3.105 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^9} \, dx\)

Optimal. Leaf size=90 \[ \frac{4 c \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{693 b^3 x^7}-\frac{2 \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{99 b^2 x^8}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9} \]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(11*b*x^9) - (2*(11*b*B - 4*A*c)*(b*x + c*x^2)^(7/2))/(99*b^2*x^8) + (4*c*(11*b*B -
4*A*c)*(b*x + c*x^2)^(7/2))/(693*b^3*x^7)

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Rubi [A]  time = 0.088501, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ \frac{4 c \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{693 b^3 x^7}-\frac{2 \left (b x+c x^2\right )^{7/2} (11 b B-4 A c)}{99 b^2 x^8}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^9,x]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(11*b*x^9) - (2*(11*b*B - 4*A*c)*(b*x + c*x^2)^(7/2))/(99*b^2*x^8) + (4*c*(11*b*B -
4*A*c)*(b*x + c*x^2)^(7/2))/(693*b^3*x^7)

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 )^{5/2}}{x^9} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}+\frac{\left (2 \left (-9 (-b B+A c)+\frac{7}{2} (-b B+2 A c)\right )\right ) \int \frac{\left (b x+c x^2\right )^{5/2}}{x^8} \, dx}{11 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}-\frac{2 (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{99 b^2 x^8}-\frac{(2 c (11 b B-4 A c)) \int \frac{\left (b x+c x^2\right )^{5/2}}{x^7} \, dx}{99 b^2}\\ &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{11 b x^9}-\frac{2 (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{99 b^2 x^8}+\frac{4 c (11 b B-4 A c) \left (b x+c x^2\right )^{7/2}}{693 b^3 x^7}\\ \end{align*}

Mathematica [A]  time = 0.0258829, size = 63, normalized size = 0.7 \[ \frac{2 (b+c x)^3 \sqrt{x (b+c x)} \left (A \left (-63 b^2+28 b c x-8 c^2 x^2\right )+11 b B x (2 c x-7 b)\right )}{693 b^3 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^9,x]

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(11*b*B*x*(-7*b + 2*c*x) + A*(-63*b^2 + 28*b*c*x - 8*c^2*x^2)))/(693*b^3*x^6)

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Maple [A]  time = 0.006, size = 62, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,A{c}^{2}{x}^{2}-22\,B{x}^{2}bc-28\,Abcx+77\,{b}^{2}Bx+63\,A{b}^{2} \right ) }{693\,{x}^{8}{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^9,x)

[Out]

-2/693*(c*x+b)*(8*A*c^2*x^2-22*B*b*c*x^2-28*A*b*c*x+77*B*b^2*x+63*A*b^2)*(c*x^2+b*x)^(5/2)/x^8/b^3

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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)^(5/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89402, size = 286, normalized size = 3.18 \begin{align*} -\frac{2 \,{\left (63 \, A b^{5} - 2 \,{\left (11 \, B b c^{4} - 4 \, A c^{5}\right )} x^{5} +{\left (11 \, B b^{2} c^{3} - 4 \, A b c^{4}\right )} x^{4} + 3 \,{\left (55 \, B b^{3} c^{2} + A b^{2} c^{3}\right )} x^{3} +{\left (209 \, B b^{4} c + 113 \, A b^{3} c^{2}\right )} x^{2} + 7 \,{\left (11 \, B b^{5} + 23 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x}}{693 \, b^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^9,x, algorithm="fricas")

[Out]

-2/693*(63*A*b^5 - 2*(11*B*b*c^4 - 4*A*c^5)*x^5 + (11*B*b^2*c^3 - 4*A*b*c^4)*x^4 + 3*(55*B*b^3*c^2 + A*b^2*c^3
)*x^3 + (209*B*b^4*c + 113*A*b^3*c^2)*x^2 + 7*(11*B*b^5 + 23*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^6)

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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{5}{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)**(5/2)/x**9,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**9, x)

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Giac [B]  time = 1.15327, size = 663, normalized size = 7.37 \begin{align*} \frac{2 \,{\left (693 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{9} B c^{\frac{7}{2}} + 3003 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{8} B b c^{3} + 924 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{8} A c^{4} + 6237 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} B b^{2} c^{\frac{5}{2}} + 4851 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} A b c^{\frac{7}{2}} + 7623 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B b^{3} c^{2} + 11781 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} A b^{2} c^{3} + 5775 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b^{4} c^{\frac{3}{2}} + 16863 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A b^{3} c^{\frac{5}{2}} + 2673 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{5} c + 15345 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b^{4} c^{2} + 693 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{6} \sqrt{c} + 9009 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{5} c^{\frac{3}{2}} + 77 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{7} + 3311 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{6} c + 693 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{7} \sqrt{c} + 63 \, A b^{8}\right )}}{693 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^9,x, algorithm="giac")

[Out]

2/693*(693*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*c^(7/2) + 3003*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b*c^3 + 924*
(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*c^4 + 6237*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b^2*c^(5/2) + 4851*(sqrt(c)
*x - sqrt(c*x^2 + b*x))^7*A*b*c^(7/2) + 7623*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^3*c^2 + 11781*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^6*A*b^2*c^3 + 5775*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^4*c^(3/2) + 16863*(sqrt(c)*x - sqr
t(c*x^2 + b*x))^5*A*b^3*c^(5/2) + 2673*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^5*c + 15345*(sqrt(c)*x - sqrt(c*x
^2 + b*x))^4*A*b^4*c^2 + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^6*sqrt(c) + 9009*(sqrt(c)*x - sqrt(c*x^2 +
b*x))^3*A*b^5*c^(3/2) + 77*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^7 + 3311*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*
b^6*c + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^7*sqrt(c) + 63*A*b^8)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^11