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

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

[Out]

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

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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} \[ \frac {a^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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 \[ \frac {-5 a^3-4 a^2 b x+4 a b^2 x^2-6 a (a+b x)^2 \log (a+b x)+2 b^3 x^3}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \]

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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fricas [A]  time = 0.98, size = 83, normalized size = 0.62 \[ \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

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 \[ \mathit {sage}_{0} x \]

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 \[ -\frac {\left (6 a \,b^{2} x^{2} \ln \left (b x +a \right )-2 b^{3} x^{3}+12 a^{2} b x \ln \left (b x +a \right )-4 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )+4 a^{2} b x +5 a^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \]

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 \[ \frac {x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {6 \, a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \]

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 \[ \int \frac {x^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

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 \[ \int \frac {x^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

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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