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

Optimal. Leaf size=162 \[ \frac {(a+b x) (A b-a B)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b \log (x) (a+b x) (A b-a B)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) (A b-a B) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 162, 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, 77} \[ \frac {(a+b x) (A b-a B)}{a^2 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b \log (x) (a+b x) (A b-a B)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) (A b-a B) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{2 a x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.49 \[ -\frac {(a+b x) \left (2 b x^2 \log (x) (a B-A b)+2 b x^2 (A b-a B) \log (a+b x)+a (a A+2 a B x-2 A b x)\right )}{2 a^3 x^2 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + b*x)*(a*(a*A - 2*A*b*x + 2*a*B*x) + 2*b*(-(A*b) + a*B)*x^2*Log[x] + 2*b*(A*b - a*B)*x^2*Log[a + b*x
]))/(a^3*x^2*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.87, size = 69, normalized size = 0.43 \[ \frac {2 \, {\left (B a b - A b^{2}\right )} x^{2} \log \left (b x + a\right ) - 2 \, {\left (B a b - A b^{2}\right )} x^{2} \log \relax (x) - A a^{2} - 2 \, {\left (B a^{2} - A a b\right )} x}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 117, normalized size = 0.72 \[ -\frac {{\left (B a b \mathrm {sgn}\left (b x + a\right ) - A b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {{\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{3} b} - \frac {A a^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (B a^{2} \mathrm {sgn}\left (b x + a\right ) - A a b \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 92, normalized size = 0.57 \[ -\frac {\left (b x +a \right ) \left (-2 A \,b^{2} x^{2} \ln \relax (x )+2 A \,b^{2} x^{2} \ln \left (b x +a \right )+2 B a b \,x^{2} \ln \relax (x )-2 B a b \,x^{2} \ln \left (b x +a \right )-2 A a b x +2 B \,a^{2} x +A \,a^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 164, normalized size = 1.01 \[ \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{2}} - \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{a^{2} x} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x}{x^3\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 131, normalized size = 0.81 \[ \frac {- A a + x \left (2 A b - 2 B a\right )}{2 a^{2} x^{2}} - \frac {b \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{2} + B a^{2} b - a b \left (- A b + B a\right )}{- 2 A b^{3} + 2 B a b^{2}} \right )}}{a^{3}} + \frac {b \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{2} + B a^{2} b + a b \left (- A b + B a\right )}{- 2 A b^{3} + 2 B a b^{2}} \right )}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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