\(\int x (a+b x^2+c x^4)^{3/2} \, dx\) [939]

Optimal result

Integrand size = 18, antiderivative size = 124 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{128 c^2}+\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2}} \]

[Out]

1/16*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(3/2)/c+3/256*(-4*a*c+b^2)^2*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^
(1/2))/c^(5/2)-3/128*(-4*a*c+b^2)*(2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1121, 626, 635, 212} \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{256 c^{5/2}}-\frac {3 \left (b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{128 c^2}+\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 c} \]

[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 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*} \text {integral}& = \frac {1}{2} \text {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 ) \text {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 ) \text {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 ) \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 )}{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] (verified)

Time = 0.80 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4} \left (-3 b^2+20 a c+8 b c x^2+8 c^2 x^4\right )}{128 c^2}+\frac {3 \left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{128 c^{5/2}} \]

[In]

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

[Out]

((b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 20*a*c + 8*b*c*x^2 + 8*c^2*x^4))/(128*c^2) + (3*(-b^2 + 4*a*c
)^2*ArcTanh[(Sqrt[c]*x^2)/(-Sqrt[a] + Sqrt[a + b*x^2 + c*x^4])])/(128*c^(5/2))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {\left (16 c^{3} x^{6}+24 b \,c^{2} x^{4}+40 a \,c^{2} x^{2}+2 b^{2} c \,x^{2}+20 a b c -3 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}\)	\(120\)
pseudoelliptic	\(\frac {\frac {3 \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )}{16}+\frac {3 \left (\frac {5 \left (\frac {b \,x^{2}}{10}+a \right ) b \,c^{\frac {3}{2}}}{6}+\left (b \,x^{4}+\frac {5}{3} a \,x^{2}\right ) c^{\frac {5}{2}}+\frac {2 c^{\frac {7}{2}} x^{6}}{3}-\frac {\sqrt {c}\, b^{3}}{8}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 \ln \left (2\right ) \left (a c -\frac {b^{2}}{4}\right )^{2}}{16}}{c^{\frac {5}{2}}}\)	\(128\)
default	\(\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}\)	\(242\)
elliptic	\(\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}\)	\(242\)

[In]

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

[Out]

1/128*(16*c^3*x^6+24*b*c^2*x^4+40*a*c^2*x^2+2*b^2*c*x^2+20*a*b*c-3*b^3)*(c*x^4+b*x^2+a)^(1/2)/c^2+3/256*(16*a^
2*c^2-8*a*b^2*c+b^4)/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.40 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, c^{3}}\right ] \]

[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]

Sympy [F]

\[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int x \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (106) = 212\).

Time = 0.32 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.51 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} c \]

[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

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (c\,x^2+\frac {b}{2}\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}+\frac {\left (3\,a\,c-\frac {3\,b^2}{4}\right )\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{8\,c} \]

[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)