3.2.75 \(\int \frac {1}{x^3 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

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

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Rubi [A]  time = 0.08, 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 = {646, 44} \begin {gather*} \frac {b^2}{2 a^3 (a+b x) \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 {a+b x}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\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}} \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 44

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] && L
tQ[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 {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 = 0.04, 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 \log (x) (a+b x)^2-12 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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IntegrateAlgebraic [B]  time = 1.32, size = 507, normalized size = 2.43 \begin {gather*} \frac {12 b^2 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^5}+\frac {2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (-5 a^8 b^2 x+160 a^6 b^4 x^3+960 a^5 b^5 x^4+2848 a^4 b^6 x^5+4992 a^3 b^7 x^6+5248 a^2 b^8 x^7+3072 a b^9 x^8+768 b^{10} x^9\right )+2 \left (a^{10} b^2+5 a^9 b^3 x+5 a^8 b^4 x^2-160 a^7 b^5 x^3-1120 a^6 b^6 x^4-3808 a^5 b^7 x^5-7840 a^4 b^8 x^6-10240 a^3 b^9 x^7-8320 a^2 b^{10} x^8-3840 a b^{11} x^9-768 b^{12} x^{10}\right )}{a^4 x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^8 b^2-30 a^7 b^3 x-196 a^6 b^4 x^2-728 a^5 b^5 x^3-1680 a^4 b^6 x^4-2464 a^3 b^7 x^5-2240 a^2 b^8 x^6-1152 a b^9 x^7-256 b^{10} x^8\right )+a^4 \sqrt {b^2} x^2 \left (2 a^9 b+32 a^8 b^2 x+226 a^7 b^3 x^2+924 a^6 b^4 x^3+2408 a^5 b^5 x^4+4144 a^4 b^6 x^5+4704 a^3 b^7 x^6+3392 a^2 b^8 x^7+1408 a b^9 x^8+256 b^{10} x^9\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-5*a^8*b^2*x + 160*a^6*b^4*x^3 + 960*a^5*b^5*x^4 + 2848*a^4*b^6*x^
5 + 4992*a^3*b^7*x^6 + 5248*a^2*b^8*x^7 + 3072*a*b^9*x^8 + 768*b^10*x^9) + 2*(a^10*b^2 + 5*a^9*b^3*x + 5*a^8*b
^4*x^2 - 160*a^7*b^5*x^3 - 1120*a^6*b^6*x^4 - 3808*a^5*b^7*x^5 - 7840*a^4*b^8*x^6 - 10240*a^3*b^9*x^7 - 8320*a
^2*b^10*x^8 - 3840*a*b^11*x^9 - 768*b^12*x^10))/(a^4*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^8*b^2 - 30*a^7*b^
3*x - 196*a^6*b^4*x^2 - 728*a^5*b^5*x^3 - 1680*a^4*b^6*x^4 - 2464*a^3*b^7*x^5 - 2240*a^2*b^8*x^6 - 1152*a*b^9*
x^7 - 256*b^10*x^8) + a^4*Sqrt[b^2]*x^2*(2*a^9*b + 32*a^8*b^2*x + 226*a^7*b^3*x^2 + 924*a^6*b^4*x^3 + 2408*a^5
*b^5*x^4 + 4144*a^4*b^6*x^5 + 4704*a^3*b^7*x^6 + 3392*a^2*b^8*x^7 + 1408*a*b^9*x^8 + 256*b^10*x^9)) + (12*b^2*
ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/a^5

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fricas [A]  time = 0.39, 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 \relax (x)}{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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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(1/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 = 134, normalized size = 0.64 \begin {gather*} -\frac {\left (-12 b^{4} x^{4} \ln \relax (x )+12 b^{4} x^{4} \ln \left (b x +a \right )-24 a \,b^{3} x^{3} \ln \relax (x )+24 a \,b^{3} x^{3} \ln \left (b x +a \right )-12 a^{2} b^{2} x^{2} \ln \relax (x )+12 a^{2} b^{2} x^{2} \ln \left (b x +a \right )-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 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{5} x^{2}} \end {gather*}

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)

[Out]

-1/2*(12*ln(b*x+a)*x^4*b^4-12*ln(x)*x^4*b^4+24*ln(b*x+a)*x^3*a*b^3-24*ln(x)*x^3*a*b^3+12*a^2*b^2*x^2*ln(b*x+a)
-12*ln(x)*x^2*a^2*b^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 = 1.44, 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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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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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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