3.2.70 \(\int \frac {x^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=99 \[ -\frac {a^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 99, 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^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(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/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(2*a)/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - a^2/(2*b^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((a + b*x)*L
og[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}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {x^2}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {a^2}{b^5 (a+b x)^3}-\frac {2 a}{b^5 (a+b x)^2}+\frac {1}{b^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 a}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(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 = 0.02, size = 51, normalized size = 0.52 \begin {gather*} \frac {a (3 a+4 b x)+2 (a+b x)^2 \log (a+b x)}{2 b^3 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 1.15, size = 1462, normalized size = 14.77 \begin {gather*} \frac {\frac {4 \sqrt {b^2} a^4}{b^4}+\frac {12 \sqrt {b^2} x a^3}{b^3}+\frac {12 \left (b^2\right )^{3/2} x^2 a^2}{b^4}-\frac {4 \left (b^2\right )^{3/2} x^2 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{b^4}-\frac {4 \left (b^2\right )^{3/2} x^2 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{b^4}-\frac {12 x \sqrt {a^2+2 b x a+b^2 x^2} a^2}{b^2}-\frac {8 \sqrt {b^2} x^3 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}+\frac {4 x^2 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}-\frac {8 \sqrt {b^2} x^3 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}+\frac {4 x^2 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}-4 \sqrt {b^2} x^4 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )+4 x^3 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )-4 \sqrt {b^2} x^4 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )+4 x^3 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2 \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2}+\frac {4 b \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^4-4 b \log \left (-a b^3-\sqrt {b^2} x b^3+\sqrt {a^2+2 b x a+b^2 x^2} b^3\right ) x^4+8 a \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^3-\frac {4 b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^3}{\sqrt {b^2}}-8 a \log \left (-a b^3-\sqrt {b^2} x b^3+\sqrt {a^2+2 b x a+b^2 x^2} b^3\right ) x^3+\frac {4 b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a b^3-\sqrt {b^2} x b^3+\sqrt {a^2+2 b x a+b^2 x^2} b^3\right ) x^3}{\sqrt {b^2}}-\frac {16 a b x^3}{\sqrt {b^2}}-\frac {4 a \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{\sqrt {b^2}}+\frac {4 a^2 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{b}+\frac {4 a \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a b^3-\sqrt {b^2} x b^3+\sqrt {a^2+2 b x a+b^2 x^2} b^3\right ) x^2}{\sqrt {b^2}}-\frac {4 a^2 \log \left (-a b^3-\sqrt {b^2} x b^3+\sqrt {a^2+2 b x a+b^2 x^2} b^3\right ) x^2}{b}+\frac {16 a \sqrt {a^2+2 b x a+b^2 x^2} x^2}{b}-\frac {24 a^2 x^2}{\sqrt {b^2}}+\frac {8 a^2 \sqrt {a^2+2 b x a+b^2 x^2} x}{b^2}-\frac {12 a^3 x}{b \sqrt {b^2}}+\frac {4 a^3 \sqrt {a^2+2 b x a+b^2 x^2}}{b^3}}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2 \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((4*a^4*Sqrt[b^2])/b^4 + (12*a^3*Sqrt[b^2]*x)/b^3 + (12*a^2*(b^2)^(3/2)*x^2)/b^4 - (12*a^2*x*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/b^2 - (4*a^2*(b^2)^(3/2)*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*S
qrt[b^2]*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b - 4*Sqrt[b^2]*x^4*Log[-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + (4*a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a
*b*x + b^2*x^2]])/b + 4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]
] - (4*a^2*(b^2)^(3/2)*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*Sqrt[b^2]*x^3*Log[
a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b - 4*Sqrt[b^2]*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]] + (4*a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b
+ 4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((-12*a^3*x)/(b*Sqr
t[b^2]) - (24*a^2*x^2)/Sqrt[b^2] - (16*a*b*x^3)/Sqrt[b^2] + (4*a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (8*a^2
*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (16*a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (4*a^2*x^2*Log[a - Sqrt[b
^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 8*a*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + 4*b
*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (4*a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqr
t[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (4*b*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (4*a^2*x^2*Log[-(a*b^3) - b^3*Sqrt[b^2]*x + b^3*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]])/b - 8*a*x^3*Log[-(a*b^3) - b^3*Sqrt[b^2]*x + b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - 4*b*x^4*
Log[-(a*b^3) - b^3*Sqrt[b^2]*x + b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + (4*a*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*L
og[-(a*b^3) - b^3*Sqrt[b^2]*x + b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (4*b*x^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]*Log[-(a*b^3) - b^3*Sqrt[b^2]*x + b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2])/((-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)

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fricas [A]  time = 0.40, size = 61, normalized size = 0.62 \begin {gather*} \frac {4 \, a b x + 3 \, a^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 67, normalized size = 0.68 \begin {gather*} \frac {\left (2 b^{2} x^{2} \ln \left (b x +a \right )+4 a b x \ln \left (b x +a \right )+2 a^{2} \ln \left (b x +a \right )+4 a b x +3 a^{2}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(2*b^2*x^2*ln(b*x+a)+4*ln(b*x+a)*x*a*b+2*a^2*ln(b*x+a)+4*a*b*x+3*a^2)*(b*x+a)/b^3/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.38, size = 46, normalized size = 0.46 \begin {gather*} \frac {\log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

log(x + a/b)/b^3 + 2*a*x/(b^4*(x + a/b)^2) + 3/2*a^2/(b^5*(x + a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(x^2/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**2/((a + b*x)**2)**(3/2), x)

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