### 3.183 $$\int \frac{x^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx$$

Optimal. Leaf size=106 $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

-((a*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (x^2*(a + b*x))/(2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(a^2*(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 = 0.0371791, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {646, 43} $-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^2}{\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+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^3}+\frac{x}{b^2}+\frac{a^2}{b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{a x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^2 (a+b x)}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^2 (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 = 0.0149632, size = 45, normalized size = 0.42 $\frac{(a+b x) \left (2 a^2 \log (a+b x)+b x (b x-2 a)\right )}{2 b^3 \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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Maple [A]  time = 0.222, size = 44, normalized size = 0.4 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,{a}^{2}\ln \left ( bx+a \right ) -2\,abx \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^2/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.25525, size = 55, normalized size = 0.52 \begin{align*} \frac{a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{x^{2}}{2 \, \sqrt{b^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65108, size = 68, normalized size = 0.64 \begin{align*} \frac{b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.08682, size = 26, normalized size = 0.25 \begin{align*} \frac{a^{2} \log{\left (a + b x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26573, size = 65, normalized size = 0.61 \begin{align*} \frac{a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{3}} + \frac{b x^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a x \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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