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

Optimal. Leaf size=108 \[ \frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{5/2}}-\frac{b \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c} \]

[Out]

-(b*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c^2) + (a + b*x^2 + c*x^4)^(3/2)/(6*c) + (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(5/2))

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Rubi [A]  time = 0.0807406, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1114, 640, 612, 621, 206} \[ \frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{5/2}}-\frac{b \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

-(b*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(16*c^2) + (a + b*x^2 + c*x^4)^(3/2)/(6*c) + (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(5/2))

Rule 1114

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

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 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 621

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 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])

Rubi steps

\begin{align*} \int x^3 \sqrt{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \sqrt{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c}-\frac{b \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{b \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c}+\frac{\left (b \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 c^2}\\ &=-\frac{b \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c}+\frac{\left (b \left (b^2-4 a c\right )\right ) \operatorname{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 )}{16 c^2}\\ &=-\frac{b \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{16 c^2}+\frac{\left (a+b x^2+c x^4\right )^{3/2}}{6 c}+\frac{b \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 )}{32 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0461451, size = 101, normalized size = 0.94 \[ \frac{2 \sqrt{c} \sqrt{a+b x^2+c x^4} \left (8 c \left (a+c x^4\right )-3 b^2+2 b c x^2\right )+3 b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{96 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 2*b*c*x^2 + 8*c*(a + c*x^4)) + 3*b*(b^2 - 4*a*c)*ArcTanh[(b + 2*c
*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(96*c^(5/2))

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Maple [A]  time = 0.158, size = 139, normalized size = 1.3 \begin{align*}{\frac{1}{6\,c} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{b{x}^{2}}{8\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{b}^{2}}{16\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{ab}{8}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{{b}^{3}}{32}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*x^4+b*x^2+a)^(3/2)/c-1/8*b/c*x^2*(c*x^4+b*x^2+a)^(1/2)-1/16*b^2/c^2*(c*x^4+b*x^2+a)^(1/2)-1/8*b/c^(3/2)
*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*a+1/32*b^3/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(
1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72775, size = 552, normalized size = 5.11 \begin{align*} \left [-\frac{3 \,{\left (b^{3} - 4 \, a b 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 \,{\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{192 \, c^{3}}, -\frac{3 \,{\left (b^{3} - 4 \, a b 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 \,{\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{96 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/192*(3*(b^3 - 4*a*b*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*(8*c^3*x^4 + 2*b*c^2*x^2 - 3*b^2*c + 8*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^3, -1/96*(3*(b^3
- 4*a*b*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*
(8*c^3*x^4 + 2*b*c^2*x^2 - 3*b^2*c + 8*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{a + b x^{2} + c x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*sqrt(a + b*x**2 + c*x**4), x)

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Giac [A]  time = 1.37998, size = 132, normalized size = 1.22 \begin{align*} \frac{1}{48} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \, x^{2} + \frac{b}{c}\right )} x^{2} - \frac{3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac{{\left (b^{3} - 4 \, a b 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 )}{32 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 1/32*(b^3 - 4*a*b*c)*log(abs(-2*(sq
rt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(5/2)