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

Optimal. Leaf size=166 \[ \frac{x^2 (a+b x) (A b-a B)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x (a+b x) (A b-a B)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) (A b-a B) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^3 (a+b x)}{3 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-((a*(A*b - a*B)*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + ((A*b - a*B)*x^2*(a + b*x))/(2*b^2*Sqrt[a
^2 + 2*a*b*x + b^2*x^2]) + (B*x^3*(a + b*x))/(3*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (a^2*(A*b - a*B)*(a + b*x)*
Log[a + b*x])/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0897315, antiderivative size = 166, normalized size of antiderivative = 1., 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} \[ \frac{x^2 (a+b x) (A b-a B)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a x (a+b x) (A b-a B)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) (A b-a B) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^3 (a+b x)}{3 b \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(A*b - a*B)*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + ((A*b - a*B)*x^2*(a + b*x))/(2*b^2*Sqrt[a
^2 + 2*a*b*x + b^2*x^2]) + (B*x^3*(a + b*x))/(3*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (a^2*(A*b - a*B)*(a + b*x)*
Log[a + b*x])/(b^4*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^2 (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^2 (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 (-A b+a B)}{b^4}+\frac{(A b-a B) x}{b^3}+\frac{B x^2}{b^2}-\frac{a^2 (-A b+a B)}{b^4 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a (A b-a B) x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^2 (a+b x)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x^3 (a+b x)}{3 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (A b-a B) (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0378173, size = 77, normalized size = 0.46 \[ \frac{(a+b x) \left (b x \left (6 a^2 B-3 a b (2 A+B x)+b^2 x (3 A+2 B x)\right )+6 a^2 (A b-a B) \log (a+b x)\right )}{6 b^4 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*x*(6*a^2*B - 3*a*b*(2*A + B*x) + b^2*x*(3*A + 2*B*x)) + 6*a^2*(A*b - a*B)*Log[a + b*x]))/(6*b^4*
Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.009, size = 90, normalized size = 0.5 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 2\,{b}^{3}B{x}^{3}+3\,A{b}^{3}{x}^{2}-3\,B{x}^{2}a{b}^{2}+6\,A\ln \left ( bx+a \right ){a}^{2}b-6\,Aa{b}^{2}x-6\,B\ln \left ( bx+a \right ){a}^{3}+6\,B{a}^{2}bx \right ) }{6\,{b}^{4}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x+a)*(2*b^3*B*x^3+3*A*b^3*x^2-3*B*x^2*a*b^2+6*A*ln(b*x+a)*a^2*b-6*A*a*b^2*x-6*B*ln(b*x+a)*a^3+6*B*a^2*b
*x)/((b*x+a)^2)^(1/2)/b^4

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Maxima [A]  time = 1.0267, size = 225, normalized size = 1.36 \begin{align*} -\frac{5 \, B a^{3} b \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{A a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{5 \, B a^{2} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{A a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{A x^{2}}{2 \, \sqrt{b^{2}}} - \frac{5 \, B a x^{2}}{6 \, \sqrt{b^{2}} b} + \frac{2 \, B a^{3} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B x^{2}}{3 \, b^{2}} - \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2}}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/3*B*a^3*b*log(x + a/b)/(b^2)^(5/2) + A*a^2*b^2*log(x + a/b)/(b^2)^(5/2) + 5/3*B*a^2*x/(b^2)^(3/2) - A*a*b*x
/(b^2)^(3/2) + 1/2*A*x^2/sqrt(b^2) - 5/6*B*a*x^2/(sqrt(b^2)*b) + 2/3*B*a^3*sqrt(b^(-2))*log(x + a/b)/b^3 + 1/3
*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*x^2/b^2 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^2/b^4

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Fricas [A]  time = 1.63262, size = 149, normalized size = 0.9 \begin{align*} \frac{2 \, B b^{3} x^{3} - 3 \,{\left (B a b^{2} - A b^{3}\right )} x^{2} + 6 \,{\left (B a^{2} b - A a b^{2}\right )} x - 6 \,{\left (B a^{3} - A a^{2} b\right )} \log \left (b x + a\right )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^3*x^3 - 3*(B*a*b^2 - A*b^3)*x^2 + 6*(B*a^2*b - A*a*b^2)*x - 6*(B*a^3 - A*a^2*b)*log(b*x + a))/b^4

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Sympy [A]  time = 0.405314, size = 58, normalized size = 0.35 \begin{align*} \frac{B x^{3}}{3 b} - \frac{a^{2} \left (- A b + B a\right ) \log{\left (a + b x \right )}}{b^{4}} - \frac{x^{2} \left (- A b + B a\right )}{2 b^{2}} + \frac{x \left (- A a b + B a^{2}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x**3/(3*b) - a**2*(-A*b + B*a)*log(a + b*x)/b**4 - x**2*(-A*b + B*a)/(2*b**2) + x*(-A*a*b + B*a**2)/b**3

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Giac [A]  time = 1.20743, size = 153, normalized size = 0.92 \begin{align*} \frac{2 \, B b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, A b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} x \mathrm{sgn}\left (b x + a\right ) - 6 \, A a b x \mathrm{sgn}\left (b x + a\right )}{6 \, b^{3}} - \frac{{\left (B a^{3} \mathrm{sgn}\left (b x + a\right ) - A a^{2} b \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^2*x^3*sgn(b*x + a) - 3*B*a*b*x^2*sgn(b*x + a) + 3*A*b^2*x^2*sgn(b*x + a) + 6*B*a^2*x*sgn(b*x + a) -
6*A*a*b*x*sgn(b*x + a))/b^3 - (B*a^3*sgn(b*x + a) - A*a^2*b*sgn(b*x + a))*log(abs(b*x + a))/b^4