3.116 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{15}} \, dx\)

Optimal. Leaf size=133 \[ -\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^{10}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^{12}}-\frac {\left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^{14}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

[Out]

-1/11*A*(c*x^4+b*x^2)^(5/2)/b/x^16-1/99*(-6*A*c+11*B*b)*(c*x^4+b*x^2)^(5/2)/b^2/x^14+4/693*c*(-6*A*c+11*B*b)*(
c*x^4+b*x^2)^(5/2)/b^3/x^12-8/3465*c^2*(-6*A*c+11*B*b)*(c*x^4+b*x^2)^(5/2)/b^4/x^10

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Rubi [A]  time = 0.28, antiderivative size = 133, normalized size of antiderivative = 1.00, 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} \[ -\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^{10}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^{12}}-\frac {\left (b x^2+c x^4\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^{14}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]

[Out]

-(A*(b*x^2 + c*x^4)^(5/2))/(11*b*x^16) - ((11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*b^2*x^14) + (4*c*(11*b*B
- 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(693*b^3*x^12) - (8*c^2*(11*b*B - 6*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*b^4*x^1
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]

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 ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^8} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac {\left (-8 (-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^7} \, dx,x,x^2\right )}{11 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac {(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}-\frac {(2 c (11 b B-6 A c)) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx,x,x^2\right )}{99 b^2}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac {(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}+\frac {4 c (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{693 b^3 x^{12}}+\frac {\left (4 c^2 (11 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx,x,x^2\right )}{693 b^3}\\ &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac {(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}+\frac {4 c (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{693 b^3 x^{12}}-\frac {8 c^2 (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 b^4 x^{10}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.67 \[ -\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (3 A \left (105 b^3-70 b^2 c x^2+40 b c^2 x^4-16 c^3 x^6\right )+11 b B x^2 \left (35 b^2-20 b c x^2+8 c^2 x^4\right )\right )}{3465 b^4 x^{16}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]

[Out]

-1/3465*((x^2*(b + c*x^2))^(5/2)*(11*b*B*x^2*(35*b^2 - 20*b*c*x^2 + 8*c^2*x^4) + 3*A*(105*b^3 - 70*b^2*c*x^2 +
40*b*c^2*x^4 - 16*c^3*x^6)))/(b^4*x^16)

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fricas [A]  time = 1.34, size = 134, normalized size = 1.01 \[ -\frac {{\left (8 \, {\left (11 \, B b c^{4} - 6 \, A c^{5}\right )} x^{10} - 4 \, {\left (11 \, B b^{2} c^{3} - 6 \, A b c^{4}\right )} x^{8} + 3 \, {\left (11 \, B b^{3} c^{2} - 6 \, A b^{2} c^{3}\right )} x^{6} + 315 \, A b^{5} + 5 \, {\left (110 \, B b^{4} c + 3 \, A b^{3} c^{2}\right )} x^{4} + 35 \, {\left (11 \, B b^{5} + 12 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, b^{4} x^{12}} \]

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^15,x, algorithm="fricas")

[Out]

-1/3465*(8*(11*B*b*c^4 - 6*A*c^5)*x^10 - 4*(11*B*b^2*c^3 - 6*A*b*c^4)*x^8 + 3*(11*B*b^3*c^2 - 6*A*b^2*c^3)*x^6
+ 315*A*b^5 + 5*(110*B*b^4*c + 3*A*b^3*c^2)*x^4 + 35*(11*B*b^5 + 12*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^4*x^
12)

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giac [B]  time = 3.23, size = 490, normalized size = 3.68 \[ \frac {16 \, {\left (2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{16} B c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} B b c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 6930 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} A c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b^{2} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 12474 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A b c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 4851 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{3} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 15246 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b^{2} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 2475 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{4} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 4950 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{3} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 495 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{5} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 990 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{4} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 605 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{6} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{5} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 121 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{7} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 66 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{6} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 11 \, B b^{8} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 6 \, A b^{7} c^{\frac {11}{2}} \mathrm {sgn}\relax (x)\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \]

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^15,x, algorithm="giac")

[Out]

16/3465*(2310*(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*c^(9/2)*sgn(x) - 1155*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*b*c^
(9/2)*sgn(x) + 6930*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*c^(11/2)*sgn(x) + 231*(sqrt(c)*x - sqrt(c*x^2 + b))^12*
B*b^2*c^(9/2)*sgn(x) + 12474*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b*c^(11/2)*sgn(x) - 4851*(sqrt(c)*x - sqrt(c*x
^2 + b))^10*B*b^3*c^(9/2)*sgn(x) + 15246*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b^2*c^(11/2)*sgn(x) + 2475*(sqrt(c
)*x - sqrt(c*x^2 + b))^8*B*b^4*c^(9/2)*sgn(x) + 4950*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^3*c^(11/2)*sgn(x) + 4
95*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^5*c^(9/2)*sgn(x) + 990*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^4*c^(11/2)*s
gn(x) + 605*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^6*c^(9/2)*sgn(x) - 330*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^5*c
^(11/2)*sgn(x) - 121*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^7*c^(9/2)*sgn(x) + 66*(sqrt(c)*x - sqrt(c*x^2 + b))^2
*A*b^6*c^(11/2)*sgn(x) + 11*B*b^8*c^(9/2)*sgn(x) - 6*A*b^7*c^(11/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 -
b)^11

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maple [A]  time = 0.05, size = 94, normalized size = 0.71 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-48 A \,c^{3} x^{6}+88 B b \,c^{2} x^{6}+120 A b \,c^{2} x^{4}-220 B \,b^{2} c \,x^{4}-210 A \,b^{2} c \,x^{2}+385 B \,b^{3} x^{2}+315 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 b^{4} x^{14}} \]

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^15,x)

[Out]

-1/3465*(c*x^2+b)*(-48*A*c^3*x^6+88*B*b*c^2*x^6+120*A*b*c^2*x^4-220*B*b^2*c*x^4-210*A*b^2*c*x^2+385*B*b^3*x^2+
315*A*b^3)*(c*x^4+b*x^2)^(3/2)/x^14/b^4

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maxima [B]  time = 1.58, size = 289, normalized size = 2.17 \[ -\frac {1}{630} \, B {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} c}{x^{8}} - \frac {35 \, \sqrt {c x^{4} + b x^{2}} b}{x^{10}} + \frac {105 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{12}}\right )} + \frac {1}{9240} \, A {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}} c}{x^{10}} + \frac {315 \, \sqrt {c x^{4} + b x^{2}} b}{x^{12}} - \frac {1155 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{14}}\right )} \]

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^15,x, algorithm="maxima")

[Out]

-1/630*B*(16*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^2) - 8*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^4) + 6*sqrt(c*x^4 + b*x^2)*c
^2/(b*x^6) - 5*sqrt(c*x^4 + b*x^2)*c/x^8 - 35*sqrt(c*x^4 + b*x^2)*b/x^10 + 105*(c*x^4 + b*x^2)^(3/2)/x^12) + 1
/9240*A*(128*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^2) - 64*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^4) + 48*sqrt(c*x^4 + b*x^2)
*c^3/(b^2*x^6) - 40*sqrt(c*x^4 + b*x^2)*c^2/(b*x^8) + 35*sqrt(c*x^4 + b*x^2)*c/x^10 + 315*sqrt(c*x^4 + b*x^2)*
b/x^12 - 1155*(c*x^4 + b*x^2)^(3/2)/x^14)

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mupad [B]  time = 1.92, size = 256, normalized size = 1.92 \[ \frac {2\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {4\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {10\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,x^8}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {8\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b\,x^6}+\frac {4\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^2\,x^4}-\frac {8\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^2} \]

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^15,x)

[Out]

(2*A*c^3*(b*x^2 + c*x^4)^(1/2))/(385*b^2*x^6) - (4*A*c*(b*x^2 + c*x^4)^(1/2))/(33*x^10) - (B*b*(b*x^2 + c*x^4)
^(1/2))/(9*x^10) - (10*B*c*(b*x^2 + c*x^4)^(1/2))/(63*x^8) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(231*b*x^8) - (A*b*
(b*x^2 + c*x^4)^(1/2))/(11*x^12) - (8*A*c^4*(b*x^2 + c*x^4)^(1/2))/(1155*b^3*x^4) + (16*A*c^5*(b*x^2 + c*x^4)^
(1/2))/(1155*b^4*x^2) - (B*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b*x^6) + (4*B*c^3*(b*x^2 + c*x^4)^(1/2))/(315*b^2*x
^4) - (8*B*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^3*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{15}}\, dx \]

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**15,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**15, x)

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