3.7.39 \(\int \frac {1}{x^3 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [639]

Optimal. Leaf size=189 \[ -\frac {b}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b \left (a+b x^2\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1126, 272, 46} \begin {gather*} -\frac {b}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b \log (x) \left (a+b x^2\right )}{a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 97, normalized size = 0.51 \begin {gather*} \frac {-a \left (2 a^2+9 a b x^2+6 b^2 x^4\right )-12 b x^2 \left (a+b x^2\right )^2 \log (x)+6 b x^2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^4 x^2 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

(-(a*(2*a^2 + 9*a*b*x^2 + 6*b^2*x^4)) - 12*b*x^2*(a + b*x^2)^2*Log[x] + 6*b*x^2*(a + b*x^2)^2*Log[a + b*x^2])/
(4*a^4*x^2*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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Maple [A]
time = 0.05, size = 133, normalized size = 0.70

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3 b^{2} x^{4}}{2 a^{3}}-\frac {9 b \,x^{2}}{4 a^{2}}-\frac {1}{2 a}\right )}{\left (b \,x^{2}+a \right )^{3} x^{2}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{4}}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (-b \,x^{2}-a \right )}{2 \left (b \,x^{2}+a \right ) a^{4}}\)	\(117\)
default	\(\frac {\left (6 \ln \left (b \,x^{2}+a \right ) b^{3} x^{6}-12 \ln \left (x \right ) b^{3} x^{6}+12 \ln \left (b \,x^{2}+a \right ) a \,b^{2} x^{4}-24 a \,b^{2} \ln \left (x \right ) x^{4}-6 a \,b^{2} x^{4}+6 \ln \left (b \,x^{2}+a \right ) a^{2} b \,x^{2}-12 a^{2} b \ln \left (x \right ) x^{2}-9 a^{2} b \,x^{2}-2 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 x^{2} a^{4} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(6*ln(b*x^2+a)*b^3*x^6-12*ln(x)*b^3*x^6+12*ln(b*x^2+a)*a*b^2*x^4-24*a*b^2*ln(x)*x^4-6*a*b^2*x^4+6*ln(b*x^2
+a)*a^2*b*x^2-12*a^2*b*ln(x)*x^2-9*a^2*b*x^2-2*a^3)*(b*x^2+a)/x^2/a^4/((b*x^2+a)^2)^(3/2)

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Maxima [A]
time = 0.28, size = 75, normalized size = 0.40 \begin {gather*} -\frac {6 \, b^{2} x^{4} + 9 \, a b x^{2} + 2 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} + \frac {3 \, b \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {3 \, b \log \left (x\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(6*b^2*x^4 + 9*a*b*x^2 + 2*a^2)/(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2) + 3/2*b*log(b*x^2 + a)/a^4 - 3*b*lo
g(x)/a^4

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Fricas [A]
time = 0.34, size = 119, normalized size = 0.63 \begin {gather*} -\frac {6 \, a b^{2} x^{4} + 9 \, a^{2} b x^{2} + 2 \, a^{3} - 6 \, {\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \log \left (b x^{2} + a\right ) + 12 \, {\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/4*(6*a*b^2*x^4 + 9*a^2*b*x^2 + 2*a^3 - 6*(b^3*x^6 + 2*a*b^2*x^4 + a^2*b*x^2)*log(b*x^2 + a) + 12*(b^3*x^6 +
2*a*b^2*x^4 + a^2*b*x^2)*log(x))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(1/(x**3*((a + b*x**2)**2)**(3/2)), x)

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Giac [A]
time = 4.49, size = 122, normalized size = 0.65 \begin {gather*} -\frac {3 \, b \log \left (x^{2}\right )}{2 \, a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {9 \, b^{3} x^{4} + 22 \, a b^{2} x^{2} + 14 \, a^{2} b}{4 \, {\left (b x^{2} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b x^{2} - a}{2 \, a^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

-3/2*b*log(x^2)/(a^4*sgn(b*x^2 + a)) + 3/2*b*log(abs(b*x^2 + a))/(a^4*sgn(b*x^2 + a)) - 1/4*(9*b^3*x^4 + 22*a*
b^2*x^2 + 14*a^2*b)/((b*x^2 + a)^2*a^4*sgn(b*x^2 + a)) + 1/2*(3*b*x^2 - a)/(a^4*x^2*sgn(b*x^2 + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)), x)

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