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

Optimal. Leaf size=120 \[ \frac{x (a+b x) (A b-a B)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a (a+b x) (A b-a B) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0609196, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {770, 77} \[ \frac{x (a+b x) (A b-a B)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a (a+b x) (A b-a B) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]))))

Rubi steps

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

Mathematica [A]  time = 0.0267332, size = 57, normalized size = 0.48 \[ \frac{(a+b x) (b x (-2 a B+2 A b+b B x)+2 a (a B-A b) \log (a+b x))}{2 b^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 66, normalized size = 0.6 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -{b}^{2}B{x}^{2}+2\,A\ln \left ( bx+a \right ) ab-2\,A{b}^{2}x-2\,B\ln \left ( bx+a \right ){a}^{2}+2\,abBx \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.998103, size = 117, normalized size = 0.98 \begin{align*} \frac{B a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{B a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{B x^{2}}{2 \, \sqrt{b^{2}}} - \frac{A 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}} A}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^2*b^2*log(x + a/b)/(b^2)^(5/2) - B*a*b*x/(b^2)^(3/2) + 1/2*B*x^2/sqrt(b^2) - A*a*sqrt(b^(-2))*log(x + a/b)
/b + sqrt(b^2*x^2 + 2*a*b*x + a^2)*A/b^2

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Fricas [A]  time = 1.6241, size = 103, normalized size = 0.86 \begin{align*} \frac{B b^{2} x^{2} - 2 \,{\left (B a b - A b^{2}\right )} x + 2 \,{\left (B a^{2} - A a b\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.375121, size = 37, normalized size = 0.31 \begin{align*} \frac{B x^{2}}{2 b} + \frac{a \left (- A b + B a\right ) \log{\left (a + b x \right )}}{b^{3}} - \frac{x \left (- A b + B a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3069, size = 101, normalized size = 0.84 \begin{align*} \frac{B b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, B a x \mathrm{sgn}\left (b x + a\right ) + 2 \, A b x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} + \frac{{\left (B a^{2} \mathrm{sgn}\left (b x + a\right ) - A a b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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