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

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 124, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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 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]

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 = 0.09, size = 126, normalized size = 1.02 \begin {gather*} \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} \end {gather*}

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)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.47, size = 130, normalized size = 1.05 \begin {gather*} \frac {\sqrt {a+b x^2+c x^4} \left (20 a b c+40 a c^2 x^2-3 b^3+2 b^2 c x^2+24 b c^2 x^4+16 c^3 x^6\right )}{128 c^2}-\frac {3 \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x^2+c x^4}+b+2 c x^2\right )}{256 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2 + c*x^4]*(-3*b^3 + 20*a*b*c + 2*b^2*c*x^2 + 40*a*c^2*x^2 + 24*b*c^2*x^4 + 16*c^3*x^6))/(128*c^
2) - (3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(256*c^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 297, normalized size = 2.40 \begin {gather*} \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 {gather*}

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]

________________________________________________________________________________________

giac [B]  time = 0.39, size = 317, normalized size = 2.56 \begin {gather*} \frac {1}{16} \, {\left (2 \, \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 )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \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 {3 \, {\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 )}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, 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 )}{c^{\frac {7}{2}}}\right )} c \end {gather*}

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]

1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (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))*a + 1/96*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2) - 3*
(b^3 - 4*a*b*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(5/2))*b + 1/768*(2*sqrt(c*
x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b/c)*x^2 - (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^3) + 3*(5*b^4
- 24*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^(7/2))*c

________________________________________________________________________________________

maple [B]  time = 0.02, size = 242, normalized size = 1.95 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{6}}{8}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{4}}{16}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{2}}{16}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x^{2}}{64 c}+\frac {3 a^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {3 a \,b^{2} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {3 b^{4} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a b}{32 c}-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{3}}{128 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^(3/2),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 details)Is 4*a*c-b^2 positive, negative or zero?

________________________________________________________________________________________

mupad [B]  time = 4.96, size = 115, normalized size = 0.93 \begin {gather*} \frac {\left (c\,x^2+\frac {b}{2}\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}+\frac {\left (3\,a\,c-\frac {3\,b^2}{4}\right )\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\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 )}{2\,c^{3/2}}\right )}{8\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \end {gather*}

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)

________________________________________________________________________________________