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]

-1/7*A*(c*x^4+b*x^2)^(3/2)/b/x^10-1/35*(-4*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/b^2/x^8+2/105*c*(-4*A*c+7*B*b)*(c*x^
4+b*x^2)^(3/2)/b^3/x^6

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Rubi [A]  time = 0.21, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 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 {\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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 1.02, size = 85, normalized size = 0.89 \[ \frac {{\left (2 \, {\left (7 \, B b c^{2} - 4 \, A c^{3}\right )} x^{6} - {\left (7 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{4} - 15 \, A b^{3} - 3 \, {\left (7 \, B b^{3} + A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, b^{3} x^{8}} \]

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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giac [B]  time = 1.36, size = 310, normalized size = 3.23 \[ \frac {4 \, {\left (105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 175 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b c^{\frac {7}{2}} \mathrm {sgn}\relax (x) - 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 49 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{3} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) - 7 \, B b^{5} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 4 \, A b^{4} c^{\frac {7}{2}} \mathrm {sgn}\relax (x)\right )}}{105 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7}} \]

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

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maple [A]  time = 0.04, size = 70, normalized size = 0.73 \[ -\frac {\left (c \,x^{2}+b \right ) \left (8 A \,c^{2} x^{4}-14 B b c \,x^{4}-12 A b c \,x^{2}+21 B \,b^{2} x^{2}+15 b^{2} A \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 b^{3} x^{8}} \]

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)/b^3/x^8

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maxima [A]  time = 1.48, size = 161, normalized size = 1.68 \[ \frac {1}{15} \, B {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{x^{6}}\right )} - \frac {1}{105} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{3} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{b x^{6}} + \frac {15 \, \sqrt {c x^{4} + b x^{2}}}{x^{8}}\right )} \]

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]

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

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mupad [B]  time = 0.68, size = 160, normalized size = 1.67 \[ \frac {4\,A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^4}-\frac {B\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {A\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^6}-\frac {B\,c\,\sqrt {c\,x^4+b\,x^2}}{15\,b\,x^4}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {8\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^2}+\frac {2\,B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*A*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^4) - (B*(b*x^2 + c*x^4)^(1/2))/(5*x^6) - (A*c*(b*x^2 + c*x^4)^(1/2)
)/(35*b*x^6) - (B*c*(b*x^2 + c*x^4)^(1/2))/(15*b*x^4) - (A*(b*x^2 + c*x^4)^(1/2))/(7*x^8) - (8*A*c^3*(b*x^2 +
c*x^4)^(1/2))/(105*b^3*x^2) + (2*B*c^2*(b*x^2 + c*x^4)^(1/2))/(15*b^2*x^2)

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

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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