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

Optimal. Leaf size=124 \[ -\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^2}+\frac{3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{5/2}}+\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c} \]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*c^2) + ((b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))
/(16*c) + (3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(256*c^(5/2))

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Rubi [A]  time = 0.0855727, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1107, 612, 621, 206} \[ -\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^2}+\frac{3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{5/2}}+\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(-3*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(128*c^2) + ((b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))
/(16*c) + (3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(256*c^(5/2))

Rule 1107

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]

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 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{32 c}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}+\frac{\left (3 \left (b^2-4 a c\right )^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{256 c^2}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}+\frac{\left (3 \left (b^2-4 a c\right )^2\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 )}{128 c^2}\\ &=-\frac{3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{128 c^2}+\frac{\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}+\frac{3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{256 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0826465, size = 126, normalized size = 1.02 \[ \frac{\frac{3 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )-2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}\right )}{8 c^{3/2}}+2 \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(2*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2) + (3*(b^2 - 4*a*c)*(-2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4
] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(8*c^(3/2)))/(32*c)

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Maple [B]  time = 0.178, size = 242, normalized size = 2. \begin{align*}{\frac{c{x}^{6}}{8}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{{b}^{2}{x}^{2}}{64\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,ab}{32\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{2}a}{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{3}{2}}}}+{\frac{3\,b{x}^{4}}{16}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{a}^{2}}{16}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{5\,a{x}^{2}}{16}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{3}}{128\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{4}}{256}\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*(c*x^4+b*x^2+a)^(3/2),x)

[Out]

1/8*c*x^6*(c*x^4+b*x^2+a)^(1/2)+1/64*b^2*x^2/c*(c*x^4+b*x^2+a)^(1/2)+5/32*a*b/c*(c*x^4+b*x^2+a)^(1/2)-3/32*a*b
^2/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/16*b*x^4*(c*x^4+b*x^2+a)^(1/2)+3/16*a^2*ln((1/2*b
+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+5/16*a*x^2*(c*x^4+b*x^2+a)^(1/2)-3/128*b^3/c^2*(c*x^4+b*x^2+a)^
(1/2)+3/256*b^4/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*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66658, size = 684, normalized size = 5.52 \begin{align*} \left [\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \,{\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{512 \, c^{3}}, -\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \,{\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{256 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*c^2 + 2*(b^2*c^2 + 20*a*c^3)*
x^2)*sqrt(c*x^4 + b*x^2 + a))/c^3, -1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(16*c^4*x^6 + 24*b*c^3*x^4 - 3*b^3*c + 20*a*b*
c^2 + 2*(b^2*c^2 + 20*a*c^3)*x^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 \left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26562, size = 182, normalized size = 1.47 \begin{align*} \frac{1}{128} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (2 \, c x^{2} + 3 \, b\right )} x^{2} + \frac{b^{2} c^{2} + 20 \, a c^{3}}{c^{3}}\right )} x^{2} - \frac{3 \, b^{3} c - 20 \, a b c^{2}}{c^{3}}\right )} - \frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(2*c*x^2 + 3*b)*x^2 + (b^2*c^2 + 20*a*c^3)/c^3)*x^2 - (3*b^3*c - 20*a*b*c^
2)/c^3) - 3/256*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b)
)/c^(5/2)