3.98 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x^{11}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{315 b^4 x^6}+\frac{4 c \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{105 b^3 x^8}-\frac{\left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{21 b^2 x^{10}}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}} \]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(9*b*x^12) - ((3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(21*b^2*x^10) + (4*c*(3*b*B -
2*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*b^3*x^8) - (8*c^2*(3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*b^4*x^6)

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Rubi [A]  time = 0.244999, antiderivative size = 133, 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, 650} \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{315 b^4 x^6}+\frac{4 c \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{105 b^3 x^8}-\frac{\left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{21 b^2 x^{10}}-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^11,x]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(9*b*x^12) - ((3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(21*b^2*x^10) + (4*c*(3*b*B -
2*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*b^3*x^8) - (8*c^2*(3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*b^4*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^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac{\left (-6 (-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^5} \, dx,x,x^2\right )}{9 b}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac{(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac{(2 c (3 b B-2 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x^4} \, dx,x,x^2\right )}{21 b^2}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac{(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac{4 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac{\left (4 c^2 (3 b B-2 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x^3} \, dx,x,x^2\right )}{105 b^3}\\ &=-\frac{A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac{(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac{4 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}-\frac{8 c^2 (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0294505, size = 88, normalized size = 0.66 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (A \left (-30 b^2 c x^2+35 b^3+24 b c^2 x^4-16 c^3 x^6\right )+3 b B x^2 \left (15 b^2-12 b c x^2+8 c^2 x^4\right )\right )}{315 b^4 x^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^11,x]

[Out]

-((x^2*(b + c*x^2))^(3/2)*(3*b*B*x^2*(15*b^2 - 12*b*c*x^2 + 8*c^2*x^4) + A*(35*b^3 - 30*b^2*c*x^2 + 24*b*c^2*x
^4 - 16*c^3*x^6)))/(315*b^4*x^12)

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Maple [A]  time = 0.006, size = 94, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -16\,A{c}^{3}{x}^{6}+24\,B{x}^{6}b{c}^{2}+24\,Ab{c}^{2}{x}^{4}-36\,B{x}^{4}{b}^{2}c-30\,A{b}^{2}c{x}^{2}+45\,B{x}^{2}{b}^{3}+35\,A{b}^{3} \right ) }{315\,{x}^{10}{b}^{4}}\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^11,x)

[Out]

-1/315*(c*x^2+b)*(-16*A*c^3*x^6+24*B*b*c^2*x^6+24*A*b*c^2*x^4-36*B*b^2*c*x^4-30*A*b^2*c*x^2+45*B*b^3*x^2+35*A*
b^3)*(c*x^4+b*x^2)^(1/2)/x^10/b^4

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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^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35793, size = 238, normalized size = 1.79 \begin{align*} -\frac{{\left (8 \,{\left (3 \, B b c^{3} - 2 \, A c^{4}\right )} x^{8} - 4 \,{\left (3 \, B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + 35 \, A b^{4} + 3 \,{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4} + 5 \,{\left (9 \, B b^{4} + A b^{3} c\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, b^{4} x^{10}} \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^11,x, algorithm="fricas")

[Out]

-1/315*(8*(3*B*b*c^3 - 2*A*c^4)*x^8 - 4*(3*B*b^2*c^2 - 2*A*b*c^3)*x^6 + 35*A*b^4 + 3*(3*B*b^3*c - 2*A*b^2*c^2)
*x^4 + 5*(9*B*b^4 + A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^4*x^10)

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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^{11}}\, 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**11,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**11, x)

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Giac [B]  time = 3.13908, size = 500, normalized size = 3.76 \begin{align*} \frac{16 \,{\left (210 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} B c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} B b c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 630 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} A c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B b^{2} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 378 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} A b c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 42 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b^{3} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 168 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} A b^{2} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 108 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{4} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 72 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b^{3} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 27 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{5} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 18 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{4} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 3 \, B b^{6} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 2 \, A b^{5} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right )\right )}}{315 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{9}} \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^11,x, algorithm="giac")

[Out]

16/315*(210*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*c^(7/2)*sgn(x) - 315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b*c^(7/
2)*sgn(x) + 630*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*c^(9/2)*sgn(x) + 63*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^2*c
^(7/2)*sgn(x) + 378*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b*c^(9/2)*sgn(x) - 42*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*
b^3*c^(7/2)*sgn(x) + 168*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^2*c^(9/2)*sgn(x) + 108*(sqrt(c)*x - sqrt(c*x^2 +
b))^4*B*b^4*c^(7/2)*sgn(x) - 72*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^3*c^(9/2)*sgn(x) - 27*(sqrt(c)*x - sqrt(c*
x^2 + b))^2*B*b^5*c^(7/2)*sgn(x) + 18*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^4*c^(9/2)*sgn(x) + 3*B*b^6*c^(7/2)*s
gn(x) - 2*A*b^5*c^(9/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9