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

Optimal. Leaf size=133 \[ -\frac {3 a^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 133, 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^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2)/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + a^3/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (x*(a + b
*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(a + b*x)*Log[a + b*x])/(b^4*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^3}{\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^3}{\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 {1}{b^6}-\frac {a^3}{b^6 (a+b x)^3}+\frac {3 a^2}{b^6 (a+b x)^2}-\frac {3 a}{b^6 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {3 a^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 0.53 \begin {gather*} \frac {-5 a^3-4 a^2 b x+4 a b^2 x^2-6 a (a+b x)^2 \log (a+b x)+2 b^3 x^3}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^3 - 4*a^2*b*x + 4*a*b^2*x^2 + 2*b^3*x^3 - 6*a*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a + b*x)*Sqrt[(a + b*x)^
2])

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IntegrateAlgebraic [B]  time = 1.05, size = 1278, normalized size = 9.61 \begin {gather*} \frac {-8 \sqrt {b^2} x^5-24 a \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^4-\frac {20 a \sqrt {b^2} x^4}{b}+8 \sqrt {a^2+2 b x a+b^2 x^2} x^4+\frac {8 a^2 \left (b^2\right )^{3/2} x^3}{b^4}+\frac {24 a \left (b^2\right )^{3/2} \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^3}{b^4}-\frac {48 a^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^3}{b}+\frac {12 a \sqrt {a^2+2 b x a+b^2 x^2} x^3}{b}+\frac {24 a^2 \sqrt {b^2} \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^2}{b^3}-\frac {24 a^3 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^2}{b^2}+\frac {32 a^3 \sqrt {b^2} x^2}{b^3}-\frac {20 a^2 \sqrt {a^2+2 b x a+b^2 x^2} x^2}{b^2}+\frac {16 a^4 \sqrt {b^2} x}{b^4}-\frac {12 a^3 \sqrt {a^2+2 b x a+b^2 x^2} x}{b^3}-\frac {4 a^4 \sqrt {a^2+2 b x a+b^2 x^2}}{b^4}}{\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 {-\frac {4 a^5}{b^3 \sqrt {b^2}}-\frac {16 x a^4}{\left (b^2\right )^{3/2}}-\frac {16 x^2 a^3}{b \sqrt {b^2}}+\frac {12 x^2 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^3}{b \sqrt {b^2}}+\frac {12 x^2 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^3}{b \sqrt {b^2}}+\frac {16 x \sqrt {a^2+2 b x a+b^2 x^2} a^3}{b^3}+\frac {24 x^3 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{\sqrt {b^2}}-\frac {12 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^2}{b^2}+\frac {24 x^3 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{\sqrt {b^2}}-\frac {12 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^2}{b^2}+\frac {12 b x^4 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{\sqrt {b^2}}-\frac {12 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 ) a}{b}+\frac {12 b x^4 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{\sqrt {b^2}}-\frac {12 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 ) a}{b}}{\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^3/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((16*a^4*Sqrt[b^2]*x)/b^4 + (32*a^3*Sqrt[b^2]*x^2)/b^3 + (8*a^2*(b^2)^(3/2)*x^3)/b^4 - (20*a*Sqrt[b^2]*x^4)/b
- 8*Sqrt[b^2]*x^5 - (4*a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (12*a^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 -
(20*a^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (12*a*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + 8*x^4*Sqrt[a^2
+ 2*a*b*x + b^2*x^2] - (24*a^3*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (48*a^2*
x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 24*a*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/a] + (24*a^2*Sqrt[b^2]*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^3 + (24*a*(b^2)^(3/2)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2
]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4)/((-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) + ((-4*a^5)/(b^3*Sqrt[b^2]) - (16*a^4*x)/(b^2)^(3/2) - (16*a^3*x^2
)/(b*Sqrt[b^2]) + (16*a^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (12*a^3*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[
b^2] + (12*a*b*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*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^2 - (12*a*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]])/b + (12*a^3*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt
[b^2] + (12*a*b*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*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^2 - (12*a*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]])/b)/((-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 = 83, normalized size = 0.62 \begin {gather*} \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3/(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 = 0.06, size = 89, normalized size = 0.67 \begin {gather*} -\frac {\left (6 a \,b^{2} x^{2} \ln \left (b x +a \right )-2 b^{3} x^{3}+12 a^{2} b x \ln \left (b x +a \right )-4 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )+4 a^{2} b x +5 a^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*ln(b*x+a)*x^2*a*b^2-2*b^3*x^3+12*ln(b*x+a)*x*a^2*b-4*a*b^2*x^2+6*a^3*ln(b*x+a)+4*a^2*b*x+5*a^3)*(b*x+a
)/b^4/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.31, size = 101, normalized size = 0.76 \begin {gather*} \frac {x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {6 \, a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*a*log(x + a/b)/b^4 + 2*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 6
*a^2*x/(b^5*(x + a/b)^2) - 11/2*a^3/(b^6*(x + a/b)^2)

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

[Out]

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

[Out]

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

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