3.97 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x^9} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 c \left (b x^2+c x^4\right )^{3/2} (7 b B-4 A c)}{105 b^3 x^6}-\frac{\left (b x^2+c x^4\right )^{3/2} (7 b B-4 A c)}{35 b^2 x^8}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}} \]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^10) - ((7*b*B - 4*A*c)*(b*x^2 + c*x^4)^(3/2))/(35*b^2*x^8) + (2*c*(7*b*B - 4
*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*b^3*x^6)

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Rubi [A]  time = 0.21089, antiderivative size = 96, normalized size of antiderivative = 1., 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} \[ \frac{2 c \left (b x^2+c x^4\right )^{3/2} (7 b B-4 A c)}{105 b^3 x^6}-\frac{\left (b x^2+c x^4\right )^{3/2} (7 b B-4 A c)}{35 b^2 x^8}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^9,x]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^10) - ((7*b*B - 4*A*c)*(b*x^2 + c*x^4)^(3/2))/(35*b^2*x^8) + (2*c*(7*b*B - 4
*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*b^3*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 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 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^9} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^5} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}+\frac{\left (-5 (-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^4} \, dx,x,x^2\right )}{7 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}-\frac{(7 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}-\frac{(c (7 b B-4 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x^3} \, dx,x,x^2\right )}{35 b^2}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}-\frac{(7 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}+\frac{2 c (7 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0247352, size = 66, normalized size = 0.69 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (A \left (-15 b^2+12 b c x^2-8 c^2 x^4\right )+7 b B x^2 \left (2 c x^2-3 b\right )\right )}{105 b^3 x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^9,x]

[Out]

((x^2*(b + c*x^2))^(3/2)*(7*b*B*x^2*(-3*b + 2*c*x^2) + A*(-15*b^2 + 12*b*c*x^2 - 8*c^2*x^4)))/(105*b^3*x^10)

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Maple [A]  time = 0.005, size = 70, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( 8\,A{c}^{2}{x}^{4}-14\,B{x}^{4}bc-12\,Abc{x}^{2}+21\,B{x}^{2}{b}^{2}+15\,A{b}^{2} \right ) }{105\,{x}^{8}{b}^{3}}\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^9,x)

[Out]

-1/105*(c*x^2+b)*(8*A*c^2*x^4-14*B*b*c*x^4-12*A*b*c*x^2+21*B*b^2*x^2+15*A*b^2)*(c*x^4+b*x^2)^(1/2)/x^8/b^3

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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^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.11355, size = 184, normalized size = 1.92 \begin{align*} \frac{{\left (2 \,{\left (7 \, B b c^{2} - 4 \, A c^{3}\right )} x^{6} -{\left (7 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{4} - 15 \, A b^{3} - 3 \,{\left (7 \, B b^{3} + A b^{2} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{105 \, b^{3} x^{8}} \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^9,x, algorithm="fricas")

[Out]

1/105*(2*(7*B*b*c^2 - 4*A*c^3)*x^6 - (7*B*b^2*c - 4*A*b*c^2)*x^4 - 15*A*b^3 - 3*(7*B*b^3 + A*b^2*c)*x^2)*sqrt(
c*x^4 + b*x^2)/(b^3*x^8)

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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^{9}}\, 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**9,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**9, x)

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Giac [B]  time = 2.31339, size = 419, normalized size = 4.36 \begin{align*} \frac{4 \,{\left (105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} B c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 175 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B b c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 280 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} A c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b^{2} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} A b c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 42 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{3} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 84 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b^{2} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 49 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{4} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 28 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{3} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 7 \, B b^{5} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 4 \, A b^{4} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right )\right )}}{105 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{7}} \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^9,x, algorithm="giac")

[Out]

4/105*(105*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*c^(5/2)*sgn(x) - 175*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b*c^(5/2)
*sgn(x) + 280*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*c^(7/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^(5
/2)*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b*c^(7/2)*sgn(x) - 42*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^3
*c^(5/2)*sgn(x) + 84*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^2*c^(7/2)*sgn(x) + 49*(sqrt(c)*x - sqrt(c*x^2 + b))^2
*B*b^4*c^(5/2)*sgn(x) - 28*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^3*c^(7/2)*sgn(x) - 7*B*b^5*c^(5/2)*sgn(x) + 4*A
*b^4*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7