3.3.62 \(\int \frac {x^3}{\sqrt {b x^2+c x^4}} \, dx\) [262]

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 = 0.05, 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 = {2043, 654, 634, 212} \begin {gather*} \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 {gather*}

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 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 634

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 654

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 2043

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} \text {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 \text {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 \text {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 = 0.04, size = 74, normalized size = 1.28 \begin {gather*} \frac {x \left (\sqrt {c} x \left (b+c x^2\right )+b \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{2 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

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

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Maple [A]
time = 0.10, size = 64, normalized size = 1.10

method	result	size
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (x \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}}-b \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) c \right )}{2 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {5}{2}}}\)	\(64\)
risch	\(\frac {x^{2} \left (c \,x^{2}+b \right )}{2 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{2 c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*(c*x^2+b)^(1/2)*(x*(c*x^2+b)^(1/2)*c^(3/2)-b*ln(x*c^(1/2)+(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 = 0.29, size = 52, normalized size = 0.90 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.42, size = 114, normalized size = 1.97 \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^{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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\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**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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Giac [A]
time = 3.04, size = 59, normalized size = 1.02 \begin {gather*} -\frac {b \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b} x}{2 \, c \mathrm {sgn}\left (x\right )} + \frac {b \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{2 \, c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \end {gather*}

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

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Mupad [B]
time = 4.30, size = 53, normalized size = 0.91 \begin {gather*} \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}} \end {gather*}

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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