3.2.95 \(\int \frac {1}{x (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [195]

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

[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 = 0.04, 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 = {660, 46} \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 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 \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 = 0.02, size = 62, normalized size = 0.49 \begin {gather*} \frac {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}} \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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Maple [A]
time = 0.52, size = 91, normalized size = 0.72

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b x}{a^{2}}+\frac {3}{2 a}\right )}{\left (b x +a \right )^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (-x \right )}{\left (b x +a \right ) a^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) a^{3}}\)	\(82\)
default	\(-\frac {\left (2 b^{2} \ln \left (b x +a \right ) x^{2}-2 b^{2} \ln \left (x \right ) x^{2}+4 \ln \left (b x +a \right ) a b x -4 \ln \left (x \right ) a b x +2 a^{2} \ln \left (b x +a \right )-2 a^{2} \ln \left (x \right )-2 a b x -3 a^{2}\right ) \left (b x +a \right )}{2 a^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(91\)

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,method=_RETURNVERBOSE)

[Out]

-1/2*(2*b^2*ln(b*x+a)*x^2-2*b^2*ln(x)*x^2+4*ln(b*x+a)*a*b*x-4*ln(x)*a*b*x+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 = 0.26, 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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Fricas [A]
time = 1.43, 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 \left (x\right )}{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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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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Giac [A]
time = 0.69, size = 67, normalized size = 0.53 \begin {gather*} -\frac {\log \left ({\left | b x + a \right |}\right )}{a^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, a b x + 3 \, a^{2}}{2 \, {\left (b x + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x + a\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="giac")

[Out]

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

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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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