3.1.92 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^{10}} \, dx\)

Optimal. Leaf size=195 \[ -\frac {256 c^4 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{45045 b^6 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{9009 b^5 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{1287 b^4 x^7}+\frac {16 c \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{429 b^3 x^8}-\frac {2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{39 b^2 x^9}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}} \]

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Rubi [A]  time = 0.18, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 658, 650} \begin {gather*} -\frac {256 c^4 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{45045 b^6 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{9009 b^5 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{1287 b^4 x^7}+\frac {16 c \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{429 b^3 x^8}-\frac {2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{39 b^2 x^9}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^10,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(15*b*x^10) - (2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(39*b^2*x^9) + (16*c*(3*b*B -
2*A*c)*(b*x + c*x^2)^(5/2))/(429*b^3*x^8) - (32*c^2*(3*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(1287*b^4*x^7) + (12
8*c^3*(3*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(9009*b^5*x^6) - (256*c^4*(3*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(450
45*b^6*x^5)

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]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^{10}} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}+\frac {\left (2 \left (-10 (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx}{15 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}-\frac {2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{39 b^2 x^9}-\frac {(8 c (3 b B-2 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^8} \, dx}{39 b^2}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}-\frac {2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{39 b^2 x^9}+\frac {16 c (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^8}+\frac {\left (16 c^2 (3 b B-2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{143 b^3}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}-\frac {2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{39 b^2 x^9}+\frac {16 c (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^8}-\frac {32 c^2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{1287 b^4 x^7}-\frac {\left (64 c^3 (3 b B-2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{1287 b^4}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}-\frac {2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{39 b^2 x^9}+\frac {16 c (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^8}-\frac {32 c^2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{1287 b^4 x^7}+\frac {128 c^3 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{9009 b^5 x^6}+\frac {\left (128 c^4 (3 b B-2 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{9009 b^5}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}}-\frac {2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{39 b^2 x^9}+\frac {16 c (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^8}-\frac {32 c^2 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{1287 b^4 x^7}+\frac {128 c^3 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{9009 b^5 x^6}-\frac {256 c^4 (3 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{45045 b^6 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 122, normalized size = 0.63 \begin {gather*} -\frac {2 (x (b+c x))^{5/2} \left (A \left (3003 b^5-2310 b^4 c x+1680 b^3 c^2 x^2-1120 b^2 c^3 x^3+640 b c^4 x^4-256 c^5 x^5\right )+3 b B x \left (1155 b^4-840 b^3 c x+560 b^2 c^2 x^2-320 b c^3 x^3+128 c^4 x^4\right )\right )}{45045 b^6 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^10,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(3*b*B*x*(1155*b^4 - 840*b^3*c*x + 560*b^2*c^2*x^2 - 320*b*c^3*x^3 + 128*c^4*x^4) + A*
(3003*b^5 - 2310*b^4*c*x + 1680*b^3*c^2*x^2 - 1120*b^2*c^3*x^3 + 640*b*c^4*x^4 - 256*c^5*x^5)))/(45045*b^6*x^1
0)

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IntegrateAlgebraic [A]  time = 0.53, size = 180, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-3003 A b^7-3696 A b^6 c x-63 A b^5 c^2 x^2+70 A b^4 c^3 x^3-80 A b^3 c^4 x^4+96 A b^2 c^5 x^5-128 A b c^6 x^6+256 A c^7 x^7-3465 b^7 B x-4410 b^6 B c x^2-105 b^5 B c^2 x^3+120 b^4 B c^3 x^4-144 b^3 B c^4 x^5+192 b^2 B c^5 x^6-384 b B c^6 x^7\right )}{45045 b^6 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^10,x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-3003*A*b^7 - 3465*b^7*B*x - 3696*A*b^6*c*x - 4410*b^6*B*c*x^2 - 63*A*b^5*c^2*x^2 - 105*
b^5*B*c^2*x^3 + 70*A*b^4*c^3*x^3 + 120*b^4*B*c^3*x^4 - 80*A*b^3*c^4*x^4 - 144*b^3*B*c^4*x^5 + 96*A*b^2*c^5*x^5
+ 192*b^2*B*c^5*x^6 - 128*A*b*c^6*x^6 - 384*b*B*c^6*x^7 + 256*A*c^7*x^7))/(45045*b^6*x^8)

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fricas [A]  time = 0.40, size = 177, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (3003 \, A b^{7} + 128 \, {\left (3 \, B b c^{6} - 2 \, A c^{7}\right )} x^{7} - 64 \, {\left (3 \, B b^{2} c^{5} - 2 \, A b c^{6}\right )} x^{6} + 48 \, {\left (3 \, B b^{3} c^{4} - 2 \, A b^{2} c^{5}\right )} x^{5} - 40 \, {\left (3 \, B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{4} + 35 \, {\left (3 \, B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x^{3} + 63 \, {\left (70 \, B b^{6} c + A b^{5} c^{2}\right )} x^{2} + 231 \, {\left (15 \, B b^{7} + 16 \, A b^{6} c\right )} x\right )} \sqrt {c x^{2} + b x}}{45045 \, b^{6} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^10,x, algorithm="fricas")

[Out]

-2/45045*(3003*A*b^7 + 128*(3*B*b*c^6 - 2*A*c^7)*x^7 - 64*(3*B*b^2*c^5 - 2*A*b*c^6)*x^6 + 48*(3*B*b^3*c^4 - 2*
A*b^2*c^5)*x^5 - 40*(3*B*b^4*c^3 - 2*A*b^3*c^4)*x^4 + 35*(3*B*b^5*c^2 - 2*A*b^4*c^3)*x^3 + 63*(70*B*b^6*c + A*
b^5*c^2)*x^2 + 231*(15*B*b^7 + 16*A*b^6*c)*x)*sqrt(c*x^2 + b*x)/(b^6*x^8)

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giac [B]  time = 0.33, size = 551, normalized size = 2.83 \begin {gather*} \frac {2 \, {\left (144144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{10} B c^{4} + 720720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} B b c^{\frac {7}{2}} + 240240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} A c^{\frac {9}{2}} + 1595880 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} B b^{2} c^{3} + 1338480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} A b c^{4} + 2027025 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B b^{3} c^{\frac {5}{2}} + 3333330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} A b^{2} c^{\frac {7}{2}} + 1606605 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b^{4} c^{2} + 4844840 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A b^{3} c^{3} + 810810 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{5} c^{\frac {3}{2}} + 4513509 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b^{4} c^{\frac {5}{2}} + 253890 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{6} c + 2788695 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{5} c^{2} + 45045 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{7} \sqrt {c} + 1141140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{6} c^{\frac {3}{2}} + 3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{8} + 297990 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{7} c + 45045 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{8} \sqrt {c} + 3003 \, A b^{9}\right )}}{45045 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^10,x, algorithm="giac")

[Out]

2/45045*(144144*(sqrt(c)*x - sqrt(c*x^2 + b*x))^10*B*c^4 + 720720*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*b*c^(7/2
) + 240240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*A*c^(9/2) + 1595880*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b^2*c^3 +
1338480*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b*c^4 + 2027025*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b^3*c^(5/2) +
3333330*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*A*b^2*c^(7/2) + 1606605*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^4*c^2
+ 4844840*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*b^3*c^3 + 810810*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^5*c^(3/2
) + 4513509*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^4*c^(5/2) + 253890*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^6*c
+ 2788695*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^5*c^2 + 45045*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^7*sqrt(c)
+ 1141140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^6*c^(3/2) + 3465*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^8 + 29
7990*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^7*c + 45045*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^8*sqrt(c) + 3003*A*
b^9)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^15

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maple [A]  time = 0.05, size = 134, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-256 A \,c^{5} x^{5}+384 B b \,c^{4} x^{5}+640 A b \,c^{4} x^{4}-960 B \,b^{2} c^{3} x^{4}-1120 A \,b^{2} c^{3} x^{3}+1680 B \,b^{3} c^{2} x^{3}+1680 A \,b^{3} c^{2} x^{2}-2520 B \,b^{4} c \,x^{2}-2310 A \,b^{4} c x +3465 B \,b^{5} x +3003 A \,b^{5}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{45045 b^{6} x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^10,x)

[Out]

-2/45045*(c*x+b)*(-256*A*c^5*x^5+384*B*b*c^4*x^5+640*A*b*c^4*x^4-960*B*b^2*c^3*x^4-1120*A*b^2*c^3*x^3+1680*B*b
^3*c^2*x^3+1680*A*b^3*c^2*x^2-2520*B*b^4*c*x^2-2310*A*b^4*c*x+3465*B*b^5*x+3003*A*b^5)*(c*x^2+b*x)^(3/2)/x^9/b
^6

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maxima [B]  time = 0.98, size = 360, normalized size = 1.85 \begin {gather*} -\frac {256 \, \sqrt {c x^{2} + b x} B c^{6}}{15015 \, b^{5} x} + \frac {512 \, \sqrt {c x^{2} + b x} A c^{7}}{45045 \, b^{6} x} + \frac {128 \, \sqrt {c x^{2} + b x} B c^{5}}{15015 \, b^{4} x^{2}} - \frac {256 \, \sqrt {c x^{2} + b x} A c^{6}}{45045 \, b^{5} x^{2}} - \frac {32 \, \sqrt {c x^{2} + b x} B c^{4}}{5005 \, b^{3} x^{3}} + \frac {64 \, \sqrt {c x^{2} + b x} A c^{5}}{15015 \, b^{4} x^{3}} + \frac {16 \, \sqrt {c x^{2} + b x} B c^{3}}{3003 \, b^{2} x^{4}} - \frac {32 \, \sqrt {c x^{2} + b x} A c^{4}}{9009 \, b^{3} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{429 \, b x^{5}} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{3}}{1287 \, b^{2} x^{5}} + \frac {3 \, \sqrt {c x^{2} + b x} B c}{715 \, x^{6}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{715 \, b x^{6}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{65 \, x^{7}} + \frac {\sqrt {c x^{2} + b x} A c}{390 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{5 \, x^{8}} + \frac {\sqrt {c x^{2} + b x} A b}{30 \, x^{8}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{6 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^10,x, algorithm="maxima")

[Out]

-256/15015*sqrt(c*x^2 + b*x)*B*c^6/(b^5*x) + 512/45045*sqrt(c*x^2 + b*x)*A*c^7/(b^6*x) + 128/15015*sqrt(c*x^2
+ b*x)*B*c^5/(b^4*x^2) - 256/45045*sqrt(c*x^2 + b*x)*A*c^6/(b^5*x^2) - 32/5005*sqrt(c*x^2 + b*x)*B*c^4/(b^3*x^
3) + 64/15015*sqrt(c*x^2 + b*x)*A*c^5/(b^4*x^3) + 16/3003*sqrt(c*x^2 + b*x)*B*c^3/(b^2*x^4) - 32/9009*sqrt(c*x
^2 + b*x)*A*c^4/(b^3*x^4) - 2/429*sqrt(c*x^2 + b*x)*B*c^2/(b*x^5) + 4/1287*sqrt(c*x^2 + b*x)*A*c^3/(b^2*x^5) +
3/715*sqrt(c*x^2 + b*x)*B*c/x^6 - 2/715*sqrt(c*x^2 + b*x)*A*c^2/(b*x^6) + 3/65*sqrt(c*x^2 + b*x)*B*b/x^7 + 1/
390*sqrt(c*x^2 + b*x)*A*c/x^7 - 1/5*(c*x^2 + b*x)^(3/2)*B/x^8 + 1/30*sqrt(c*x^2 + b*x)*A*b/x^8 - 1/6*(c*x^2 +
b*x)^(3/2)*A/x^9

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mupad [B]  time = 3.78, size = 326, normalized size = 1.67 \begin {gather*} \frac {4\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{1287\,b^2\,x^5}-\frac {32\,A\,c\,\sqrt {c\,x^2+b\,x}}{195\,x^7}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{13\,x^7}-\frac {28\,B\,c\,\sqrt {c\,x^2+b\,x}}{143\,x^6}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{715\,b\,x^6}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{15\,x^8}-\frac {32\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{9009\,b^3\,x^4}+\frac {64\,A\,c^5\,\sqrt {c\,x^2+b\,x}}{15015\,b^4\,x^3}-\frac {256\,A\,c^6\,\sqrt {c\,x^2+b\,x}}{45045\,b^5\,x^2}+\frac {512\,A\,c^7\,\sqrt {c\,x^2+b\,x}}{45045\,b^6\,x}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{429\,b\,x^5}+\frac {16\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{3003\,b^2\,x^4}-\frac {32\,B\,c^4\,\sqrt {c\,x^2+b\,x}}{5005\,b^3\,x^3}+\frac {128\,B\,c^5\,\sqrt {c\,x^2+b\,x}}{15015\,b^4\,x^2}-\frac {256\,B\,c^6\,\sqrt {c\,x^2+b\,x}}{15015\,b^5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^10,x)

[Out]

(4*A*c^3*(b*x + c*x^2)^(1/2))/(1287*b^2*x^5) - (32*A*c*(b*x + c*x^2)^(1/2))/(195*x^7) - (2*B*b*(b*x + c*x^2)^(
1/2))/(13*x^7) - (28*B*c*(b*x + c*x^2)^(1/2))/(143*x^6) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(715*b*x^6) - (2*A*b*(
b*x + c*x^2)^(1/2))/(15*x^8) - (32*A*c^4*(b*x + c*x^2)^(1/2))/(9009*b^3*x^4) + (64*A*c^5*(b*x + c*x^2)^(1/2))/
(15015*b^4*x^3) - (256*A*c^6*(b*x + c*x^2)^(1/2))/(45045*b^5*x^2) + (512*A*c^7*(b*x + c*x^2)^(1/2))/(45045*b^6
*x) - (2*B*c^2*(b*x + c*x^2)^(1/2))/(429*b*x^5) + (16*B*c^3*(b*x + c*x^2)^(1/2))/(3003*b^2*x^4) - (32*B*c^4*(b
*x + c*x^2)^(1/2))/(5005*b^3*x^3) + (128*B*c^5*(b*x + c*x^2)^(1/2))/(15015*b^4*x^2) - (256*B*c^6*(b*x + c*x^2)
^(1/2))/(15015*b^5*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{10}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**10,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**10, x)

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