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*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])

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Rubi [A]  time = 0.0579414, antiderivative size = 126, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.022777, 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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Maple [A]  time = 0.226, size = 91, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\,{b}^{2}\ln \left ( x \right ){x}^{2}-2\,{b}^{2}\ln \left ( bx+a \right ){x}^{2}+4\,\ln \left ( x \right ) xab-4\,\ln \left ( bx+a \right ) xab+2\,\ln \left ( x \right ){a}^{2}-2\,{a}^{2}\ln \left ( bx+a \right ) +2\,abx+3\,{a}^{2} \right ) \left ( bx+a \right ) }{2\,{a}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.69607, size = 182, normalized size = 1.44 \begin{align*} \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 \left (x\right )}{2 \,{\left (a^{3} b^{2} x^{2} + 2 \, a^{4} b x + a^{5}\right )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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