3.91 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=160 \[ \frac {32 c^3 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^6}+\frac {4 c \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^7}-\frac {2 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^8}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

[Out]

-2/13*A*(c*x^2+b*x)^(5/2)/b/x^9-2/143*(-8*A*c+13*B*b)*(c*x^2+b*x)^(5/2)/b^2/x^8+4/429*c*(-8*A*c+13*B*b)*(c*x^2
+b*x)^(5/2)/b^3/x^7-16/3003*c^2*(-8*A*c+13*B*b)*(c*x^2+b*x)^(5/2)/b^4/x^6+32/15015*c^3*(-8*A*c+13*B*b)*(c*x^2+
b*x)^(5/2)/b^5/x^5

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Rubi [A]  time = 0.17, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ \frac {32 c^3 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{15015 b^5 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{3003 b^4 x^6}+\frac {4 c \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{429 b^3 x^7}-\frac {2 \left (b x+c x^2\right )^{5/2} (13 b B-8 A c)}{143 b^2 x^8}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^9,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(13*b*x^9) - (2*(13*b*B - 8*A*c)*(b*x + c*x^2)^(5/2))/(143*b^2*x^8) + (4*c*(13*b*B
- 8*A*c)*(b*x + c*x^2)^(5/2))/(429*b^3*x^7) - (16*c^2*(13*b*B - 8*A*c)*(b*x + c*x^2)^(5/2))/(3003*b^4*x^6) + (
32*c^3*(13*b*B - 8*A*c)*(b*x + c*x^2)^(5/2))/(15015*b^5*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^9} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {\left (2 \left (-9 (-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^8} \, dx}{13 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac {2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {(6 c (13 b B-8 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{143 b^2}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac {2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}+\frac {4 c (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {\left (8 c^2 (13 b B-8 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{429 b^3}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac {2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}+\frac {4 c (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}-\frac {16 c^2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac {\left (16 c^3 (13 b B-8 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{3003 b^4}\\ &=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac {2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}+\frac {4 c (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}-\frac {16 c^2 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}+\frac {32 c^3 (13 b B-8 A c) \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 100, normalized size = 0.62 \[ \frac {2 (x (b+c x))^{5/2} \left (A \left (-1155 b^4+840 b^3 c x-560 b^2 c^2 x^2+320 b c^3 x^3-128 c^4 x^4\right )+13 b B x \left (-105 b^3+70 b^2 c x-40 b c^2 x^2+16 c^3 x^3\right )\right )}{15015 b^5 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^9,x]

[Out]

(2*(x*(b + c*x))^(5/2)*(13*b*B*x*(-105*b^3 + 70*b^2*c*x - 40*b*c^2*x^2 + 16*c^3*x^3) + A*(-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)))/(15015*b^5*x^9)

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fricas [A]  time = 0.89, size = 153, normalized size = 0.96 \[ -\frac {2 \, {\left (1155 \, A b^{6} - 16 \, {\left (13 \, B b c^{5} - 8 \, A c^{6}\right )} x^{6} + 8 \, {\left (13 \, B b^{2} c^{4} - 8 \, A b c^{5}\right )} x^{5} - 6 \, {\left (13 \, B b^{3} c^{3} - 8 \, A b^{2} c^{4}\right )} x^{4} + 5 \, {\left (13 \, B b^{4} c^{2} - 8 \, A b^{3} c^{3}\right )} x^{3} + 35 \, {\left (52 \, B b^{5} c + A b^{4} c^{2}\right )} x^{2} + 105 \, {\left (13 \, B b^{6} + 14 \, A b^{5} c\right )} x\right )} \sqrt {c x^{2} + b x}}{15015 \, b^{5} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-2/15015*(1155*A*b^6 - 16*(13*B*b*c^5 - 8*A*c^6)*x^6 + 8*(13*B*b^2*c^4 - 8*A*b*c^5)*x^5 - 6*(13*B*b^3*c^3 - 8*
A*b^2*c^4)*x^4 + 5*(13*B*b^4*c^2 - 8*A*b^3*c^3)*x^3 + 35*(52*B*b^5*c + A*b^4*c^2)*x^2 + 105*(13*B*b^6 + 14*A*b
^5*c)*x)*sqrt(c*x^2 + b*x)/(b^5*x^7)

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giac [B]  time = 0.21, size = 491, normalized size = 3.07 \[ \frac {2 \, {\left (30030 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9} B c^{\frac {7}{2}} + 132132 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} B b c^{3} + 48048 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} A c^{4} + 255255 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B b^{2} c^{\frac {5}{2}} + 240240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} A b c^{\frac {7}{2}} + 276705 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b^{3} c^{2} + 531960 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A b^{2} c^{3} + 180180 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{4} c^{\frac {3}{2}} + 675675 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b^{3} c^{\frac {5}{2}} + 70070 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{5} c + 535535 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{4} c^{2} + 15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{6} \sqrt {c} + 270270 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{5} c^{\frac {3}{2}} + 1365 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{7} + 84630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{6} c + 15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{7} \sqrt {c} + 1155 \, A b^{8}\right )}}{15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{13}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^9,x, algorithm="giac")

[Out]

2/15015*(30030*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*c^(7/2) + 132132*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b*c^3
+ 48048*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*c^4 + 255255*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b^2*c^(5/2) + 240
240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*A*b*c^(7/2) + 276705*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^3*c^2 + 53196
0*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*b^2*c^3 + 180180*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^4*c^(3/2) + 67567
5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^3*c^(5/2) + 70070*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^5*c + 535535*(
sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^4*c^2 + 15015*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^6*sqrt(c) + 270270*(s
qrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^5*c^(3/2) + 1365*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^7 + 84630*(sqrt(c)*
x - sqrt(c*x^2 + b*x))^2*A*b^6*c + 15015*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^7*sqrt(c) + 1155*A*b^8)/(sqrt(c)*
x - sqrt(c*x^2 + b*x))^13

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maple [A]  time = 0.04, size = 110, normalized size = 0.69 \[ -\frac {2 \left (c x +b \right ) \left (128 A \,c^{4} x^{4}-208 B b \,c^{3} x^{4}-320 A b \,c^{3} x^{3}+520 B \,b^{2} c^{2} x^{3}+560 A \,b^{2} c^{2} x^{2}-910 B \,b^{3} c \,x^{2}-840 A \,b^{3} c x +1365 b^{4} B x +1155 A \,b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 b^{5} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^9,x)

[Out]

-2/15015*(c*x+b)*(128*A*c^4*x^4-208*B*b*c^3*x^4-320*A*b*c^3*x^3+520*B*b^2*c^2*x^3+560*A*b^2*c^2*x^2-910*B*b^3*
c*x^2-840*A*b^3*c*x+1365*B*b^4*x+1155*A*b^4)*(c*x^2+b*x)^(3/2)/b^5/x^8

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maxima [B]  time = 1.01, size = 314, normalized size = 1.96 \[ \frac {32 \, \sqrt {c x^{2} + b x} B c^{5}}{1155 \, b^{4} x} - \frac {256 \, \sqrt {c x^{2} + b x} A c^{6}}{15015 \, b^{5} x} - \frac {16 \, \sqrt {c x^{2} + b x} B c^{4}}{1155 \, b^{3} x^{2}} + \frac {128 \, \sqrt {c x^{2} + b x} A c^{5}}{15015 \, b^{4} x^{2}} + \frac {4 \, \sqrt {c x^{2} + b x} B c^{3}}{385 \, b^{2} x^{3}} - \frac {32 \, \sqrt {c x^{2} + b x} A c^{4}}{5005 \, b^{3} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{231 \, b x^{4}} + \frac {16 \, \sqrt {c x^{2} + b x} A c^{3}}{3003 \, b^{2} x^{4}} + \frac {\sqrt {c x^{2} + b x} B c}{132 \, x^{5}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{429 \, b x^{5}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{44 \, x^{6}} + \frac {3 \, \sqrt {c x^{2} + b x} A c}{715 \, x^{6}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{4 \, x^{7}} + \frac {3 \, \sqrt {c x^{2} + b x} A b}{65 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{5 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^9,x, algorithm="maxima")

[Out]

32/1155*sqrt(c*x^2 + b*x)*B*c^5/(b^4*x) - 256/15015*sqrt(c*x^2 + b*x)*A*c^6/(b^5*x) - 16/1155*sqrt(c*x^2 + b*x
)*B*c^4/(b^3*x^2) + 128/15015*sqrt(c*x^2 + b*x)*A*c^5/(b^4*x^2) + 4/385*sqrt(c*x^2 + b*x)*B*c^3/(b^2*x^3) - 32
/5005*sqrt(c*x^2 + b*x)*A*c^4/(b^3*x^3) - 2/231*sqrt(c*x^2 + b*x)*B*c^2/(b*x^4) + 16/3003*sqrt(c*x^2 + b*x)*A*
c^3/(b^2*x^4) + 1/132*sqrt(c*x^2 + b*x)*B*c/x^5 - 2/429*sqrt(c*x^2 + b*x)*A*c^2/(b*x^5) + 3/44*sqrt(c*x^2 + b*
x)*B*b/x^6 + 3/715*sqrt(c*x^2 + b*x)*A*c/x^6 - 1/4*(c*x^2 + b*x)^(3/2)*B/x^7 + 3/65*sqrt(c*x^2 + b*x)*A*b/x^7
- 1/5*(c*x^2 + b*x)^(3/2)*A/x^8

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mupad [B]  time = 3.27, size = 280, normalized size = 1.75 \[ \frac {16\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{3003\,b^2\,x^4}-\frac {28\,A\,c\,\sqrt {c\,x^2+b\,x}}{143\,x^6}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {8\,B\,c\,\sqrt {c\,x^2+b\,x}}{33\,x^5}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{429\,b\,x^5}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{13\,x^7}-\frac {32\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{5005\,b^3\,x^3}+\frac {128\,A\,c^5\,\sqrt {c\,x^2+b\,x}}{15015\,b^4\,x^2}-\frac {256\,A\,c^6\,\sqrt {c\,x^2+b\,x}}{15015\,b^5\,x}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{231\,b\,x^4}+\frac {4\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{385\,b^2\,x^3}-\frac {16\,B\,c^4\,\sqrt {c\,x^2+b\,x}}{1155\,b^3\,x^2}+\frac {32\,B\,c^5\,\sqrt {c\,x^2+b\,x}}{1155\,b^4\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^9,x)

[Out]

(16*A*c^3*(b*x + c*x^2)^(1/2))/(3003*b^2*x^4) - (28*A*c*(b*x + c*x^2)^(1/2))/(143*x^6) - (2*B*b*(b*x + c*x^2)^
(1/2))/(11*x^6) - (8*B*c*(b*x + c*x^2)^(1/2))/(33*x^5) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(429*b*x^5) - (2*A*b*(b
*x + c*x^2)^(1/2))/(13*x^7) - (32*A*c^4*(b*x + c*x^2)^(1/2))/(5005*b^3*x^3) + (128*A*c^5*(b*x + c*x^2)^(1/2))/
(15015*b^4*x^2) - (256*A*c^6*(b*x + c*x^2)^(1/2))/(15015*b^5*x) - (2*B*c^2*(b*x + c*x^2)^(1/2))/(231*b*x^4) +
(4*B*c^3*(b*x + c*x^2)^(1/2))/(385*b^2*x^3) - (16*B*c^4*(b*x + c*x^2)^(1/2))/(1155*b^3*x^2) + (32*B*c^5*(b*x +
c*x^2)^(1/2))/(1155*b^4*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**9,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**9, x)

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