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

Optimal. Leaf size=105 \[ \frac{x \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{a A \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0400829, antiderivative size = 105, 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} \[ \frac{x \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{a A \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{b B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0213964, size = 44, normalized size = 0.42 \[ \frac{\sqrt{(a+b x)^2} (x (2 a B+2 A b+b B x)+2 a A \log (x))}{2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.036, size = 53, normalized size = 0.5 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ({b}^{2}B{x}^{2}+2\,Aab\ln \left ( bx \right ) +2\,A{b}^{2}x+2\,abBx+2\,Aab+B{a}^{2} \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*csgn(b*x+a)*(b^2*B*x^2+2*A*a*b*ln(b*x)+2*A*b^2*x+2*a*b*B*x+2*A*a*b+B*a^2)/b

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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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35756, size = 57, normalized size = 0.54 \begin{align*} \frac{1}{2} \, B b x^{2} + A a \log \left (x\right ) +{\left (B a + A b\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,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.457625, size = 22, normalized size = 0.21 \begin{align*} A a \log{\left (x \right )} + \frac{B b x^{2}}{2} + x \left (A b + B a\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,x)

[Out]

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

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Giac [A]  time = 1.21522, size = 62, normalized size = 0.59 \begin{align*} \frac{1}{2} \, B b x^{2} \mathrm{sgn}\left (b x + a\right ) + B a x \mathrm{sgn}\left (b x + a\right ) + A b x \mathrm{sgn}\left (b x + a\right ) + A a \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\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,x, algorithm="giac")

[Out]

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