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

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

[Out]

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

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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} \[ \frac {1}{2 a (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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

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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fricas [A]  time = 1.08, size = 80, normalized size = 0.63 \[ \frac {2 \, a b x + 3 \, a^{2} - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}} \]

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

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

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 \[ -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} + \frac {1}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} \]

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 \[ \int \frac {1}{x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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 \[ \int \frac {1}{x \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

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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