\(\int x^3 \sqrt {a+b x^2+c x^4} \, dx\) [922]

Optimal result

Integrand size = 20, antiderivative size = 108 \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=-\frac {b \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c^2}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{5/2}} \]

[Out]

1/6*(c*x^4+b*x^2+a)^(3/2)/c+1/32*b*(-4*a*c+b^2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(5/2)
-1/16*b*(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 = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1128, 654, 626, 635, 212} \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{32 c^{5/2}}-\frac {b \left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{16 c^2}+\frac {\left (a+b x^2+c x^4\right )^{3/2}}{6 c} \]

[In]

Int[x^3*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

-1/16*(b*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/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 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 654

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 1128

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {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 \text {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 ) \text {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 ) \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 )}{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] (verified)

Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95 \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\frac {2 \sqrt {c} \sqrt {a+b x^2+c x^4} \left (-3 b^2+2 b c x^2+8 c \left (a+c x^4\right )\right )-3 \left (b^3-4 a b c\right ) \log \left (c^2 \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right )}{96 c^{5/2}} \]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {\left (8 c^{2} x^{4}+2 b c \,x^{2}+8 a c -3 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{48 c^{2}}-\frac {b \left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {5}{2}}}\)	\(91\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\)	\(99\)
elliptic	\(\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{6 c}-\frac {b \left (\frac {\left (2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\)	\(99\)
pseudoelliptic	\(\frac {16 c^{\frac {5}{2}} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}+4 b \,c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+12 \ln \left (2\right ) a b c -3 \ln \left (2\right ) b^{3}-12 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) a b c +3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{3}+16 a \,c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}-6 b^{2} \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{96 c^{\frac {5}{2}}}\)	\(182\)

[In]

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

[Out]

1/48*(8*c^2*x^4+2*b*c*x^2+8*a*c-3*b^2)*(c*x^4+b*x^2+a)^(1/2)/c^2-1/32*b*(4*a*c-b^2)/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 = 237, normalized size of antiderivative = 2.19 \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{3}}, -\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{3}}\right ] \]

[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]

Sympy [F]

\[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\int x^{3} \sqrt {a + b x^{2} + c x^{4}}\, dx \]

[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)

Maxima [F(-2)]

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

[In]

integrate(x^3*(c*x^4+b*x^2+a)^(1/2),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 [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89 \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\frac {1}{48} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {{\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{32 \, c^{\frac {5}{2}}} \]

[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*(sqr
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(5/2)

Mupad [B] (verification not implemented)

Time = 13.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int x^3 \sqrt {a+b x^2+c x^4} \, dx=\frac {\left (8\,c\,\left (c\,x^4+a\right )-3\,b^2+2\,b\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2+a}}{48\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^4+b\,x^2+a}+\frac {2\,c\,x^2+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{32\,c^{5/2}} \]

[In]

int(x^3*(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

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