3.2.89 \(\int \frac {1}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [189]

Optimal. Leaf size=184 \[ \frac {-a-b x}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 181, normalized size of antiderivative = 0.98, 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 {b (a+b x)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 \log (x) (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

-1/3*(a + b*x)/(a*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b*(a + b*x))/(2*a^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
- (b^2*(a + b*x))/(a^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b^3*(a + b*x)*Log[x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^
2*x^2]) + (b^3*(a + b*x)*Log[a + b*x])/(a^4*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^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{x^4 \left (a b+b^2 x\right )} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {1}{a b x^4}-\frac {1}{a^2 x^3}+\frac {b}{a^3 x^2}-\frac {b^2}{a^4 x}+\frac {b^3}{a^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {a+b x}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 72, normalized size = 0.39 \begin {gather*} -\frac {(a+b x) \left (a \left (2 a^2-3 a b x+6 b^2 x^2\right )+6 b^3 x^3 \log (x)-6 b^3 x^3 \log (a+b x)\right )}{6 a^4 x^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

-1/6*((a + b*x)*(a*(2*a^2 - 3*a*b*x + 6*b^2*x^2) + 6*b^3*x^3*Log[x] - 6*b^3*x^3*Log[a + b*x]))/(a^4*x^3*Sqrt[(
a + b*x)^2])

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Maple [A]
time = 0.41, size = 69, normalized size = 0.38

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+6 a \,b^{2} x^{2}-3 a^{2} b x +2 a^{3}\right )}{6 \sqrt {\left (b x +a \right )^{2}}\, a^{4} x^{3}}\)	\(69\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{3 a}+\frac {b x}{2 a^{2}}-\frac {b^{2} x^{2}}{a^{3}}\right )}{\left (b x +a \right ) x^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \ln \left (x \right )}{\left (b x +a \right ) a^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{4}}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(b*x+a)*(6*b^3*ln(x)*x^3-6*b^3*ln(b*x+a)*x^3+6*a*b^2*x^2-3*a^2*b*x+2*a^3)/((b*x+a)^2)^(1/2)/a^4/x^3

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Maxima [A]
time = 0.26, size = 123, normalized size = 0.67 \begin {gather*} \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {11 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}}{6 \, a^{4} x} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b}{6 \, a^{3} x^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

(-1)^(2*a*b*x + 2*a^2)*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 - 11/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2/(a^
4*x) + 5/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b/(a^3*x^2) - 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)/(a^2*x^3)

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Fricas [A]
time = 1.26, size = 54, normalized size = 0.29 \begin {gather*} \frac {6 \, b^{3} x^{3} \log \left (b x + a\right ) - 6 \, b^{3} x^{3} \log \left (x\right ) - 6 \, a b^{2} x^{2} + 3 \, a^{2} b x - 2 \, a^{3}}{6 \, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(b*x + a) - 6*b^3*x^3*log(x) - 6*a*b^2*x^2 + 3*a^2*b*x - 2*a^3)/(a^4*x^3)

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Sympy [A]
time = 0.09, size = 44, normalized size = 0.24 \begin {gather*} \frac {- 2 a^{2} + 3 a b x - 6 b^{2} x^{2}}{6 a^{3} x^{3}} + \frac {b^{3} \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/((b*x+a)**2)**(1/2),x)

[Out]

(-2*a**2 + 3*a*b*x - 6*b**2*x**2)/(6*a**3*x**3) + b**3*(-log(x) + log(a/b + x))/a**4

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Giac [A]
time = 1.22, size = 65, normalized size = 0.35 \begin {gather*} \frac {1}{6} \, {\left (\frac {6 \, b^{3} \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac {6 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{2} - 3 \, a^{2} b x + 2 \, a^{3}}{a^{4} x^{3}}\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*(6*b^3*log(abs(b*x + a))/a^4 - 6*b^3*log(abs(x))/a^4 - (6*a*b^2*x^2 - 3*a^2*b*x + 2*a^3)/(a^4*x^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^4\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*((a + b*x)^2)^(1/2)),x)

[Out]

int(1/(x^4*((a + b*x)^2)^(1/2)), x)

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