3.74 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{x^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 \left (b x+c x^2\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^3}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(5*b*x^4) - (2*(5*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

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Rubi [A]  time = 0.0450751, 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} \[ -\frac{2 \left (b x+c x^2\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^3}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^4,x]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(5*b*x^4) - (2*(5*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

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) \sqrt{b x+c x^2}}{x^4} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac{\left (2 \left (-4 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{\sqrt{b x+c x^2}}{x^3} \, dx}{5 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4}-\frac{2 (5 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0142855, size = 36, normalized size = 0.63 \[ -\frac{2 (x (b+c x))^{3/2} (3 A b-2 A c x+5 b B x)}{15 b^2 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^4,x]

[Out]

(-2*(x*(b + c*x))^(3/2)*(3*A*b + 5*b*B*x - 2*A*c*x))/(15*b^2*x^4)

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Maple [A]  time = 0.005, size = 40, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -2\,Acx+5\,bBx+3\,Ab \right ) }{15\,{x}^{3}{b}^{2}}\sqrt{c{x}^{2}+bx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x)

[Out]

-2/15*(c*x+b)*(-2*A*c*x+5*B*b*x+3*A*b)*(c*x^2+b*x)^(1/2)/x^3/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)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83955, size = 126, normalized size = 2.21 \begin{align*} -\frac{2 \,{\left (3 \, A b^{2} +{\left (5 \, B b c - 2 \, A c^{2}\right )} x^{2} +{\left (5 \, B b^{2} + A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{15 \, b^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x, algorithm="fricas")

[Out]

-2/15*(3*A*b^2 + (5*B*b*c - 2*A*c^2)*x^2 + (5*B*b^2 + A*b*c)*x)*sqrt(c*x^2 + b*x)/(b^2*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )} \left (A + B x\right )}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**4,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**4, x)

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Giac [B]  time = 1.14373, size = 258, normalized size = 4.53 \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B c + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b \sqrt{c} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A c^{\frac{3}{2}} + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{2} + 25 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b c + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{2} \sqrt{c} + 3 \, A b^{3}\right )}}{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b*sqrt(c) + 15*(sqrt(c
)*x - sqrt(c*x^2 + b*x))^3*A*c^(3/2) + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2 + 25*(sqrt(c)*x - sqrt(c*x^2
+ b*x))^2*A*b*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*sqrt(c) + 3*A*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^
5