3.88 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 \left (b x+c x^2\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^5}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6} \]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(7*b*x^6) - (2*(7*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(35*b^2*x^5)

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Rubi [A]  time = 0.0482936, 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} \[ -\frac{2 \left (b x+c x^2\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^5}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(7*b*x^6) - (2*(7*b*B - 2*A*c)*(b*x + c*x^2)^(5/2))/(35*b^2*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 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^6} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}+\frac{\left (2 \left (-6 (-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^5} \, dx}{7 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}-\frac{2 (7 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{35 b^2 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0191591, size = 36, normalized size = 0.63 \[ -\frac{2 (x (b+c x))^{5/2} (5 A b-2 A c x+7 b B x)}{35 b^2 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(5*A*b + 7*b*B*x - 2*A*c*x))/(35*b^2*x^6)

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Maple [A]  time = 0.004, size = 40, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -2\,Acx+7\,bBx+5\,Ab \right ) }{35\,{x}^{5}{b}^{2}} \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^6,x)

[Out]

-2/35*(c*x+b)*(-2*A*c*x+7*B*b*x+5*A*b)*(c*x^2+b*x)^(3/2)/x^5/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)^(3/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66996, size = 173, normalized size = 3.04 \begin{align*} -\frac{2 \,{\left (5 \, A b^{3} +{\left (7 \, B b c^{2} - 2 \, A c^{3}\right )} x^{3} +{\left (14 \, B b^{2} c + A b c^{2}\right )} x^{2} +{\left (7 \, B b^{3} + 8 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x}}{35 \, b^{2} x^{4}} \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^6,x, algorithm="fricas")

[Out]

-2/35*(5*A*b^3 + (7*B*b*c^2 - 2*A*c^3)*x^3 + (14*B*b^2*c + A*b*c^2)*x^2 + (7*B*b^3 + 8*A*b^2*c)*x)*sqrt(c*x^2
+ b*x)/(b^2*x^4)

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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^{6}}\, 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**6,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**6, x)

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Giac [B]  time = 1.17444, size = 420, normalized size = 7.37 \begin{align*} \frac{2 \,{\left (35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B c^{2} + 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b c^{\frac{3}{2}} + 35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A c^{\frac{5}{2}} + 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{2} c + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b c^{2} + 35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{3} \sqrt{c} + 140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{2} c^{\frac{3}{2}} + 7 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{4} + 98 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{3} c + 35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{4} \sqrt{c} + 5 \, A b^{5}\right )}}{35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{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^6,x, algorithm="giac")

[Out]

2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*c^2 + 70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b*c^(3/2) + 35*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^5*A*c^(5/2) + 70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c + 105*(sqrt(c)*x - sqrt(
c*x^2 + b*x))^4*A*b*c^2 + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*sqrt(c) + 140*(sqrt(c)*x - sqrt(c*x^2 + b
*x))^3*A*b^2*c^(3/2) + 7*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4 + 98*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*
c + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^4*sqrt(c) + 5*A*b^5)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^7