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

Optimal. Leaf size=57 \[ -\frac {2 \sqrt {b x+c x^2} (3 b B-2 A c)}{3 b^2 x}-\frac {2 A \sqrt {b x+c x^2}}{3 b x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {2 \sqrt {b x+c x^2} (3 b B-2 A c)}{3 b^2 x}-\frac {2 A \sqrt {b x+c x^2}}{3 b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^2 \sqrt {b x+c x^2}} \, dx &=-\frac {2 A \sqrt {b x+c x^2}}{3 b x^2}+\frac {\left (2 \left (-2 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{3 b}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{3 b x^2}-\frac {2 (3 b B-2 A c) \sqrt {b x+c x^2}}{3 b^2 x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.61 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} (A (b-2 c x)+3 b B x)}{3 b^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.30, size = 38, normalized size = 0.67 \begin {gather*} \frac {2 \sqrt {b x+c x^2} (-A b+2 A c x-3 b B x)}{3 b^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 34, normalized size = 0.60 \begin {gather*} -\frac {2 \, \sqrt {c x^{2} + b x} {\left (A b + {\left (3 \, B b - 2 \, A c\right )} x\right )}}{3 \, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 76, normalized size = 1.33 \begin {gather*} \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A \sqrt {c} + A b\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*sqrt(c) + A*b)/(sqrt(c)*x - s
qrt(c*x^2 + b*x))^3

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maple [A]  time = 0.05, size = 39, normalized size = 0.68 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-2 A c x +3 B b x +A b \right )}{3 \sqrt {c \,x^{2}+b x}\, b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(c*x+b)*(-2*A*c*x+3*B*b*x+A*b)/x/b^2/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 0.87, size = 62, normalized size = 1.09 \begin {gather*} -\frac {2 \, \sqrt {c x^{2} + b x} B}{b x} + \frac {4 \, \sqrt {c x^{2} + b x} A c}{3 \, b^{2} x} - \frac {2 \, \sqrt {c x^{2} + b x} A}{3 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.09, size = 33, normalized size = 0.58 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (A\,b-2\,A\,c\,x+3\,B\,b\,x\right )}{3\,b^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(b*x + c*x^2)^(1/2)*(A*b - 2*A*c*x + 3*B*b*x))/(3*b^2*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{2} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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