3.99 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{13}} \, dx\)

Optimal. Leaf size=170 \[ \frac {16 c^3 \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{3465 b^5 x^6}-\frac {8 c^2 \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{1155 b^4 x^8}+\frac {2 c \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{231 b^3 x^{10}}-\frac {\left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{99 b^2 x^{12}}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}} \]

[Out]

-1/11*A*(c*x^4+b*x^2)^(3/2)/b/x^14-1/99*(-8*A*c+11*B*b)*(c*x^4+b*x^2)^(3/2)/b^2/x^12+2/231*c*(-8*A*c+11*B*b)*(
c*x^4+b*x^2)^(3/2)/b^3/x^10-8/1155*c^2*(-8*A*c+11*B*b)*(c*x^4+b*x^2)^(3/2)/b^4/x^8+16/3465*c^3*(-8*A*c+11*B*b)
*(c*x^4+b*x^2)^(3/2)/b^5/x^6

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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} \[ \frac {16 c^3 \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{3465 b^5 x^6}-\frac {8 c^2 \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{1155 b^4 x^8}+\frac {2 c \left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{231 b^3 x^{10}}-\frac {\left (b x^2+c x^4\right )^{3/2} (11 b B-8 A c)}{99 b^2 x^{12}}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^13,x]

[Out]

-(A*(b*x^2 + c*x^4)^(3/2))/(11*b*x^14) - ((11*b*B - 8*A*c)*(b*x^2 + c*x^4)^(3/2))/(99*b^2*x^12) + (2*c*(11*b*B
- 8*A*c)*(b*x^2 + c*x^4)^(3/2))/(231*b^3*x^10) - (8*c^2*(11*b*B - 8*A*c)*(b*x^2 + c*x^4)^(3/2))/(1155*b^4*x^8
) + (16*c^3*(11*b*B - 8*A*c)*(b*x^2 + c*x^4)^(3/2))/(3465*b^5*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^{13}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {\left (-7 (-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^6} \, dx,x,x^2\right )}{11 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}-\frac {(11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {(c (11 b B-8 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^5} \, dx,x,x^2\right )}{33 b^2}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}-\frac {(11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}+\frac {2 c (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {\left (4 c^2 (11 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^4} \, dx,x,x^2\right )}{231 b^3}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}-\frac {(11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}+\frac {2 c (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}-\frac {8 c^2 (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}-\frac {\left (8 c^3 (11 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^3} \, dx,x,x^2\right )}{1155 b^4}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}-\frac {(11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}+\frac {2 c (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}-\frac {8 c^2 (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}+\frac {16 c^3 (11 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{3465 b^5 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 94, normalized size = 0.55 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (x^2 \left (\frac {c x^2}{b}+1\right ) \left (35 b^3-30 b^2 c x^2+24 b c^2 x^4-16 c^3 x^6\right ) (8 A c-11 b B)-315 A b^3 \left (b+c x^2\right )\right )}{3465 b^4 x^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^13,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-315*A*b^3*(b + c*x^2) + (-11*b*B + 8*A*c)*x^2*(1 + (c*x^2)/b)*(35*b^3 - 30*b^2*c*x^2
+ 24*b*c^2*x^4 - 16*c^3*x^6)))/(3465*b^4*x^12)

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fricas [A]  time = 1.34, size = 133, normalized size = 0.78 \[ \frac {{\left (16 \, {\left (11 \, B b c^{4} - 8 \, A c^{5}\right )} x^{10} - 8 \, {\left (11 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} x^{8} + 6 \, {\left (11 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} x^{6} - 315 \, A b^{5} - 5 \, {\left (11 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} x^{4} - 35 \, {\left (11 \, B b^{5} + A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, b^{5} x^{12}} \]

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^13,x, algorithm="fricas")

[Out]

1/3465*(16*(11*B*b*c^4 - 8*A*c^5)*x^10 - 8*(11*B*b^2*c^3 - 8*A*b*c^4)*x^8 + 6*(11*B*b^3*c^2 - 8*A*b^2*c^3)*x^6
- 315*A*b^5 - 5*(11*B*b^4*c - 8*A*b^3*c^2)*x^4 - 35*(11*B*b^5 + A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*x^12)

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giac [B]  time = 2.97, size = 430, normalized size = 2.53 \[ \frac {32 \, {\left (3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} B c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 4851 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 11088 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{2} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 7392 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{3} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 2640 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{2} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 1815 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{4} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 1320 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{3} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 605 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{5} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 440 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{4} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 121 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{6} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 88 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{5} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 11 \, B b^{7} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 8 \, A b^{6} c^{\frac {11}{2}} \mathrm {sgn}\relax (x)\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="giac")

[Out]

32/3465*(3465*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*c^(9/2)*sgn(x) - 4851*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*b*c^
(9/2)*sgn(x) + 11088*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*c^(11/2)*sgn(x) + 231*(sqrt(c)*x - sqrt(c*x^2 + b))^10
*B*b^2*c^(9/2)*sgn(x) + 7392*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b*c^(11/2)*sgn(x) - 165*(sqrt(c)*x - sqrt(c*x^
2 + b))^8*B*b^3*c^(9/2)*sgn(x) + 2640*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^2*c^(11/2)*sgn(x) + 1815*(sqrt(c)*x
- sqrt(c*x^2 + b))^6*B*b^4*c^(9/2)*sgn(x) - 1320*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^3*c^(11/2)*sgn(x) - 605*(
sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^5*c^(9/2)*sgn(x) + 440*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^4*c^(11/2)*sgn(x
) + 121*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^6*c^(9/2)*sgn(x) - 88*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^5*c^(11/
2)*sgn(x) - 11*B*b^7*c^(9/2)*sgn(x) + 8*A*b^6*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 = 118, normalized size = 0.69 \[ -\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-176 B b \,c^{3} x^{8}-192 A b \,c^{3} x^{6}+264 B \,b^{2} c^{2} x^{6}+240 A \,b^{2} c^{2} x^{4}-330 B \,b^{3} c \,x^{4}-280 A \,b^{3} c \,x^{2}+385 B \,b^{4} x^{2}+315 A \,b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 b^{5} x^{12}} \]

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^13,x)

[Out]

-1/3465*(c*x^2+b)*(128*A*c^4*x^8-176*B*b*c^3*x^8-192*A*b*c^3*x^6+264*B*b^2*c^2*x^6+240*A*b^2*c^2*x^4-330*B*b^3
*c*x^4-280*A*b^3*c*x^2+385*B*b^4*x^2+315*A*b^4)*(c*x^4+b*x^2)^(1/2)/b^5/x^12

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maxima [A]  time = 1.57, size = 257, normalized size = 1.51 \[ \frac {1}{315} \, B {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{3} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} c}{b x^{8}} - \frac {35 \, \sqrt {c x^{4} + b x^{2}}}{x^{10}}\right )} - \frac {1}{3465} \, A {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{5} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{4} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{3} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}} c}{b x^{10}} + \frac {315 \, \sqrt {c x^{4} + b x^{2}}}{x^{12}}\right )} \]

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^13,x, algorithm="maxima")

[Out]

1/315*B*(16*sqrt(c*x^4 + b*x^2)*c^4/(b^4*x^2) - 8*sqrt(c*x^4 + b*x^2)*c^3/(b^3*x^4) + 6*sqrt(c*x^4 + b*x^2)*c^
2/(b^2*x^6) - 5*sqrt(c*x^4 + b*x^2)*c/(b*x^8) - 35*sqrt(c*x^4 + b*x^2)/x^10) - 1/3465*A*(128*sqrt(c*x^4 + b*x^
2)*c^5/(b^5*x^2) - 64*sqrt(c*x^4 + b*x^2)*c^4/(b^4*x^4) + 48*sqrt(c*x^4 + b*x^2)*c^3/(b^3*x^6) - 40*sqrt(c*x^4
+ b*x^2)*c^2/(b^2*x^8) + 35*sqrt(c*x^4 + b*x^2)*c/(b*x^10) + 315*sqrt(c*x^4 + b*x^2)/x^12)

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mupad [B]  time = 1.41, size = 260, normalized size = 1.53 \[ \frac {8\,A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{693\,b^2\,x^8}-\frac {B\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {A\,c\,\sqrt {c\,x^4+b\,x^2}}{99\,b\,x^{10}}-\frac {B\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,b\,x^8}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {16\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^6}+\frac {64\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^4\,x^4}-\frac {128\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^5\,x^2}+\frac {2\,B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^6}-\frac {8\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^4}+\frac {16\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^4\,x^2} \]

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^13,x)

[Out]

(8*A*c^2*(b*x^2 + c*x^4)^(1/2))/(693*b^2*x^8) - (B*(b*x^2 + c*x^4)^(1/2))/(9*x^10) - (A*c*(b*x^2 + c*x^4)^(1/2
))/(99*b*x^10) - (B*c*(b*x^2 + c*x^4)^(1/2))/(63*b*x^8) - (A*(b*x^2 + c*x^4)^(1/2))/(11*x^12) - (16*A*c^3*(b*x
^2 + c*x^4)^(1/2))/(1155*b^3*x^6) + (64*A*c^4*(b*x^2 + c*x^4)^(1/2))/(3465*b^4*x^4) - (128*A*c^5*(b*x^2 + c*x^
4)^(1/2))/(3465*b^5*x^2) + (2*B*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^6) - (8*B*c^3*(b*x^2 + c*x^4)^(1/2))/(31
5*b^3*x^4) + (16*B*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{13}}\, dx \]

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**13,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**13, x)

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