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

Optimal. Leaf size=103 \[ \frac{\log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b B x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.042111, antiderivative size = 103, 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} \[ \frac{\log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{b B x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0185067, size = 44, normalized size = 0.43 \[ \frac{\sqrt{(a+b x)^2} \left (x \log (x) (a B+A b)-a A+b B x^2\right )}{x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.011, size = 42, normalized size = 0.4 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ( A\ln \left ( bx \right ) xb+B\ln \left ( bx \right ) xa+Bb{x}^{2}+aBx-aA \right ) }{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.31371, size = 57, normalized size = 0.55 \begin{align*} \frac{B b x^{2} +{\left (B a + A b\right )} x \log \left (x\right ) - A a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.549306, size = 19, normalized size = 0.18 \begin{align*} - \frac{A a}{x} + B b x + \left (A b + B a\right ) \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a/x + B*b*x + (A*b + B*a)*log(x)

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Giac [A]  time = 1.31414, size = 63, normalized size = 0.61 \begin{align*} B b x \mathrm{sgn}\left (b x + a\right ) +{\left (B a \mathrm{sgn}\left (b x + a\right ) + A b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac{A a \mathrm{sgn}\left (b x + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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