\(\int x (a+b x^2+c x^4)^{3/2} \, dx\) [967]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4} \left (-3 b^2+20 a c+8 b c x^2+8 c^2 x^4\right )}{128 c^2}+\frac {3 \left (-b^2+4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2+c x^4}}\right )}{128 c^{5/2}} \] Input:

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

Output:

((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))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1432, 1087, 1087, 1092, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1432

\(\displaystyle \frac {1}{2} \int \left (c x^4+b x^2+a\right )^{3/2}dx^2\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^4+b x^2+a}dx^2}{16 c}\right )\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c}\right )}{16 c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c}\right )}{16 c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{3/2}}\right )}{16 c}\right )\)

Input:

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

Output:

(((b + 2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 
 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(4*c) - ((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))))/(16*c))/2
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

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] - Simp[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] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

Maple [A] (verified)

Time = 0.21 (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	\(-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}\)	\(242\)
elliptic	\(-\frac {3 b^{2} a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {3}{2}}}+\frac {5 b a \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 c}+\frac {b^{2} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 c}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{256 c^{\frac {5}{2}}}-\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 c^{2}}+\frac {3 b \,x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {5 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{16}+\frac {3 a^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}+\frac {c \,x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}\)	\(242\)

Input:

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

Output:

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)

Time = 0.09 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.40 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, c^{3}}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{6} + 24 \, b c^{3} x^{4} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, c^{3}}\right ] \] Input:

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

Output:

[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]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 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 = 0.15 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.51 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} b + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} c \] Input:

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

Output:

1/16*(2*sqrt(c*x^4 + b*x^2 + a)*(2*x^2 + b/c) + (b^2 - 4*a*c)*log(abs(2*(s 
qrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(3/2))*a + 1/96*(2*s 
qrt(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))*sq 
rt(c) + b))/c^(7/2))*c
 

Mupad [B] (verification not implemented)

Time = 17.71 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (c\,x^2+\frac {b}{2}\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{8\,c}+\frac {\left (3\,a\,c-\frac {3\,b^2}{4}\right )\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2+a}+\frac {\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{8\,c} \] Input:

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

Output:

((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)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 2869, normalized size of antiderivative = 23.14 \[ \int x \left (a+b x^2+c x^4\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

(1536*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c 
*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**3*b*c**3 + 3072*sqrt(c)*sqrt 
(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x 
**2)/sqrt(4*a*c - b**2))*a**3*c**4*x**2 - 384*sqrt(c)*sqrt(a + b*x**2 + c* 
x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c 
- b**2))*a**2*b**3*c**2 + 2304*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sq 
rt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b 
**2*c**3*x**2 + 9216*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt 
(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b*c**4*x**4 
 + 6144*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + 
 c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*c**5*x**6 - 96*sqrt(c)*s 
qrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2* 
c*x**2)/sqrt(4*a*c - b**2))*a*b**5*c - 1728*sqrt(c)*sqrt(a + b*x**2 + c*x* 
*4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - 
b**2))*a*b**4*c**2*x**2 - 4608*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sq 
rt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b**3 
*c**3*x**4 - 3072*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a 
+ b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a*b**2*c**4*x**6 + 
24*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x* 
*4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**7 + 240*sqrt(c)*sqrt(a + b*x...