3.96 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x^7} \, dx\)

Optimal. Leaf size=61 \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^6}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \]

[Out]

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

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Rubi [A]  time = 0.161813, antiderivative size = 61, normalized size of antiderivative = 1., 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} \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^6}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]

[Out]

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \sqrt{b x^2+c x^4}}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}+\frac{\left (-4 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x^3} \, dx,x,x^2\right )}{5 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}-\frac{(5 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0176179, size = 44, normalized size = 0.72 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (3 A b-2 A c x^2+5 b B x^2\right )}{15 b^2 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]

[Out]

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

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Maple [A]  time = 0.006, size = 48, normalized size = 0.8 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -2\,A{x}^{2}c+5\,B{x}^{2}b+3\,Ab \right ) }{15\,{b}^{2}{x}^{6}}\sqrt{c{x}^{4}+b{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x)

[Out]

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

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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)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.919044, size = 131, normalized size = 2.15 \begin{align*} -\frac{{\left ({\left (5 \, B b c - 2 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} +{\left (5 \, B b^{2} + A b c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15 \, b^{2} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{7}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**7,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**7, x)

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Giac [B]  time = 1.8069, size = 338, normalized size = 5.54 \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) - 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} A c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 20 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{2} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{3} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 10 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{2} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 5 \, B b^{4} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) - 2 \, A b^{3} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right )\right )}}{15 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*c^(3/2)*sgn(x) - 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b*c^(3/2)*sgn
(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*c^(5/2)*sgn(x) + 20*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^2*c^(3/2)*s
gn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b*c^(5/2)*sgn(x) - 10*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^3*c^(3/
2)*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^2*c^(5/2)*sgn(x) + 5*B*b^4*c^(3/2)*sgn(x) - 2*A*b^3*c^(5/2)
*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5