3.2.6 \(\int \frac {\sqrt {b x^2+c x^4}}{x} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{2} \sqrt {b x^2+c x^4}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2018, 664, 620, 206} \begin {gather*} \frac {1}{2} \sqrt {b x^2+c x^4}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x,x]

[Out]

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

Rule 206

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 64, normalized size = 1.16 \begin {gather*} \frac {1}{2} \sqrt {x^2 \left (b+c x^2\right )} \left (\frac {b \log \left (\sqrt {c} \sqrt {b+c x^2}+c x\right )}{\sqrt {c} x \sqrt {b+c x^2}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 61, normalized size = 1.11 \begin {gather*} \frac {1}{2} \sqrt {b x^2+c x^4}-\frac {b \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{4 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.10, size = 115, normalized size = 2.09 \begin {gather*} \left [\frac {b \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} c}{4 \, c}, -\frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} c}{2 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 52, normalized size = 0.95 \begin {gather*} \frac {b \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{4 \, \sqrt {c}} + \frac {1}{2} \, {\left (\sqrt {c x^{2} + b} x - \frac {b \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{\sqrt {c}}\right )} \mathrm {sgn}\relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*log(abs(b))*sgn(x)/sqrt(c) + 1/2*(sqrt(c*x^2 + b)*x - b*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/sqrt(c))*
sgn(x)

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maple [A]  time = 0.00, size = 64, normalized size = 1.16 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (b \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+\sqrt {c \,x^{2}+b}\, \sqrt {c}\, x \right )}{2 \sqrt {c \,x^{2}+b}\, \sqrt {c}\, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.43, size = 49, normalized size = 0.89 \begin {gather*} \frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{4 \, \sqrt {c}} + \frac {1}{2} \, \sqrt {c x^{4} + b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 1/2*sqrt(c*x^4 + b*x^2)

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mupad [B]  time = 4.21, size = 50, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}}{2}+\frac {b\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{4\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x,x)

[Out]

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

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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}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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