3.90 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=125 \[ -\frac{16 c^2 \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^5}+\frac{8 c \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^6}-\frac{2 \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^7}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8} \]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(11*b*x^8) - (2*(11*b*B - 6*A*c)*(b*x + c*x^2)^(5/2))/(99*b^2*x^7) + (8*c*(11*b*B -
6*A*c)*(b*x + c*x^2)^(5/2))/(693*b^3*x^6) - (16*c^2*(11*b*B - 6*A*c)*(b*x + c*x^2)^(5/2))/(3465*b^4*x^5)

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Rubi [A]  time = 0.118227, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ -\frac{16 c^2 \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{3465 b^4 x^5}+\frac{8 c \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{693 b^3 x^6}-\frac{2 \left (b x+c x^2\right )^{5/2} (11 b B-6 A c)}{99 b^2 x^7}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^8,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(11*b*x^8) - (2*(11*b*B - 6*A*c)*(b*x + c*x^2)^(5/2))/(99*b^2*x^7) + (8*c*(11*b*B -
6*A*c)*(b*x + c*x^2)^(5/2))/(693*b^3*x^6) - (16*c^2*(11*b*B - 6*A*c)*(b*x + c*x^2)^(5/2))/(3465*b^4*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^8} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac{\left (2 \left (-8 (-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^7} \, dx}{11 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8}-\frac{2 (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{99 b^2 x^7}-\frac{(4 c (11 b B-6 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{99 b^2}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8}-\frac{2 (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{99 b^2 x^7}+\frac{8 c (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{693 b^3 x^6}+\frac{\left (8 c^2 (11 b B-6 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{693 b^3}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{11 b x^8}-\frac{2 (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{99 b^2 x^7}+\frac{8 c (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{693 b^3 x^6}-\frac{16 c^2 (11 b B-6 A c) \left (b x+c x^2\right )^{5/2}}{3465 b^4 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0338788, size = 79, normalized size = 0.63 \[ -\frac{2 (x (b+c x))^{5/2} \left (3 A \left (-70 b^2 c x+105 b^3+40 b c^2 x^2-16 c^3 x^3\right )+11 b B x \left (35 b^2-20 b c x+8 c^2 x^2\right )\right )}{3465 b^4 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^8,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(11*b*B*x*(35*b^2 - 20*b*c*x + 8*c^2*x^2) + 3*A*(105*b^3 - 70*b^2*c*x + 40*b*c^2*x^2 -
16*c^3*x^3)))/(3465*b^4*x^8)

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Maple [A]  time = 0.007, size = 86, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -48\,A{x}^{3}{c}^{3}+88\,B{x}^{3}b{c}^{2}+120\,A{x}^{2}b{c}^{2}-220\,B{x}^{2}{b}^{2}c-210\,A{b}^{2}cx+385\,{b}^{3}Bx+315\,A{b}^{3} \right ) }{3465\,{x}^{7}{b}^{4}} \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^8,x)

[Out]

-2/3465*(c*x+b)*(-48*A*c^3*x^3+88*B*b*c^2*x^3+120*A*b*c^2*x^2-220*B*b^2*c*x^2-210*A*b^2*c*x+385*B*b^3*x+315*A*
b^3)*(c*x^2+b*x)^(3/2)/x^7/b^4

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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^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84999, size = 296, normalized size = 2.37 \begin{align*} -\frac{2 \,{\left (315 \, A b^{5} + 8 \,{\left (11 \, B b c^{4} - 6 \, A c^{5}\right )} x^{5} - 4 \,{\left (11 \, B b^{2} c^{3} - 6 \, A b c^{4}\right )} x^{4} + 3 \,{\left (11 \, B b^{3} c^{2} - 6 \, A b^{2} c^{3}\right )} x^{3} + 5 \,{\left (110 \, B b^{4} c + 3 \, A b^{3} c^{2}\right )} x^{2} + 35 \,{\left (11 \, B b^{5} + 12 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x}}{3465 \, b^{4} x^{6}} \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^8,x, algorithm="fricas")

[Out]

-2/3465*(315*A*b^5 + 8*(11*B*b*c^4 - 6*A*c^5)*x^5 - 4*(11*B*b^2*c^3 - 6*A*b*c^4)*x^4 + 3*(11*B*b^3*c^2 - 6*A*b
^2*c^3)*x^3 + 5*(110*B*b^4*c + 3*A*b^3*c^2)*x^2 + 35*(11*B*b^5 + 12*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^4*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{3}{2}} \left (A + B x\right )}{x^{8}}\, 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**8,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**8, x)

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Giac [B]  time = 1.17536, size = 582, normalized size = 4.66 \begin{align*} \frac{2 \,{\left (4620 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{8} B c^{3} + 17325 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} B b c^{\frac{5}{2}} + 6930 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} A c^{\frac{7}{2}} + 28413 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B b^{2} c^{2} + 30492 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} A b c^{3} + 25410 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b^{3} c^{\frac{3}{2}} + 58905 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A b^{2} c^{\frac{5}{2}} + 12870 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{4} c + 63855 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b^{3} c^{2} + 3465 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{5} \sqrt{c} + 41580 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{4} c^{\frac{3}{2}} + 385 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{6} + 16170 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{5} c + 3465 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{6} \sqrt{c} + 315 \, A b^{7}\right )}}{3465 \,{\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)^(3/2)/x^8,x, algorithm="giac")

[Out]

2/3465*(4620*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*c^3 + 17325*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b*c^(5/2) + 6
930*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*A*c^(7/2) + 28413*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^2*c^2 + 30492*(s
qrt(c)*x - sqrt(c*x^2 + b*x))^6*A*b*c^3 + 25410*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^3*c^(3/2) + 58905*(sqrt(
c)*x - sqrt(c*x^2 + b*x))^5*A*b^2*c^(5/2) + 12870*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^4*c + 63855*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^4*A*b^3*c^2 + 3465*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^5*sqrt(c) + 41580*(sqrt(c)*x -
sqrt(c*x^2 + b*x))^3*A*b^4*c^(3/2) + 385*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^6 + 16170*(sqrt(c)*x - sqrt(c*x
^2 + b*x))^2*A*b^5*c + 3465*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^6*sqrt(c) + 315*A*b^7)/(sqrt(c)*x - sqrt(c*x^2
+ b*x))^11