3.2.97 \(\int \frac {1}{x^3 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [197]

Optimal. Leaf size=209 \[ \frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 209, 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 = {660, 46} \begin {gather*} \frac {6 b^2 \log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

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 {1}{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 {1}{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}{a^3 b^3 x^3}-\frac {3}{a^4 b^2 x^2}+\frac {6}{a^5 b x}-\frac {1}{a^3 (a+b x)^3}-\frac {3}{a^4 (a+b x)^2}-\frac {6}{a^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 b^2}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 b^2 (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {6 b^2 (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 99, normalized size = 0.47 \begin {gather*} \frac {a \left (-a^3+4 a^2 b x+18 a b^2 x^2+12 b^3 x^3\right )+12 b^2 x^2 (a+b x)^2 \log (x)-12 b^2 x^2 (a+b x)^2 \log (a+b x)}{2 a^5 x^2 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(a*(-a^3 + 4*a^2*b*x + 18*a*b^2*x^2 + 12*b^3*x^3) + 12*b^2*x^2*(a + b*x)^2*Log[x] - 12*b^2*x^2*(a + b*x)^2*Log
[a + b*x])/(2*a^5*x^2*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.52, size = 134, normalized size = 0.64

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {6 b^{3} x^{3}}{a^{4}}+\frac {9 b^{2} x^{2}}{a^{3}}+\frac {2 b x}{a^{2}}-\frac {1}{2 a}\right )}{\left (b x +a \right )^{3} x^{2}}-\frac {6 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}+\frac {6 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}\)	\(115\)
default	\(-\frac {\left (12 \ln \left (b x +a \right ) b^{4} x^{4}-12 \ln \left (x \right ) b^{4} x^{4}+24 \ln \left (b x +a \right ) a \,b^{3} x^{3}-24 \ln \left (x \right ) a \,b^{3} x^{3}+12 \ln \left (b x +a \right ) a^{2} b^{2} x^{2}-12 \ln \left (x \right ) a^{2} b^{2} x^{2}-12 a \,b^{3} x^{3}-18 a^{2} b^{2} x^{2}-4 a^{3} b x +a^{4}\right ) \left (b x +a \right )}{2 x^{2} a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(12*ln(b*x+a)*b^4*x^4-12*ln(x)*b^4*x^4+24*ln(b*x+a)*a*b^3*x^3-24*ln(x)*a*b^3*x^3+12*ln(b*x+a)*a^2*b^2*x^2
-12*ln(x)*a^2*b^2*x^2-12*a*b^3*x^3-18*a^2*b^2*x^2-4*a^3*b*x+a^4)*(b*x+a)/x^2/a^5/((b*x+a)^2)^(3/2)

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Maxima [A]
time = 0.28, size = 135, normalized size = 0.65 \begin {gather*} -\frac {6 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {6 \, b^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {5 \, b}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} x} - \frac {1}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x^{2}} + \frac {1}{2 \, a^{3} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.73, size = 130, normalized size = 0.62 \begin {gather*} \frac {12 \, a b^{3} x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x - a^{4} - 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(12*a*b^3*x^3 + 18*a^2*b^2*x^2 + 4*a^3*b*x - a^4 - 12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(b*x + a) +
12*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*log(x))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{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(1/x**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/(x**3*((a + b*x)**2)**(3/2)), x)

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Giac [A]
time = 0.57, size = 97, normalized size = 0.46 \begin {gather*} -\frac {6 \, b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {6 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, b^{3} x^{3} + 18 \, a b^{2} x^{2} + 4 \, a^{2} b x - a^{3}}{2 \, {\left (b x^{2} + a x\right )}^{2} a^{4} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

-6*b^2*log(abs(b*x + a))/(a^5*sgn(b*x + a)) + 6*b^2*log(abs(x))/(a^5*sgn(b*x + a)) + 1/2*(12*b^3*x^3 + 18*a*b^
2*x^2 + 4*a^2*b*x - a^3)/((b*x^2 + a*x)^2*a^4*sgn(b*x + a))

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

[Out]

int(1/(x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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