3.4.73 \(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\)

Optimal. Leaf size=75 \[ \frac {b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 14} \begin {gather*} \frac {b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1112

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.52 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (a-2 b x^2 \log (x)\right )}{2 x^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

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

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IntegrateAlgebraic [B]  time = 0.51, size = 734, normalized size = 9.79 \begin {gather*} \frac {-\frac {1}{2} \left (b^2\right )^{3/2} x^4 \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )-\frac {1}{2} \left (b^2\right )^{3/2} x^4 \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )-\frac {1}{2} a b \sqrt {b^2} x^2 \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )+\frac {1}{2} b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )-\frac {1}{2} a b \sqrt {b^2} x^2 \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )+\frac {1}{2} b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )+a^2 \sqrt {b^2}}{\left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right ) \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )}+\frac {-a b \sqrt {a^2+2 a b x^2+b^2 x^4}-a b^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}-\sqrt {b^2} x^2}{a}\right )+b \sqrt {b^2} x^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}-\sqrt {b^2} x^2}{a}\right )-b^3 x^4 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}-\sqrt {b^2} x^2}{a}\right )+a b \sqrt {b^2} x^2}{\left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right ) \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

(a*b*Sqrt[b^2]*x^2 - a*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4] - a*b^2*x^2*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4])/a] - b^3*x^4*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a] + b*Sqrt[b^2]
*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/a])/((-a - S
qrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])) + (a^2*
Sqrt[b^2] - (a*b*Sqrt[b^2]*x^2*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/2 - ((b^2)^(3/2)*x^4
*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/2 + (b^2*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[-
a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/2 - (a*b*Sqrt[b^2]*x^2*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]])/2 - ((b^2)^(3/2)*x^4*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/2 + (b
^2*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/2)/((-a - Sqr
t[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]))

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fricas [A]  time = 1.97, size = 17, normalized size = 0.23 \begin {gather*} \frac {2 \, b x^{2} \log \relax (x) - a}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^2*log(x) - a)/x^2

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giac [A]  time = 0.23, size = 45, normalized size = 0.60 \begin {gather*} \frac {1}{2} \, b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 38, normalized size = 0.51 \begin {gather*} \frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (2 b \,x^{2} \ln \relax (x )-a \right )}{2 \left (b \,x^{2}+a \right ) x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, size = 14, normalized size = 0.19 \begin {gather*} \frac {1}{2} \, b \log \left (x^{2}\right ) - \frac {a}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*log(x^2) - 1/2*a/x^2

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mupad [B]  time = 4.45, size = 112, normalized size = 1.49 \begin {gather*} \frac {\ln \left (a\,b+\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {b^2}+b^2\,x^2\right )\,\sqrt {b^2}}{2}-\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^2}-\frac {a\,b\,\ln \left (a\,b+\frac {a^2}{x^2}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^2}\right )}{2\,\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2)^(1/2)/x^3,x)

[Out]

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

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sympy [A]  time = 0.15, size = 10, normalized size = 0.13 \begin {gather*} - \frac {a}{2 x^{2}} + b \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(2*x**2) + b*log(x)

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