3.7.38 \(\int \frac {1}{x (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [638]

Optimal. Leaf size=147 \[ \frac {1}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 147, 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 {1}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\log (x) \left (a+b x^2\right )}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.02, size = 74, normalized size = 0.50 \begin {gather*} \frac {a \left (3 a+2 b x^2\right )+4 \left (a+b x^2\right )^2 \log (x)-2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 107, normalized size = 0.73

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \,x^{2}}{2 a^{2}}+\frac {3}{4 a}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\)	\(97\)
default	\(-\frac {\left (2 \ln \left (b \,x^{2}+a \right ) b^{2} x^{4}-4 \ln \left (x \right ) b^{2} x^{4}+4 \ln \left (b \,x^{2}+a \right ) a b \,x^{2}-8 \ln \left (x \right ) a b \,x^{2}-2 a b \,x^{2}+2 a^{2} \ln \left (b \,x^{2}+a \right )-4 a^{2} \ln \left (x \right )-3 a^{2}\right ) \left (b \,x^{2}+a \right )}{4 a^{3} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*ln(b*x^2+a)*b^2*x^4-4*ln(x)*b^2*x^4+4*ln(b*x^2+a)*a*b*x^2-8*ln(x)*a*b*x^2-2*a*b*x^2+2*a^2*ln(b*x^2+a)-
4*a^2*ln(x)-3*a^2)*(b*x^2+a)/a^3/((b*x^2+a)^2)^(3/2)

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Maxima [A]
time = 0.29, size = 57, normalized size = 0.39 \begin {gather*} \frac {2 \, b x^{2} + 3 \, a}{4 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {\log \left (x\right )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 90, normalized size = 0.61 \begin {gather*} \frac {2 \, a b x^{2} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b*x^2 + 3*a^2 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(b*x^2 + a) + 4*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(x))/
(a^3*b^2*x^4 + 2*a^4*b*x^2 + 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^{2}\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**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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Giac [A]
time = 4.63, size = 89, normalized size = 0.61 \begin {gather*} \frac {\log \left (x^{2}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b^{2} x^{4} + 8 \, a b x^{2} + 6 \, a^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x^2)/(a^3*sgn(b*x^2 + a)) - 1/2*log(abs(b*x^2 + a))/(a^3*sgn(b*x^2 + a)) + 1/4*(3*b^2*x^4 + 8*a*b*x^2
+ 6*a^2)/((b*x^2 + a)^2*a^3*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\,{\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*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

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

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