3.114 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {\left (b x^2+c x^4\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^{10}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}} \]

[Out]

-1/7*A*(c*x^4+b*x^2)^(5/2)/b/x^12-1/35*(-2*A*c+7*B*b)*(c*x^4+b*x^2)^(5/2)/b^2/x^10

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Rubi [A]  time = 0.17, antiderivative size = 61, normalized size of antiderivative = 1.00, 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} \[ -\frac {\left (b x^2+c x^4\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^{10}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^11,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(7*b*x^12) - ((7*b*B - 2*A*c)*(b*x^2 + c*x^4)^(5/2))/(35*b^2*x^10)

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 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 ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}+\frac {\left (-6 (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{7 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}-\frac {(7 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.72 \[ -\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (5 A b-2 A c x^2+7 b B x^2\right )}{35 b^2 x^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^11,x]

[Out]

-1/35*((x^2*(b + c*x^2))^(5/2)*(5*A*b + 7*b*B*x^2 - 2*A*c*x^2))/(b^2*x^12)

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fricas [A]  time = 1.07, size = 82, normalized size = 1.34 \[ -\frac {{\left ({\left (7 \, B b c^{2} - 2 \, A c^{3}\right )} x^{6} + {\left (14 \, B b^{2} c + A b c^{2}\right )} x^{4} + 5 \, A b^{3} + {\left (7 \, B b^{3} + 8 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, b^{2} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="fricas")

[Out]

-1/35*((7*B*b*c^2 - 2*A*c^3)*x^6 + (14*B*b^2*c + A*b*c^2)*x^4 + 5*A*b^3 + (7*B*b^3 + 8*A*b^2*c)*x^2)*sqrt(c*x^
4 + b*x^2)/(b^2*x^8)

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giac [B]  time = 1.89, size = 370, normalized size = 6.07 \[ \frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{2} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b c^{\frac {7}{2}} \mathrm {sgn}\relax (x) - 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{3} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{2} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 77 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{4} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{3} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) - 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{5} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) + 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{4} c^{\frac {7}{2}} \mathrm {sgn}\relax (x) + 7 \, B b^{6} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 2 \, A b^{5} c^{\frac {7}{2}} \mathrm {sgn}\relax (x)\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="giac")

[Out]

2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*c^(5/2)*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b*c^(5/2)*s
gn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*c^(7/2)*sgn(x) + 105*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^2*c^(5/
2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b*c^(7/2)*sgn(x) - 140*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^3*
c^(5/2)*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^2*c^(7/2)*sgn(x) + 77*(sqrt(c)*x - sqrt(c*x^2 + b))^4
*B*b^4*c^(5/2)*sgn(x) + 28*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^3*c^(7/2)*sgn(x) - 14*(sqrt(c)*x - sqrt(c*x^2 +
b))^2*B*b^5*c^(5/2)*sgn(x) + 14*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^4*c^(7/2)*sgn(x) + 7*B*b^6*c^(5/2)*sgn(x)
- 2*A*b^5*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7

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maple [A]  time = 0.05, size = 48, normalized size = 0.79 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+7 B b \,x^{2}+5 A b \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^11,x)

[Out]

-1/35*(c*x^2+b)*(-2*A*c*x^2+7*B*b*x^2+5*A*b)*(c*x^4+b*x^2)^(3/2)/b^2/x^10

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maxima [B]  time = 1.56, size = 193, normalized size = 3.16 \[ -\frac {1}{10} \, B {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{x^{6}} + \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\right )} + \frac {1}{140} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{x^{6}} + \frac {15 \, \sqrt {c x^{4} + b x^{2}} b}{x^{8}} - \frac {35 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{10}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^11,x, algorithm="maxima")

[Out]

-1/10*B*(2*sqrt(c*x^4 + b*x^2)*c^2/(b*x^2) - sqrt(c*x^4 + b*x^2)*c/x^4 - 3*sqrt(c*x^4 + b*x^2)*b/x^6 + 5*(c*x^
4 + b*x^2)^(3/2)/x^8) + 1/140*A*(8*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^2) - 4*sqrt(c*x^4 + b*x^2)*c^2/(b*x^4) + 3*s
qrt(c*x^4 + b*x^2)*c/x^6 + 15*sqrt(c*x^4 + b*x^2)*b/x^8 - 35*(c*x^4 + b*x^2)^(3/2)/x^10)

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mupad [B]  time = 1.03, size = 156, normalized size = 2.56 \[ \frac {2\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^2}-\frac {8\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,x^6}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {2\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{5\,x^4}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^4}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{5\,b\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^11,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{11}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**11,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**11, x)

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