \right ) }{35\,{x}^{4}{b}^{5}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/35*(c*x^2+b)*(-128*A*c^4*x^8+112*B*b*c^3*x^8-64*A*b*c^3*x^6+56*B*b^2*c^2*x^6+16*A*b^2*c^2*x^4-14*B*b^3*c*x^
4-8*A*b^3*c*x^2+7*B*b^4*x^2+5*A*b^4)/x^4/b^5/(c*x^4+b*x^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5006, size = 252, normalized size = 1.83 \begin{align*} -\frac{{\left (16 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{8} + 8 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{6} + 5 \, A b^{4} - 2 \,{\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{4} +{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{35 \,{\left (b^{5} c x^{10} + b^{6} x^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/35*(16*(7*B*b*c^3 - 8*A*c^4)*x^8 + 8*(7*B*b^2*c^2 - 8*A*b*c^3)*x^6 + 5*A*b^4 - 2*(7*B*b^3*c - 8*A*b^2*c^2)*
x^4 + (7*B*b^4 - 8*A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*c*x^10 + b^6*x^8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x^{2}}{x^{5} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral((A + B*x**2)/(x**5*(x**2*(b + c*x**2))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(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^5), x)