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

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

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Rubi [A]  time = 0.04, antiderivative size = 106, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 {gather*}

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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 45, normalized size = 0.42 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.27, size = 168, normalized size = 1.58 \begin {gather*} -\frac {a^2 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b^4}-\frac {a^2 \left (\sqrt {b^2}-b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b^4}+\frac {(b x-3 a) \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}+\frac {2 a x-b x^2}{4 b \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 29, normalized size = 0.27 \begin {gather*} \frac {b^{2} x^{2} - 2 \, a b x + 2 \, a^{2} \log \left (b x + a\right )}{2 \, b^{3}} \end {gather*}

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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giac [A]  time = 0.18, size = 48, normalized size = 0.45 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.05, size = 44, normalized size = 0.42 \begin {gather*} \frac {\left (b x +a \right ) \left (b^{2} x^{2}+2 a^{2} \ln \left (b x +a \right )-2 a b x \right )}{2 \sqrt {\left (b x +a \right )^{2}}\, b^{3}} \end {gather*}

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 = 1.40, size = 31, normalized size = 0.29 \begin {gather*} \frac {x^{2}}{2 \, b} - \frac {a x}{b^{2}} + \frac {a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x)^2)^(1/2),x)

[Out]

int(x^2/((a + b*x)^2)^(1/2), x)

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sympy [A]  time = 0.14, size = 26, normalized size = 0.25 \begin {gather*} \frac {a^{2} \log {\left (a + b x \right )}}{b^{3}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b} \end {gather*}

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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