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

Optimal. Leaf size=69 \[ \frac{(a+b x) (A b-a B) \log (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0247329, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {640, 608, 31} \[ \frac{(a+b x) (A b-a B) \log (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B \sqrt{a^2+2 a b x+b^2 x^2}}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 640

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

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A]  time = 0.0168415, size = 40, normalized size = 0.58 \[ \frac{(a+b x) ((A b-a B) \log (a+b x)+b B x)}{b^2 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 43, normalized size = 0.6 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( A\ln \left ( bx+a \right ) b-B\ln \left ( bx+a \right ) a+bBx \right ) }{{b}^{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)/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.04469, size = 80, normalized size = 1.16 \begin{align*} A \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right ) - \frac{B a \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{b} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6324, size = 54, normalized size = 0.78 \begin{align*} \frac{B b x -{\left (B a - A b\right )} \log \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.354707, size = 20, normalized size = 0.29 \begin{align*} \frac{B x}{b} - \frac{\left (- A b + B a\right ) \log{\left (a + b x \right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15999, size = 61, normalized size = 0.88 \begin{align*} \frac{B x \mathrm{sgn}\left (b x + a\right )}{b} - \frac{{\left (B a \mathrm{sgn}\left (b x + a\right ) - A b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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