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

Optimal. Leaf size=126 \[ \frac {1}{2 a (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 126, 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 {1}{2 a (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 62, normalized size = 0.49 \begin {gather*} \frac {a (3 a+2 b x)+2 \log (x) (a+b x)^2-2 (a+b x)^2 \log (a+b x)}{2 a^3 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 1.29, size = 588, normalized size = 4.67 \begin {gather*} \frac {2 \left (\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x\right )^4 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^3 \left (a^4+4 a^3 b x+12 a^2 b^2 x^2-4 a^2 \sqrt {b^2} x \sqrt {a^2+2 a b x+b^2 x^2}-8 a b \sqrt {b^2} x^2 \sqrt {a^2+2 a b x+b^2 x^2}-8 \left (b^2\right )^{3/2} x^3 \sqrt {a^2+2 a b x+b^2 x^2}+16 a b^3 x^3+8 b^4 x^4\right )}+\frac {a^8 b-3 a^6 b^3 x^2-32 a^5 b^4 x^3-140 a^4 b^5 x^4-320 a^3 b^6 x^5-400 a^2 b^7 x^6+\sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^7-a^6 b x+4 a^5 b^2 x^2+28 a^4 b^3 x^3+112 a^3 b^4 x^4+208 a^2 b^5 x^5+192 a b^6 x^6+64 b^7 x^7\right )-256 a b^8 x^7-64 b^9 x^8}{a^2 b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^6 b^2-22 a^5 b^3 x-100 a^4 b^4 x^2-240 a^3 b^5 x^3-320 a^2 b^6 x^4-224 a b^7 x^5-64 b^8 x^6\right )+a^2 b \sqrt {b^2} x^2 \left (2 a^7 b+24 a^6 b^2 x+122 a^5 b^3 x^2+340 a^4 b^4 x^3+560 a^3 b^5 x^4+544 a^2 b^6 x^5+288 a b^7 x^6+64 b^8 x^7\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^8*b - 3*a^6*b^3*x^2 - 32*a^5*b^4*x^3 - 140*a^4*b^5*x^4 - 320*a^3*b^6*x^5 - 400*a^2*b^7*x^6 - 256*a*b^8*x^7
- 64*b^9*x^8 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a^7 - a^6*b*x + 4*a^5*b^2*x^2 + 28*a^4*b^3*x^3 + 112*a
^3*b^4*x^4 + 208*a^2*b^5*x^5 + 192*a*b^6*x^6 + 64*b^7*x^7))/(a^2*b*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^6*b
^2 - 22*a^5*b^3*x - 100*a^4*b^4*x^2 - 240*a^3*b^5*x^3 - 320*a^2*b^6*x^4 - 224*a*b^7*x^5 - 64*b^8*x^6) + a^2*b*
Sqrt[b^2]*x^2*(2*a^7*b + 24*a^6*b^2*x + 122*a^5*b^3*x^2 + 340*a^4*b^4*x^3 + 560*a^3*b^5*x^4 + 544*a^2*b^6*x^5
+ 288*a*b^7*x^6 + 64*b^8*x^7)) + (2*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4*ArcTanh[(Sqrt[b^2]*x)/a
- Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/(a^3*(a^4 + 4*a^3*b*x + 12*a^2*b^2*x^2 + 16*a*b^3*x^3 + 8*b^4*x^4 - 4*a^2
*Sqrt[b^2]*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 8*a*b*Sqrt[b^2]*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 8*(b^2)^(3/2)
*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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fricas [A]  time = 0.41, size = 80, normalized size = 0.63 \begin {gather*} \frac {2 \, 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^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(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 = 91, normalized size = 0.72 \begin {gather*} -\frac {\left (-2 b^{2} x^{2} \ln \relax (x )+2 b^{2} x^{2} \ln \left (b x +a \right )-4 a b x \ln \relax (x )+4 a b x \ln \left (b x +a \right )-2 a^{2} \ln \relax (x )+2 a^{2} \ln \left (b x +a \right )-2 a b x -3 a^{2}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 1.32, size = 78, normalized size = 0.62 \begin {gather*} -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} + \frac {1}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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