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

Optimal. Leaf size=170 \[ \frac {16 c^3 \sqrt {b x^2+c x^4} (9 b B-8 A c)}{315 b^5 x^2}-\frac {8 c^2 \sqrt {b x^2+c x^4} (9 b B-8 A c)}{315 b^4 x^4}+\frac {2 c \sqrt {b x^2+c x^4} (9 b B-8 A c)}{105 b^3 x^6}-\frac {\sqrt {b x^2+c x^4} (9 b B-8 A c)}{63 b^2 x^8}-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}} \]

[Out]

-1/9*A*(c*x^4+b*x^2)^(1/2)/b/x^10-1/63*(-8*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^8+2/105*c*(-8*A*c+9*B*b)*(c*x^
4+b*x^2)^(1/2)/b^3/x^6-8/315*c^2*(-8*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b^4/x^4+16/315*c^3*(-8*A*c+9*B*b)*(c*x^4+b
*x^2)^(1/2)/b^5/x^2

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Rubi [A]  time = 0.30, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 792, 658, 650} \[ \frac {16 c^3 \sqrt {b x^2+c x^4} (9 b B-8 A c)}{315 b^5 x^2}-\frac {8 c^2 \sqrt {b x^2+c x^4} (9 b B-8 A c)}{315 b^4 x^4}+\frac {2 c \sqrt {b x^2+c x^4} (9 b B-8 A c)}{105 b^3 x^6}-\frac {\sqrt {b x^2+c x^4} (9 b B-8 A c)}{63 b^2 x^8}-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(9*b*x^10) - ((9*b*B - 8*A*c)*Sqrt[b*x^2 + c*x^4])/(63*b^2*x^8) + (2*c*(9*b*B - 8*A*c
)*Sqrt[b*x^2 + c*x^4])/(105*b^3*x^6) - (8*c^2*(9*b*B - 8*A*c)*Sqrt[b*x^2 + c*x^4])/(315*b^4*x^4) + (16*c^3*(9*
b*B - 8*A*c)*Sqrt[b*x^2 + c*x^4])/(315*b^5*x^2)

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]

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^9 \sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^5 \sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}+\frac {\left (-5 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{9 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}-\frac {(9 b B-8 A c) \sqrt {b x^2+c x^4}}{63 b^2 x^8}-\frac {(c (9 b B-8 A c)) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{21 b^2}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}-\frac {(9 b B-8 A c) \sqrt {b x^2+c x^4}}{63 b^2 x^8}+\frac {2 c (9 b B-8 A c) \sqrt {b x^2+c x^4}}{105 b^3 x^6}+\frac {\left (4 c^2 (9 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{105 b^3}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}-\frac {(9 b B-8 A c) \sqrt {b x^2+c x^4}}{63 b^2 x^8}+\frac {2 c (9 b B-8 A c) \sqrt {b x^2+c x^4}}{105 b^3 x^6}-\frac {8 c^2 (9 b B-8 A c) \sqrt {b x^2+c x^4}}{315 b^4 x^4}-\frac {\left (8 c^3 (9 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b x+c x^2}} \, dx,x,x^2\right )}{315 b^4}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}-\frac {(9 b B-8 A c) \sqrt {b x^2+c x^4}}{63 b^2 x^8}+\frac {2 c (9 b B-8 A c) \sqrt {b x^2+c x^4}}{105 b^3 x^6}-\frac {8 c^2 (9 b B-8 A c) \sqrt {b x^2+c x^4}}{315 b^4 x^4}+\frac {16 c^3 (9 b B-8 A c) \sqrt {b x^2+c x^4}}{315 b^5 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 94, normalized size = 0.55 \[ \frac {x^2 \left (\frac {c x^2}{b}+1\right ) \left (5 b^3-6 b^2 c x^2+8 b c^2 x^4-16 c^3 x^6\right ) (8 A c-9 b B)-35 A b^3 \left (b+c x^2\right )}{315 b^4 x^8 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*A*b^3*(b + c*x^2) + (-9*b*B + 8*A*c)*x^2*(1 + (c*x^2)/b)*(5*b^3 - 6*b^2*c*x^2 + 8*b*c^2*x^4 - 16*c^3*x^6)
)/(315*b^4*x^8*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 1.24, size = 110, normalized size = 0.65 \[ \frac {{\left (16 \, {\left (9 \, B b c^{3} - 8 \, A c^{4}\right )} x^{8} - 8 \, {\left (9 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{6} - 35 \, A b^{4} + 6 \, {\left (9 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{4} - 5 \, {\left (9 \, B b^{4} - 8 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{5} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/315*(16*(9*B*b*c^3 - 8*A*c^4)*x^8 - 8*(9*B*b^2*c^2 - 8*A*b*c^3)*x^6 - 35*A*b^4 + 6*(9*B*b^3*c - 8*A*b^2*c^2)
*x^4 - 5*(9*B*b^4 - 8*A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*x^10)

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giac [A]  time = 0.27, size = 287, normalized size = 1.69 \[ \frac {630 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{5} B c^{\frac {3}{2}} + 756 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{4} B b c + 1008 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{4} A c^{2} + 315 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3} B b^{2} \sqrt {c} + 1680 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3} A b c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} B b^{3} + 1080 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{2} A b^{2} c + 315 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} A b^{3} \sqrt {c} + 35 \, A b^{4}}{315 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/315*(630*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^5*B*c^(3/2) + 756*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^4*B*b*c +
1008*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^4*A*c^2 + 315*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^3*B*b^2*sqrt(c) +
1680*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^3*A*b*c^(3/2) + 45*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*B*b^3 + 1080
*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^2*A*b^2*c + 315*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*A*b^3*sqrt(c) + 35*A*
b^4)/(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))^9

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maple [A]  time = 0.05, size = 118, normalized size = 0.69 \[ -\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{5} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/315*(c*x^2+b)*(128*A*c^4*x^8-144*B*b*c^3*x^8-64*A*b*c^3*x^6+72*B*b^2*c^2*x^6+48*A*b^2*c^2*x^4-54*B*b^3*c*x^
4-40*A*b^3*c*x^2+45*B*b^4*x^2+35*A*b^4)/x^8/b^5/(c*x^4+b*x^2)^(1/2)

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maxima [A]  time = 1.54, size = 215, normalized size = 1.26 \[ \frac {1}{35} \, B {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}}}{b x^{8}}\right )} - \frac {1}{315} \, A {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{5} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{4} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}}}{b x^{10}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/35*B*(16*sqrt(c*x^4 + b*x^2)*c^3/(b^4*x^2) - 8*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^4) + 6*sqrt(c*x^4 + b*x^2)*c/(
b^2*x^6) - 5*sqrt(c*x^4 + b*x^2)/(b*x^8)) - 1/315*A*(128*sqrt(c*x^4 + b*x^2)*c^4/(b^5*x^2) - 64*sqrt(c*x^4 + b
*x^2)*c^3/(b^4*x^4) + 48*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^6) - 40*sqrt(c*x^4 + b*x^2)*c/(b^2*x^8) + 35*sqrt(c*x^
4 + b*x^2)/(b*x^10))

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mupad [B]  time = 0.32, size = 156, normalized size = 0.92 \[ \frac {\left (8\,A\,c-9\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{63\,b^2\,x^8}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{9\,b\,x^{10}}-\frac {\left (16\,A\,c^2-18\,B\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^6}+\frac {\left (64\,A\,c^3-72\,B\,b\,c^2\right )\,\sqrt {c\,x^4+b\,x^2}}{315\,b^4\,x^4}-\frac {\left (128\,A\,c^4-144\,B\,b\,c^3\right )\,\sqrt {c\,x^4+b\,x^2}}{315\,b^5\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^9*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

((8*A*c - 9*B*b)*(b*x^2 + c*x^4)^(1/2))/(63*b^2*x^8) - (A*(b*x^2 + c*x^4)^(1/2))/(9*b*x^10) - ((16*A*c^2 - 18*
B*b*c)*(b*x^2 + c*x^4)^(1/2))/(105*b^3*x^6) + ((64*A*c^3 - 72*B*b*c^2)*(b*x^2 + c*x^4)^(1/2))/(315*b^4*x^4) -
((128*A*c^4 - 144*B*b*c^3)*(b*x^2 + c*x^4)^(1/2))/(315*b^5*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x^{2}}{x^{9} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**9/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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