3.10.23 \(\int x \sqrt {a+b x^2+c x^4} \, dx\) [923]

Optimal. Leaf size=83 \[ \frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1121, 626, 635, 212} \begin {gather*} \frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

((b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2
+ c*x^4])])/(16*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 626

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 635

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

Rule 1121

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

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 85, normalized size = 1.02 \begin {gather*} \frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c}+\frac {\left (b^2-4 a c\right ) \log \left (b c+2 c^2 x^2-2 c^{3/2} \sqrt {a+b x^2+c x^4}\right )}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 101, normalized size = 1.22

method	result	size
default	\(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a}{4 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b^{2}}{16 c^{\frac {3}{2}}}\)	\(101\)
risch	\(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a}{4 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b^{2}}{16 c^{\frac {3}{2}}}\)	\(101\)
elliptic	\(\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) a}{4 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) b^{2}}{16 c^{\frac {3}{2}}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.37, size = 197, normalized size = 2.37 \begin {gather*} \left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + b c\right )}}{32 \, c^{2}}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + b c\right )}}{16 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((b^2 - 4*a*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(
c) - 4*a*c) - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c^2*x^2 + b*c))/c^2, 1/16*((b^2 - 4*a*c)*sqrt(-c)*arctan(1/2*sqrt(c
*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*sqrt(c*x^4 + b*x^2 + a)*(2*c^2*x^2 + b
*c))/c^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.71, size = 76, normalized size = 0.92 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{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+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.62, size = 72, normalized size = 0.87 \begin {gather*} \frac {\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}}{2}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{4\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

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

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