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

Optimal. Leaf size=140 \[ \frac {A b-a B}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 77} \begin {gather*} \frac {A b-a B}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 80, normalized size = 0.57 \begin {gather*} \frac {a \left (a^2 (-B)+3 a A b+2 A b^2 x\right )+2 A b \log (x) (a+b x)^2-2 A b (a+b x)^2 \log (a+b x)}{2 a^3 b (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.76, size = 241, normalized size = 1.72 \begin {gather*} \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^3}+\frac {a^4 b B-a^3 A b^2+a^2 b^3 B x^2+\sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^3 B-a^2 A b-a^2 b B x+a A b^2 x+2 A b^3 x^2\right )-3 a A b^4 x^2-2 A b^5 x^3}{a^2 b \sqrt {b^2} x^2 \left (2 a^2 b^2+4 a b^3 x+2 b^4 x^2\right )+a^2 b x^2 \left (-2 a b^3-2 b^4 x\right ) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^3*A*b^2) + a^4*b*B - 3*a*A*b^4*x^2 + a^2*b^3*B*x^2 - 2*A*b^5*x^3 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]*(-(a^2*A*b) + a^3*B + a*A*b^2*x - a^2*b*B*x + 2*A*b^3*x^2))/(a^2*b*x^2*(-2*a*b^3 - 2*b^4*x)*Sqrt[a^2 + 2*a*b
*x + b^2*x^2] + a^2*b*Sqrt[b^2]*x^2*(2*a^2*b^2 + 4*a*b^3*x + 2*b^4*x^2)) + (2*A*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt
[a^2 + 2*a*b*x + b^2*x^2]/a])/a^3

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 109, normalized size = 0.78 \begin {gather*} \frac {2 \, A a b^{2} x - B a^{3} + 3 \, A a^{2} b - 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \left (b x + a\right ) + 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*a*b^2*x - B*a^3 + 3*A*a^2*b - 2*(A*b^3*x^2 + 2*A*a*b^2*x + A*a^2*b)*log(b*x + a) + 2*(A*b^3*x^2 + 2*A
*a*b^2*x + A*a^2*b)*log(x))/(a^3*b^3*x^2 + 2*a^4*b^2*x + a^5*b)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.07, size = 116, normalized size = 0.83 \begin {gather*} -\frac {\left (-2 A \,b^{3} x^{2} \ln \relax (x )+2 A \,b^{3} x^{2} \ln \left (b x +a \right )-4 A a \,b^{2} x \ln \relax (x )+4 A a \,b^{2} x \ln \left (b x +a \right )-2 A \,a^{2} b \ln \relax (x )+2 A \,a^{2} b \ln \left (b x +a \right )-2 A a \,b^{2} x -3 A \,a^{2} b +B \,a^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*A*b^3*x^2*ln(b*x+a)-2*A*ln(x)*x^2*b^3+4*A*a*b^2*x*ln(b*x+a)-4*A*ln(x)*x*a*b^2+2*A*a^2*b*ln(b*x+a)-2*A*
ln(x)*a^2*b-2*A*a*b^2*x-3*A*a^2*b+B*a^3)*(b*x+a)/b/a^3/((b*x+a)^2)^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 96, normalized size = 0.69 \begin {gather*} -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} - \frac {B}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {A}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-1)^(2*a*b*x + 2*a^2)*A*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 + A/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2) - 1/2
*B/(b^3*(x + a/b)^2) + 1/2*A/(a*b^2*(x + a/b)^2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________