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

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Rubi [A]  time = 0.04, antiderivative size = 105, 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, 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 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])

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) \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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 1.12, size = 22, normalized size = 0.21 \[ \frac {1}{2} \, B b x^{2} + A a \log \relax (x) + {\left (B a + A b\right )} x \]

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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giac [A]  time = 0.16, size = 46, normalized size = 0.44 \[ \frac {1}{2} \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + B a x \mathrm {sgn}\left (b x + a\right ) + A b x \mathrm {sgn}\left (b x + a\right ) + A a \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \]

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)

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maple [C]  time = 0.07, size = 53, normalized size = 0.50 \[ \frac {\left (B \,b^{2} x^{2}+2 A a b \ln \left (b x \right )+2 A \,b^{2} x +2 B a b x +2 A a b +B \,a^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{2 b} \]

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

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maxima [A]  time = 0.51, size = 133, normalized size = 1.27 \[ \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B x + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a}{2 \, b} \]

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]

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

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mupad [B]  time = 1.37, size = 122, normalized size = 1.16 \[ A\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}-A\,\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )\,\sqrt {a^2}+\frac {B\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{2\,b}+\frac {A\,a\,b\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )}{\sqrt {b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 22, normalized size = 0.21 \[ A a \log {\relax (x )} + \frac {B b x^{2}}{2} + x \left (A b + B a\right ) \]

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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