3.206 \(\int \frac {1}{x^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=235 \[ -\frac {b}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 b \log (x) (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 b}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-4*b/a^5/((b*x+a)^2)^(1/2)-1/4*b/a^2/(b*x+a)^3/((b*x+a)^2)^(1/2)-2/3*b/a^3/(b*x+a)^2/((b*x+a)^2)^(1/2)-3/2*b/a
^4/(b*x+a)/((b*x+a)^2)^(1/2)+(-b*x-a)/a^5/x/((b*x+a)^2)^(1/2)-5*b*(b*x+a)*ln(x)/a^6/((b*x+a)^2)^(1/2)+5*b*(b*x
+a)*ln(b*x+a)/a^6/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 235, 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} \[ -\frac {3 b}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 b}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 b \log (x) (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-4*b)/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(4*a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*b)/(3*a^
3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b)/(2*a^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*
x)/(a^5*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*b*(a + b*x)*Log[x])/(a^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*b*(
a + b*x)*Log[a + b*x])/(a^6*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^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{a^5 b^5 x^2}-\frac {5}{a^6 b^4 x}+\frac {1}{a^2 b^3 (a+b x)^5}+\frac {2}{a^3 b^3 (a+b x)^4}+\frac {3}{a^4 b^3 (a+b x)^3}+\frac {4}{a^5 b^3 (a+b x)^2}+\frac {5}{a^6 b^3 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 b (a+b x) \log (x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b (a+b x) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 103, normalized size = 0.44 \[ \frac {-a \left (12 a^4+125 a^3 b x+260 a^2 b^2 x^2+210 a b^3 x^3+60 b^4 x^4\right )-60 b x \log (x) (a+b x)^4+60 b x (a+b x)^4 \log (a+b x)}{12 a^6 x (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-(a*(12*a^4 + 125*a^3*b*x + 260*a^2*b^2*x^2 + 210*a*b^3*x^3 + 60*b^4*x^4)) - 60*b*x*(a + b*x)^4*Log[x] + 60*b
*x*(a + b*x)^4*Log[a + b*x])/(12*a^6*x*(a + b*x)^3*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.89, size = 197, normalized size = 0.84 \[ -\frac {60 \, a b^{4} x^{4} + 210 \, a^{2} b^{3} x^{3} + 260 \, a^{3} b^{2} x^{2} + 125 \, a^{4} b x + 12 \, a^{5} - 60 \, {\left (b^{5} x^{5} + 4 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} + 4 \, a^{3} b^{2} x^{2} + a^{4} b x\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{5} x^{5} + 4 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} + 4 \, a^{3} b^{2} x^{2} + a^{4} b x\right )} \log \relax (x)}{12 \, {\left (a^{6} b^{4} x^{5} + 4 \, a^{7} b^{3} x^{4} + 6 \, a^{8} b^{2} x^{3} + 4 \, a^{9} b x^{2} + a^{10} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(60*a*b^4*x^4 + 210*a^2*b^3*x^3 + 260*a^3*b^2*x^2 + 125*a^4*b*x + 12*a^5 - 60*(b^5*x^5 + 4*a*b^4*x^4 + 6
*a^2*b^3*x^3 + 4*a^3*b^2*x^2 + a^4*b*x)*log(b*x + a) + 60*(b^5*x^5 + 4*a*b^4*x^4 + 6*a^2*b^3*x^3 + 4*a^3*b^2*x
^2 + a^4*b*x)*log(x))/(a^6*b^4*x^5 + 4*a^7*b^3*x^4 + 6*a^8*b^2*x^3 + 4*a^9*b*x^2 + a^10*x)

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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^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.15, size = 199, normalized size = 0.85 \[ \frac {\left (-60 b^{5} x^{5} \ln \relax (x )+60 b^{5} x^{5} \ln \left (b x +a \right )-240 a \,b^{4} x^{4} \ln \relax (x )+240 a \,b^{4} x^{4} \ln \left (b x +a \right )-360 a^{2} b^{3} x^{3} \ln \relax (x )+360 a^{2} b^{3} x^{3} \ln \left (b x +a \right )-60 a \,b^{4} x^{4}-240 a^{3} b^{2} x^{2} \ln \relax (x )+240 a^{3} b^{2} x^{2} \ln \left (b x +a \right )-210 a^{2} b^{3} x^{3}-60 a^{4} b x \ln \relax (x )+60 a^{4} b x \ln \left (b x +a \right )-260 a^{3} b^{2} x^{2}-125 a^{4} b x -12 a^{5}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} a^{6} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/12*(60*ln(b*x+a)*x^5*b^5-60*b^5*x^5*ln(x)+240*a*b^4*x^4*ln(b*x+a)-240*a*b^4*x^4*ln(x)+360*a^2*b^3*x^3*ln(b*x
+a)-360*a^2*b^3*x^3*ln(x)-60*a*b^4*x^4+240*a^3*b^2*x^2*ln(b*x+a)-240*a^3*b^2*x^2*ln(x)-210*a^2*b^3*x^3+60*a^4*
b*x*ln(b*x+a)-60*a^4*b*x*ln(x)-260*a^3*b^2*x^2-125*a^4*b*x-12*a^5)*(b*x+a)/x/a^6/((b*x+a)^2)^(5/2)

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maxima [A]  time = 1.40, size = 148, normalized size = 0.63 \[ \frac {5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{6}} - \frac {5 \, b}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}} - \frac {5 \, b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5}} - \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x} - \frac {5}{2 \, a^{4} b {\left (x + \frac {a}{b}\right )}^{2}} - \frac {1}{4 \, a^{2} b^{3} {\left (x + \frac {a}{b}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int(1/(x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(1/(x**2*((a + b*x)**2)**(5/2)), x)

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