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

Optimal. Leaf size=108 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{x (a+b x)}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}+\frac{b B \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.043048, antiderivative size = 108, 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, 76} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{x (a+b x)}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}+\frac{b B \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0184731, size = 48, normalized size = 0.44 \[ -\frac{\sqrt{(a+b x)^2} \left (a (A+2 B x)+2 A b x-2 b B x^2 \log (x)\right )}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.011, size = 37, normalized size = 0.3 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) \left ( -2\,B\ln \left ( bx \right ) b{x}^{2}+2\,Abx+2\,aBx+aA \right ) }{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x)

[Out]

-1/2*csgn(b*x+a)*(-2*B*ln(b*x)*b*x^2+2*A*b*x+2*a*B*x+a*A)/x^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)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27811, size = 70, normalized size = 0.65 \begin{align*} \frac{2 \, B b x^{2} \log \left (x\right ) - A a - 2 \,{\left (B a + A b\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.450945, size = 26, normalized size = 0.24 \begin{align*} B b \log{\left (x \right )} - \frac{A a + x \left (2 A b + 2 B a\right )}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**3,x)

[Out]

B*b*log(x) - (A*a + x*(2*A*b + 2*B*a))/(2*x**2)

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Giac [A]  time = 1.18057, size = 68, normalized size = 0.63 \begin{align*} B b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{A a \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (B a \mathrm{sgn}\left (b x + a\right ) + A b \mathrm{sgn}\left (b x + a\right )\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

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