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*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)

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Rubi [A]  time = 0.16537, antiderivative size = 160, normalized size of antiderivative = 1., 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 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^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*}

Mathematica [A]  time = 0.0401204, size = 100, normalized size = 0.62 \[ \frac{2 (x (b+c x))^{5/2} \left (A \left (-560 b^2 c^2 x^2+840 b^3 c x-1155 b^4+320 b c^3 x^3-128 c^4 x^4\right )+13 b B x \left (70 b^2 c x-105 b^3-40 b c^2 x^2+16 c^3 x^3\right )\right )}{15015 b^5 x^9} \]

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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Maple [A]  time = 0.006, size = 110, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 128\,A{c}^{4}{x}^{4}-208\,Bb{c}^{3}{x}^{4}-320\,Ab{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}cx+1365\,{b}^{4}Bx+1155\,A{b}^{4} \right ) }{15015\,{x}^{8}{b}^{5}} \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^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)/x^8/b^5

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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^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.90158, size = 348, normalized size = 2.17 \begin{align*} -\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}} \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^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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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^{9}}\, 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**9,x)

[Out]

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

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Giac [B]  time = 1.19638, size = 663, normalized size = 4.14 \begin{align*} \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}} \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^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