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

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

[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)

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Rubi [A]  time = 0.184824, antiderivative size = 195, normalized size of antiderivative = 1., 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} \[ -\frac{256 c^4 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{45045 b^6 x^5}+\frac{128 c^3 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{9009 b^5 x^6}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{1287 b^4 x^7}+\frac{16 c \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{429 b^3 x^8}-\frac{2 \left (b x+c x^2\right )^{5/2} (3 b B-2 A c)}{39 b^2 x^9}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{15 b x^{10}} \]

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 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 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 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]

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*}

Mathematica [A]  time = 0.0467471, size = 122, normalized size = 0.63 \[ -\frac{2 (x (b+c x))^{5/2} \left (A \left (1680 b^3 c^2 x^2-1120 b^2 c^3 x^3-2310 b^4 c x+3003 b^5+640 b c^4 x^4-256 c^5 x^5\right )+3 b B x \left (560 b^2 c^2 x^2-840 b^3 c x+1155 b^4-320 b c^3 x^3+128 c^4 x^4\right )\right )}{45045 b^6 x^{10}} \]

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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Maple [A]  time = 0.007, size = 134, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -256\,A{c}^{5}{x}^{5}+384\,Bb{c}^{4}{x}^{5}+640\,Ab{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}cx+3465\,B{b}^{5}x+3003\,A{b}^{5} \right ) }{45045\,{x}^{9}{b}^{6}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.75464, size = 398, normalized size = 2.04 \begin{align*} -\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{align*}

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

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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Giac [B]  time = 1.17663, size = 744, normalized size = 3.82 \begin{align*} \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{align*}

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