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

Optimal. Leaf size=113 \[ -\frac{\log (x) (a+b x) (A b-a B)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-a B) \log (a+b x)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x} \]

[Out]

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

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Rubi [A]  time = 0.0648089, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {769, 646, 36, 29, 31} \[ -\frac{\log (x) (a+b x) (A b-a B)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-a B) \log (a+b x)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{A+B x}{x^2 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x}-\frac{\left (2 A b^2-2 a b B\right ) \int \frac{1}{x \sqrt{a^2+2 a b x+b^2 x^2}} \, dx}{2 a b}\\ &=-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x}-\frac{\left (\left (2 A b^2-2 a b B\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{x \left (a b+b^2 x\right )} \, dx}{2 a b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x}+\frac{\left (\left (2 A b^2-2 a b B\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{a b+b^2 x} \, dx}{2 a^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (\left (2 A b^2-2 a b B\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{x} \, dx}{2 a^2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{A \sqrt{a^2+2 a b x+b^2 x^2}}{a^2 x}-\frac{(A b-a B) (a+b x) \log (x)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) (a+b x) \log (a+b x)}{a^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0292133, size = 57, normalized size = 0.5 \[ \frac{(a+b x) (\log (x) (a B x-A b x)+x (A b-a B) \log (a+b x)-a A)}{a^2 x \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 61, normalized size = 0.5 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( A\ln \left ( x \right ) xb-A\ln \left ( bx+a \right ) xb-B\ln \left ( x \right ) xa+B\ln \left ( bx+a \right ) xa+aA \right ) }{x{a}^{2}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64044, size = 92, normalized size = 0.81 \begin{align*} -\frac{{\left (B a - A b\right )} x \log \left (b x + a\right ) -{\left (B a - A b\right )} x \log \left (x\right ) + A a}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.561037, size = 95, normalized size = 0.84 \begin{align*} - \frac{A}{a x} + \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b + B a^{2} - a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} - \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b + B a^{2} + a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22742, size = 109, normalized size = 0.96 \begin{align*} \frac{{\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 )}{a^{2}} - \frac{A \mathrm{sgn}\left (b x + a\right )}{a x} - \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 | b x + a \right |}\right )}{a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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