3.104 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 \left (b x+c x^2\right )^{7/2} (9 b B-2 A c)}{63 b^2 x^7}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8} \]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(9*b*x^8) - (2*(9*b*B - 2*A*c)*(b*x + c*x^2)^(7/2))/(63*b^2*x^7)

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Rubi [A]  time = 0.0492487, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {792, 650} \[ -\frac{2 \left (b x+c x^2\right )^{7/2} (9 b B-2 A c)}{63 b^2 x^7}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^8,x]

[Out]

(-2*A*(b*x + c*x^2)^(7/2))/(9*b*x^8) - (2*(9*b*B - 2*A*c)*(b*x + c*x^2)^(7/2))/(63*b^2*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 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^8} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8}+\frac{\left (2 \left (-8 (-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^7} \, dx}{9 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{9 b x^8}-\frac{2 (9 b B-2 A c) \left (b x+c x^2\right )^{7/2}}{63 b^2 x^7}\\ \end{align*}

Mathematica [A]  time = 0.015944, size = 43, normalized size = 0.75 \[ -\frac{2 (b+c x)^3 \sqrt{x (b+c x)} (7 A b-2 A c x+9 b B x)}{63 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^8,x]

[Out]

(-2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b + 9*b*B*x - 2*A*c*x))/(63*b^2*x^5)

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Maple [A]  time = 0.006, size = 40, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -2\,Acx+9\,bBx+7\,Ab \right ) }{63\,{x}^{7}{b}^{2}} \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^8,x)

[Out]

-2/63*(c*x+b)*(-2*A*c*x+9*B*b*x+7*A*b)*(c*x^2+b*x)^(5/2)/x^7/b^2

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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^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.91573, size = 223, normalized size = 3.91 \begin{align*} -\frac{2 \,{\left (7 \, A b^{4} +{\left (9 \, B b c^{3} - 2 \, A c^{4}\right )} x^{4} +{\left (27 \, B b^{2} c^{2} + A b c^{3}\right )} x^{3} + 3 \,{\left (9 \, B b^{3} c + 5 \, A b^{2} c^{2}\right )} x^{2} +{\left (9 \, B b^{4} + 19 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x}}{63 \, b^{2} x^{5}} \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^8,x, algorithm="fricas")

[Out]

-2/63*(7*A*b^4 + (9*B*b*c^3 - 2*A*c^4)*x^4 + (27*B*b^2*c^2 + A*b*c^3)*x^3 + 3*(9*B*b^3*c + 5*A*b^2*c^2)*x^2 +
(9*B*b^4 + 19*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^2*x^5)

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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^{8}}\, 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**8,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**8, x)

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Giac [B]  time = 1.13519, size = 582, normalized size = 10.21 \begin{align*} \frac{2 \,{\left (63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{8} B c^{3} + 189 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} B b c^{\frac{5}{2}} + 63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} A c^{\frac{7}{2}} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B b^{2} c^{2} + 273 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} A b c^{3} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b^{3} c^{\frac{3}{2}} + 567 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A b^{2} c^{\frac{5}{2}} + 189 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{4} c + 693 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b^{3} c^{2} + 63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{5} \sqrt{c} + 525 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{4} c^{\frac{3}{2}} + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{6} + 243 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{5} c + 63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{6} \sqrt{c} + 7 \, A b^{7}\right )}}{63 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{9}} \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^8,x, algorithm="giac")

[Out]

2/63*(63*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*c^3 + 189*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*b*c^(5/2) + 63*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^7*A*c^(7/2) + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b^2*c^2 + 273*(sqrt(c)*x - s
qrt(c*x^2 + b*x))^6*A*b*c^3 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^3*c^(3/2) + 567*(sqrt(c)*x - sqrt(c*x^
2 + b*x))^5*A*b^2*c^(5/2) + 189*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^4*c + 693*(sqrt(c)*x - sqrt(c*x^2 + b*x)
)^4*A*b^3*c^2 + 63*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^5*sqrt(c) + 525*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b
^4*c^(3/2) + 9*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^6 + 243*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^5*c + 63*(s
qrt(c)*x - sqrt(c*x^2 + b*x))*A*b^6*sqrt(c) + 7*A*b^7)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9