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

Optimal. Leaf size=58 \[ \frac {\sqrt {b x^2+c x^4}}{2 c}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 58, 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, 640, 620, 206} \[ \frac {\sqrt {b x^2+c x^4}}{2 c}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[b*x^2 + c*x^4],x]

[Out]

Sqrt[b*x^2 + c*x^4]/(2*c) - (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(2*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 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 640

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

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

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Mathematica [A]  time = 0.03, size = 73, normalized size = 1.26 \[ \frac {x \left (\sqrt {c} x \left (b+c x^2\right )-b \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )\right )}{2 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

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fricas [A]  time = 0.47, size = 114, normalized size = 1.97 \[ \left [\frac {b \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} c}{4 \, c^{2}}, \frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} c}{2 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^4+b*x^2)^(1/2),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^2, 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^2]

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giac [A]  time = 0.20, size = 59, normalized size = 1.02 \[ \frac {b \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{4} + b x^{2}}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 64, normalized size = 1.10 \[ -\frac {\sqrt {c \,x^{2}+b}\, \left (b c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-\sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x \right ) x}{2 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/2*x*(c*x^2+b)^(1/2)*(-x*(c*x^2+b)^(1/2)*c^(3/2)+b*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*c)/(c*x^4+b*x^2)^(1/2)/c^(5
/2)

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maxima [A]  time = 1.45, size = 52, normalized size = 0.90 \[ -\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{4 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{4} + b x^{2}}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.30, size = 53, normalized size = 0.91 \[ \frac {\sqrt {c\,x^4+b\,x^2}}{2\,c}-\frac {b\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{4\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2 + c*x^4)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**3/sqrt(x**2*(b + c*x**2)), x)

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