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

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

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Rubi [A]  time = 0.04, antiderivative size = 108, 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} \begin {gather*} -\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} \end {gather*}

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 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^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*}

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.44 \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[(a + b*x)^2]*(2*A*b*x + a*(A + 2*B*x) - 2*b*B*x^2*Log[x]))/(x^2*(a + b*x))

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IntegrateAlgebraic [B]  time = 1.22, size = 270, normalized size = 2.50 \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-a A b-2 a b B x-2 A b^2 x\right )+\sqrt {b^2} \left (a^2 A+2 a^2 B x+3 a A b x+2 a b B x^2+2 A b^2 x^2\right )}{2 x^2 \left (a b+b^2 x\right )-2 \sqrt {b^2} x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2} \sqrt {b^2} B \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {1}{2} \sqrt {b^2} B \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+b B \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(a*A*b) - 2*A*b^2*x - 2*a*b*B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + Sqrt[b^2]*(a^2*A + 3*a*A*b*x + 2*a^2*B*x +
2*A*b^2*x^2 + 2*a*b*B*x^2))/(2*x^2*(a*b + b^2*x) - 2*Sqrt[b^2]*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + b*B*ArcTa
nh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - (Sqrt[b^2]*B*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/2 - (Sqrt[b^2]*B*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2

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fricas [A]  time = 0.41, size = 29, normalized size = 0.27 \begin {gather*} \frac {2 \, B b x^{2} \log \relax (x) - A a - 2 \, {\left (B a + A b\right )} x}{2 \, x^{2}} \end {gather*}

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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giac [A]  time = 0.19, size = 50, normalized size = 0.46 \begin {gather*} 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 {gather*}

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

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maple [C]  time = 0.06, size = 37, normalized size = 0.34 \begin {gather*} -\frac {\left (-2 B b \,x^{2} \ln \left (b x \right )+2 A b x +2 B a x +A a \right ) \mathrm {csgn}\left (b x +a \right )}{2 x^{2}} \end {gather*}

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

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maxima [B]  time = 0.55, size = 172, normalized size = 1.59 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{2 \, a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{x} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{2 \, a^{2} x^{2}} \end {gather*}

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]

(-1)^(2*b^2*x + 2*a*b)*B*b*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*B*b*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)*A*b^2/a^2 - sqrt(b^2*x^2 + 2*a*b*x + a^2)*B/x + 1/2*sqrt(b^2*x^2 + 2*a*b
*x + a^2)*A*b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/(a^2*x^2)

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mupad [B]  time = 1.33, size = 134, normalized size = 1.24 \begin {gather*} B\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )\,\sqrt {b^2}-\frac {B\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}-\frac {A\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+2\,b\,x\right )}{2\,x^2\,\left (a+b\,x\right )}-\frac {B\,a\,b\,\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.25, size = 27, normalized size = 0.25 \begin {gather*} B b \log {\relax (x )} + \frac {- A a + x \left (- 2 A b - 2 B a\right )}{2 x^{2}} \end {gather*}

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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