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

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

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Rubi [A]  time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.50 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.88, size = 480, normalized size = 4.25 \begin {gather*} \frac {\frac {2 \sqrt {b^2} B x \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )}{a}-\frac {2 B \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )}{a}-\frac {2 A b}{a}}{\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}-\frac {2 A b \left (\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x\right )^2 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^2 \left (-2 \sqrt {b^2} x \sqrt {a^2+2 a b x+b^2 x^2}+a^2+2 a b x+2 b^2 x^2\right )}+\frac {-2 A \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^2+4 a b x+4 b^2 x^2\right )-2 A \left (-a^3 b-5 a^2 b^2 x-8 a b^3 x^2-4 b^4 x^3\right )}{\sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^3 b x-12 a^2 b^2 x^2-24 a b^3 x^3-16 b^4 x^4\right )+\sqrt {b^2} \left (2 a^4 x+14 a^3 b x^2+36 a^2 b^2 x^3+40 a b^3 x^4+16 b^4 x^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a^2 + 4*a*b*x + 4*b^2*x^2) - 2*A*(-(a^3*b) - 5*a^2*b^2*x - 8*a*
b^3*x^2 - 4*b^4*x^3))/(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^3*b*x - 12*a^2*b^2*x^2 - 24*a*b^3*x^3 - 16*b^4*x^4)
+ Sqrt[b^2]*(2*a^4*x + 14*a^3*b*x^2 + 36*a^2*b^2*x^3 + 40*a*b^3*x^4 + 16*b^4*x^5)) + ((-2*A*b)/a + (2*Sqrt[b^
2]*B*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/a - (2*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Arc
Tanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/a)/(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])
- (2*A*b*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2
*x^2]/a])/(a^2*(a^2 + 2*a*b*x + 2*b^2*x^2 - 2*Sqrt[b^2]*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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fricas [A]  time = 0.43, size = 41, normalized size = 0.36 \begin {gather*} -\frac {{\left (B a - A b\right )} x \log \left (b x + a\right ) - {\left (B a - A b\right )} x \log \relax (x) + A a}{a^{2} x} \end {gather*}

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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giac [A]  time = 0.16, size = 81, normalized size = 0.72 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 61, normalized size = 0.54 \begin {gather*} \frac {\left (b x +a \right ) \left (-A b x \ln \relax (x )+A b x \ln \left (b x +a \right )+B a x \ln \relax (x )-B a x \ln \left (b x +a \right )-A a \right )}{\sqrt {\left (b x +a \right )^{2}}\, a^{2} x} \end {gather*}

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

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maxima [A]  time = 0.51, size = 106, normalized size = 0.94 \begin {gather*} -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a} + \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{a^{2} x} \end {gather*}

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]

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

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mupad [B]  time = 1.49, size = 117, normalized size = 1.04 \begin {gather*} \frac {A\,a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}\right )}{{\left (a^2\right )}^{3/2}}-\frac {A\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{a^2\,x}-\frac {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)/(x^2*((a + b*x)^2)^(1/2)),x)

[Out]

(A*a*b*atanh((a^2 + a*b*x)/((a^2)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))))/(a^2)^(3/2) - (A*(a^2 + b^2*x^2 + 2
*a*b*x)^(1/2))/(a^2*x) - (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 = 0.40, size = 95, normalized size = 0.84 \begin {gather*} - \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 {gather*}

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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