3.1.98 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{11}} \, dx\) [98]

Optimal. Leaf size=133 \[ -\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac {(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac {4 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}-\frac {8 c^2 (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6} \]

[Out]

-1/9*A*(c*x^4+b*x^2)^(3/2)/b/x^12-1/21*(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/b^2/x^10+4/105*c*(-2*A*c+3*B*b)*(c*x
^4+b*x^2)^(3/2)/b^3/x^8-8/315*c^2*(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/b^4/x^6

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Rubi [A]
time = 0.16, 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 = {2059, 806, 672, 664} \begin {gather*} -\frac {8 c^2 \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{315 b^4 x^6}+\frac {4 c \left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{105 b^3 x^8}-\frac {\left (b x^2+c x^4\right )^{3/2} (3 b B-2 A c)}{21 b^2 x^{10}}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*(A*(b*x^2 + c*x^4)^(3/2))/(b*x^12) - ((3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(21*b^2*x^10) + (4*c*(3*b*B
- 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*b^3*x^8) - (8*c^2*(3*b*B - 2*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*b^4*x^6)

Rule 664

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 672

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 806

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 2059

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^{11}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac {\left (-6 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \text {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^5} \, dx,x,x^2\right )}{9 b}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac {(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac {(2 c (3 b B-2 A c)) \text {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^4} \, dx,x,x^2\right )}{21 b^2}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac {(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac {4 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac {\left (4 c^2 (3 b B-2 A c)\right ) \text {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^3} \, dx,x,x^2\right )}{105 b^3}\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac {(3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac {4 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}-\frac {8 c^2 (3 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 88, normalized size = 0.66 \begin {gather*} -\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (3 b B x^2 \left (15 b^2-12 b c x^2+8 c^2 x^4\right )+A \left (35 b^3-30 b^2 c x^2+24 b c^2 x^4-16 c^3 x^6\right )\right )}{315 b^4 x^{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/315*((x^2*(b + c*x^2))^(3/2)*(3*b*B*x^2*(15*b^2 - 12*b*c*x^2 + 8*c^2*x^4) + A*(35*b^3 - 30*b^2*c*x^2 + 24*b
*c^2*x^4 - 16*c^3*x^6)))/(b^4*x^12)

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Maple [A]
time = 0.38, size = 94, normalized size = 0.71

method	result	size
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-16 A \,c^{3} x^{6}+24 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-36 x^{4} B \,b^{2} c -30 A \,b^{2} c \,x^{2}+45 x^{2} B \,b^{3}+35 A \,b^{3}\right ) \sqrt {x^{4} c +b \,x^{2}}}{315 b^{4} x^{10}}\)	\(94\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-16 A \,c^{3} x^{6}+24 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-36 x^{4} B \,b^{2} c -30 A \,b^{2} c \,x^{2}+45 x^{2} B \,b^{3}+35 A \,b^{3}\right ) \sqrt {x^{4} c +b \,x^{2}}}{315 b^{4} x^{10}}\)	\(94\)
trager	\(-\frac {\left (-16 A \,c^{4} x^{8}+24 B b \,c^{3} x^{8}+8 A b \,c^{3} x^{6}-12 B \,b^{2} c^{2} x^{6}-6 A \,b^{2} c^{2} x^{4}+9 B \,b^{3} c \,x^{4}+5 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right ) \sqrt {x^{4} c +b \,x^{2}}}{315 b^{4} x^{10}}\)	\(111\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-16 A \,c^{4} x^{8}+24 B b \,c^{3} x^{8}+8 A b \,c^{3} x^{6}-12 B \,b^{2} c^{2} x^{6}-6 A \,b^{2} c^{2} x^{4}+9 B \,b^{3} c \,x^{4}+5 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{10} b^{4}}\)	\(111\)

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^11,x,method=_RETURNVERBOSE)

[Out]

-1/315*(c*x^2+b)*(-16*A*c^3*x^6+24*B*b*c^2*x^6+24*A*b*c^2*x^4-36*B*b^2*c*x^4-30*A*b^2*c*x^2+45*B*b^3*x^2+35*A*
b^3)*(c*x^4+b*x^2)^(1/2)/b^4/x^10

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Maxima [A]
time = 0.29, size = 209, normalized size = 1.57 \begin {gather*} -\frac {1}{105} \, B {\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 )} + \frac {1}{315} \, A {\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 )} \end {gather*}

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^11,x, algorithm="maxima")

[Out]

-1/105*B*(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) + 1/315*A*(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)

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Fricas [A]
time = 1.29, size = 109, normalized size = 0.82 \begin {gather*} -\frac {{\left (8 \, {\left (3 \, B b c^{3} - 2 \, A c^{4}\right )} x^{8} - 4 \, {\left (3 \, B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + 35 \, A b^{4} + 3 \, {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4} + 5 \, {\left (9 \, B b^{4} + A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{4} x^{10}} \end {gather*}

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^11,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{11}}\, dx \end {gather*}

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**11,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (117) = 234\).
time = 1.74, size = 370, normalized size = 2.78 \begin {gather*} \frac {16 \, {\left (210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{2} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 378 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{3} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 168 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{2} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 108 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{4} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{3} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 27 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{5} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{4} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 3 \, B b^{6} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 2 \, A b^{5} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9}} \end {gather*}

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^11,x, algorithm="giac")

[Out]

16/315*(210*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*c^(7/2)*sgn(x) - 315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b*c^(7/
2)*sgn(x) + 630*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*c^(9/2)*sgn(x) + 63*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^2*c
^(7/2)*sgn(x) + 378*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b*c^(9/2)*sgn(x) - 42*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*
b^3*c^(7/2)*sgn(x) + 168*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^2*c^(9/2)*sgn(x) + 108*(sqrt(c)*x - sqrt(c*x^2 +
b))^4*B*b^4*c^(7/2)*sgn(x) - 72*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^3*c^(9/2)*sgn(x) - 27*(sqrt(c)*x - sqrt(c*
x^2 + b))^2*B*b^5*c^(7/2)*sgn(x) + 18*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^4*c^(9/2)*sgn(x) + 3*B*b^6*c^(7/2)*s
gn(x) - 2*A*b^5*c^(9/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9

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Mupad [B]
time = 1.04, size = 210, normalized size = 1.58 \begin {gather*} \frac {2\,A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^6}-\frac {B\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {A\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,b\,x^8}-\frac {B\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^6}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {8\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^4}+\frac {16\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^4\,x^2}+\frac {4\,B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^4}-\frac {8\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^2} \end {gather*}

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^11,x)

[Out]

(2*A*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^6) - (B*(b*x^2 + c*x^4)^(1/2))/(7*x^8) - (A*c*(b*x^2 + c*x^4)^(1/2)
)/(63*b*x^8) - (B*c*(b*x^2 + c*x^4)^(1/2))/(35*b*x^6) - (A*(b*x^2 + c*x^4)^(1/2))/(9*x^10) - (8*A*c^3*(b*x^2 +
c*x^4)^(1/2))/(315*b^3*x^4) + (16*A*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^4*x^2) + (4*B*c^2*(b*x^2 + c*x^4)^(1/2)
)/(105*b^2*x^4) - (8*B*c^3*(b*x^2 + c*x^4)^(1/2))/(105*b^3*x^2)

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