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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.0885879, antiderivative size = 140, normalized size of antiderivative = 1., 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} \[ \frac{A b-a B}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A \log (x) (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x) \log (a+b x)}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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]

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]))))

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 = 0.0362897, size = 80, normalized size = 0.57 \[ \frac{a \left (a^2 (-B)+3 a A b+2 A b^2 x\right )+2 A b \log (x) (a+b x)^2-2 A b (a+b x)^2 \log (a+b x)}{2 a^3 b (a+b x) \sqrt{(a+b x)^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 117, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,A\ln \left ( x \right ){x}^{2}{b}^{3}-2\,A\ln \left ( bx+a \right ){x}^{2}{b}^{3}+4\,A\ln \left ( x \right ) xa{b}^{2}-4\,A\ln \left ( bx+a \right ) xa{b}^{2}+2\,A\ln \left ( x \right ){a}^{2}b-2\,A\ln \left ( bx+a \right ){a}^{2}b+2\,Aa{b}^{2}x+3\,Ab{a}^{2}-B{a}^{3} \right ) \left ( bx+a \right ) }{2\,b{a}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.52404, size = 236, normalized size = 1.69 \begin{align*} \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 \left (x\right )}{2 \,{\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \end{align*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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