3.3.35 \(\int \frac {\sqrt {b x^2+c x^4}}{x^4} \, dx\) [235]

Optimal. Leaf size=56 \[ -\frac {\sqrt {b x^2+c x^4}}{2 x^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {b}} \]

[Out]

-1/2*c*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(1/2)-1/2*(c*x^4+b*x^2)^(1/2)/x^3

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Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2045, 2033, 212} \begin {gather*} -\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {b}}-\frac {\sqrt {b x^2+c x^4}}{2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^4,x]

[Out]

-1/2*Sqrt[b*x^2 + c*x^4]/x^3 - (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*Sqrt[b])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2045

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*
x^n)^p/(c*(m + j*p + 1))), x] - Dist[b*p*((n - j)/(c^n*(m + j*p + 1))), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 63, normalized size = 1.12 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (1+\frac {c x^2 \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {b+c x^2}}\right )}{2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^4,x]

[Out]

-1/2*(Sqrt[x^2*(b + c*x^2)]*(1 + (c*x^2*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]])/(Sqrt[b]*Sqrt[b + c*x^2])))/x^3

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Maple [A]
time = 0.10, size = 85, normalized size = 1.52

method	result	size
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{3}}-\frac {c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 \sqrt {b}\, x \sqrt {c \,x^{2}+b}}\)	\(74\)
default	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (\sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{2}-\sqrt {c \,x^{2}+b}\, c \,x^{2}+\left (c \,x^{2}+b \right )^{\frac {3}{2}}\right )}{2 x^{3} \sqrt {c \,x^{2}+b}\, b}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^4, x)

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Fricas [A]
time = 0.37, size = 134, normalized size = 2.39 \begin {gather*} \left [\frac {\sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} b}{4 \, b x^{3}}, \frac {\sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) - \sqrt {c x^{4} + b x^{2}} b}{2 \, b x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(b)*c*x^3*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*sqrt(c*x^4 + b*x^2)*b)/(b*x^
3), 1/2*(sqrt(-b)*c*x^3*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) - sqrt(c*x^4 + b*x^2)*b)/(b*x^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**4, x)

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Giac [A]
time = 8.09, size = 50, normalized size = 0.89 \begin {gather*} \frac {\frac {c^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} - \frac {\sqrt {c x^{2} + b} c \mathrm {sgn}\left (x\right )}{x^{2}}}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c^2*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - sqrt(c*x^2 + b)*c*sgn(x)/x^2)/c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^4,x)

[Out]

int((b*x^2 + c*x^4)^(1/2)/x^4, x)

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