3.6.53 \(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx\) [553]

Optimal. Leaf size=72 \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {b x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 72, 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 = {1126, 14} \begin {gather*} \frac {b 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}}{x \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^2,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 35, normalized size = 0.49 \begin {gather*} \frac {\left (-a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}}{x \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^2,x]

[Out]

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

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Maple [A]
time = 0.03, size = 34, normalized size = 0.47

method	result	size
gosper	\(-\frac {\left (-b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}\)	\(34\)
default	\(-\frac {\left (-b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}\)	\(34\)
risch	\(-\frac {a \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}+\frac {b x \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 10, normalized size = 0.14 \begin {gather*} b x - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x - a/x

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Fricas [A]
time = 0.34, size = 13, normalized size = 0.18 \begin {gather*} \frac {b x^{2} - a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^2 - a)/x

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Sympy [A]
time = 0.02, size = 5, normalized size = 0.07 \begin {gather*} - \frac {a}{x} + b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/x + b*x

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Giac [A]
time = 4.02, size = 26, normalized size = 0.36 \begin {gather*} b x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {a \mathrm {sgn}\left (b x^{2} + a\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2)^(1/2)/x^2,x)

[Out]

int(((a + b*x^2)^2)^(1/2)/x^2, x)

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