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

Optimal. Leaf size=60 \[ -\frac {2 (b+2 c x) (3 b B-4 A c)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2 A}{3 b x \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 60, 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, 613} \begin {gather*} -\frac {2 (b+2 c x) (3 b B-4 A c)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2 A}{3 b x \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(3*b*x*Sqrt[b*x + c*x^2]) - (2*(3*b*B - 4*A*c)*(b + 2*c*x))/(3*b^3*Sqrt[b*x + c*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 A}{3 b x \sqrt {b x+c x^2}}+\frac {\left (2 \left (b B-A c+\frac {1}{2} (b B-2 A c)\right )\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {2 A}{3 b x \sqrt {b x+c x^2}}-\frac {2 (3 b B-4 A c) (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.87 \begin {gather*} -\frac {2 \left (A \left (b^2-4 b c x-8 c^2 x^2\right )+3 b B x (b+2 c x)\right )}{3 b^3 x \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*b*B*x*(b + 2*c*x) + A*(b^2 - 4*b*c*x - 8*c^2*x^2)))/(3*b^3*x*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.34, size = 66, normalized size = 1.10 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (A b^2-4 A b c x-8 A c^2 x^2+3 b^2 B x+6 b B c x^2\right )}{3 b^3 x^2 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*x + c*x^2]*(A*b^2 + 3*b^2*B*x - 4*A*b*c*x + 6*b*B*c*x^2 - 8*A*c^2*x^2))/(3*b^3*x^2*(b + c*x))

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fricas [A]  time = 0.40, size = 68, normalized size = 1.13 \begin {gather*} -\frac {2 \, {\left (A b^{2} + 2 \, {\left (3 \, B b c - 4 \, A c^{2}\right )} x^{2} + {\left (3 \, B b^{2} - 4 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x), x)

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maple [A]  time = 0.05, size = 58, normalized size = 0.97 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-8 A \,c^{2} x^{2}+6 B b c \,x^{2}-4 A b c x +3 B \,b^{2} x +A \,b^{2}\right )}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x/(c*x^2+b*x)^(3/2),x)

[Out]

-2/3*(c*x+b)*(-8*A*c^2*x^2+6*B*b*c*x^2-4*A*b*c*x+3*B*b^2*x+A*b^2)/b^3/(c*x^2+b*x)^(3/2)

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maxima [A]  time = 0.91, size = 96, normalized size = 1.60 \begin {gather*} -\frac {4 \, B c x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {16 \, A c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, B}{\sqrt {c x^{2} + b x} b} + \frac {8 \, A c}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {2 \, A}{3 \, \sqrt {c x^{2} + b x} b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

-4*B*c*x/(sqrt(c*x^2 + b*x)*b^2) + 16/3*A*c^2*x/(sqrt(c*x^2 + b*x)*b^3) - 2*B/(sqrt(c*x^2 + b*x)*b) + 8/3*A*c/
(sqrt(c*x^2 + b*x)*b^2) - 2/3*A/(sqrt(c*x^2 + b*x)*b*x)

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mupad [B]  time = 1.16, size = 62, normalized size = 1.03 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,B\,b^2\,x+A\,b^2+6\,B\,b\,c\,x^2-4\,A\,b\,c\,x-8\,A\,c^2\,x^2\right )}{3\,b^3\,x^2\,\left (b+c\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x*(b*x + c*x^2)^(3/2)),x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(A*b^2 - 8*A*c^2*x^2 + 3*B*b^2*x + 6*B*b*c*x^2 - 4*A*b*c*x))/(3*b^3*x^2*(b + c*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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