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

Optimal. Leaf size=68 \[ \frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2013, 612, 620, 206} \begin {gather*} \frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[b*x^2 + c*x^4],x]

[Out]

((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(8*c) - (b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*c^(3/2))

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 612

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

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 2013

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(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]] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int x \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{16 c}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c}\\ &=\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 1.32 \begin {gather*} \frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (b+2 c x^2\right )-b^2 \log \left (\sqrt {c} \sqrt {b+c x^2}+c x\right )\right )}{8 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(b + 2*c*x^2) - b^2*Log[c*x + Sqrt[c]*Sqrt[b + c*x^2]]))/(8*c^(3
/2)*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.19, size = 78, normalized size = 1.15 \begin {gather*} \frac {b^2 \log \left (-2 c^{3/2} \sqrt {b x^2+c x^4}+b c+2 c^2 x^2\right )}{16 c^{3/2}}+\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(8*c) + (b^2*Log[b*c + 2*c^2*x^2 - 2*c^(3/2)*Sqrt[b*x^2 + c*x^4]])/(16*c^(
3/2))

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fricas [A]  time = 0.87, size = 140, normalized size = 2.06 \begin {gather*} \left [\frac {b^{2} \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}} {\left (2 \, c^{2} x^{2} + b c\right )}}{16 \, c^{2}}, \frac {b^{2} \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}} {\left (2 \, c^{2} x^{2} + b c\right )}}{8 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(b^2*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)*(2*c^2*x^2 + b*c)
)/c^2, 1/8*(b^2*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 + b
*c))/c^2]

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giac [A]  time = 0.19, size = 69, normalized size = 1.01 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{2} + b} {\left (2 \, x^{2} \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x + \frac {b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{8 \, c^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{16 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(c*x^2 + b)*(2*x^2*sgn(x) + b*sgn(x)/c)*x + 1/8*b^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(3
/2) - 1/16*b^2*log(abs(b))*sgn(x)/c^(3/2)

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maple [A]  time = 0.01, size = 84, normalized size = 1.24 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-b^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-\sqrt {c \,x^{2}+b}\, b \sqrt {c}\, x +2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, x \right )}{8 \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.43, size = 73, normalized size = 1.07 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{4} + b x^{2}} x^{2} - \frac {b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{4} + b x^{2}} b}{8 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^4 + b*x^2)*x^2 - 1/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(3/2) + 1/8*sqrt(c*x
^4 + b*x^2)*b/c

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mupad [B]  time = 4.37, size = 64, normalized size = 0.94 \begin {gather*} \frac {\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}}{2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2 + c*x^4)^(1/2),x)

[Out]

((b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2))/2 - (b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(16*c^(
3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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