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

Optimal. Leaf size=150 \[ \frac{3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3}-\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c} \]

[Out]

(3*b*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^3) - (b*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/
2))/(32*c^2) + (a + b*x^2 + c*x^4)^(5/2)/(10*c) - (3*b*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a
+ b*x^2 + c*x^4])])/(512*c^(7/2))

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Rubi [A]  time = 0.116978, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3}-\frac{3 b \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(3*b*(b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^3) - (b*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/
2))/(32*c^2) + (a + b*x^2 + c*x^4)^(5/2)/(10*c) - (3*b*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a
+ b*x^2 + c*x^4])])/(512*c^(7/2))

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 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac{b \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c}+\frac{\left (3 b \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{64 c^2}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac{\left (3 b \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 )}{512 c^3}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac{\left (3 b \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 )}{256 c^3}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^3}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c}-\frac{3 b \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 )}{512 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.142122, size = 149, normalized size = 0.99 \[ -\frac{3 b \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )-2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}\right )}{512 c^{7/2}}-\frac{b \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac{\left (a+b x^2+c x^4\right )^{5/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(32*c^2) + (a + b*x^2 + c*x^4)^(5/2)/(10*c) - (3*b*(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])]))/(512*c^(7/2))

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Maple [B]  time = 0.167, size = 316, normalized size = 2.1 \begin{align*} -{\frac{5\,{b}^{2}a}{64\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,a{b}^{3}}{64}\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}}}}-{\frac{3\,b{a}^{2}}{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{c{x}^{8}}{10}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{a{x}^{4}}{5}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{4}}{256\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{5}}{512}\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{7}{2}}}}+{\frac{{a}^{2}}{10\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{11\,b{x}^{6}}{80}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{{b}^{2}{x}^{4}}{160\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{b}^{3}{x}^{2}}{128\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{7\,ab{x}^{2}}{160\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*a*b^2/c^2*(c*x^4+b*x^2+a)^(1/2)+3/64*a*b^3/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-3/32*
a^2*b/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/10*c*x^8*(c*x^4+b*x^2+a)^(1/2)+1/5*a*x^4*(c*x^
4+b*x^2+a)^(1/2)+3/256*b^4/c^3*(c*x^4+b*x^2+a)^(1/2)-3/512*b^5/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a
)^(1/2))+1/10*a^2/c*(c*x^4+b*x^2+a)^(1/2)+11/80*b*x^6*(c*x^4+b*x^2+a)^(1/2)+1/160*b^2*x^4/c*(c*x^4+b*x^2+a)^(1
/2)-1/128*b^3/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)+7/160*a*b*x^2/c*(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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68222, size = 837, normalized size = 5.58 \begin{align*} \left [\frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b 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 (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \,{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \,{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{5120 \, c^{4}}, \frac{15 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b 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 (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \,{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \,{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{2560 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 +
8*(b^2*c^3 + 32*a*c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/2560*(15*(b^5 -
8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x
^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]

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

[Out]

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

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Giac [A]  time = 1.22629, size = 232, normalized size = 1.55 \begin{align*} \frac{1}{1280} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, c x^{2} + 11 \, b\right )} x^{2} + \frac{b^{2} c^{3} + 32 \, a c^{4}}{c^{4}}\right )} x^{2} - \frac{5 \, b^{3} c^{2} - 28 \, a b c^{3}}{c^{4}}\right )} x^{2} + \frac{15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}}{c^{4}}\right )} + \frac{3 \,{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b 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 )}{512 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(2*(8*c*x^2 + 11*b)*x^2 + (b^2*c^3 + 32*a*c^4)/c^4)*x^2 - (5*b^3*c^2 - 28
*a*b*c^3)/c^4)*x^2 + (15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3)/c^4) + 3/512*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*lo
g(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(7/2)