\(\int \frac {x^3}{(d+e x^4) (a+b x^4+c x^8)^2} \, dx\) [107]

Optimal result

Integrand size = 27, antiderivative size = 243 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {-b c d+b^2 e-2 a c e-c (2 c d-b e) x^4}{4 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x^4+c x^8\right )}+\frac {(2 c d-b e) \left (2 c^2 d^2-b^2 e^2-2 c e (b d-3 a e)\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 \left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^2}+\frac {e^3 \log \left (d+e x^4\right )}{4 \left (c d^2-b d e+a e^2\right )^2}-\frac {e^3 \log \left (a+b x^4+c x^8\right )}{8 \left (c d^2-b d e+a e^2\right )^2} \] Output:

1/4*(-b*c*d+b^2*e-2*a*c*e-c*(-b*e+2*c*d)*x^4)/(-4*a*c+b^2)/(a*e^2-b*d*e+c* 
d^2)/(c*x^8+b*x^4+a)+1/4*(-b*e+2*c*d)*(2*c^2*d^2-b^2*e^2-2*c*e*(-3*a*e+b*d 
))*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)/(a*e^2-b*d*e 
+c*d^2)^2+1/4*e^3*ln(e*x^4+d)/(a*e^2-b*d*e+c*d^2)^2-1/8*e^3*ln(c*x^8+b*x^4 
+a)/(a*e^2-b*d*e+c*d^2)^2
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {1}{8} \left (\frac {-2 b^2 e+4 c \left (a e+c d x^4\right )+2 b c \left (d-e x^4\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right ) \left (a+b x^4+c x^8\right )}+\frac {2 (-2 c d+b e) \left (-2 c^2 d^2+b^2 e^2+2 c e (b d-3 a e)\right ) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2} \left (c d^2+e (-b d+a e)\right )^2}+\frac {2 e^3 \log \left (d+e x^4\right )}{\left (c d^2+e (-b d+a e)\right )^2}-\frac {e^3 \log \left (a+b x^4+c x^8\right )}{\left (c d^2+e (-b d+a e)\right )^2}\right ) \] Input:

Integrate[x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

((-2*b^2*e + 4*c*(a*e + c*d*x^4) + 2*b*c*(d - e*x^4))/((b^2 - 4*a*c)*(-(c* 
d^2) + e*(b*d - a*e))*(a + b*x^4 + c*x^8)) + (2*(-2*c*d + b*e)*(-2*c^2*d^2 
 + b^2*e^2 + 2*c*e*(b*d - 3*a*e))*ArcTan[(b + 2*c*x^4)/Sqrt[-b^2 + 4*a*c]] 
)/((-b^2 + 4*a*c)^(3/2)*(c*d^2 + e*(-(b*d) + a*e))^2) + (2*e^3*Log[d + e*x 
^4])/(c*d^2 + e*(-(b*d) + a*e))^2 - (e^3*Log[a + b*x^4 + c*x^8])/(c*d^2 + 
e*(-(b*d) + a*e))^2)/8
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1798, 1165, 1200, 2009}

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.

\(\displaystyle \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx\)

\(\Big \downarrow \) 1798

\(\displaystyle \frac {1}{4} \int \frac {1}{\left (e x^4+d\right ) \left (c x^8+b x^4+a\right )^2}dx^4\)

\(\Big \downarrow \) 1165

\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {c e (2 c d-b e) x^4+2 c^2 d^2-b^2 e^2-c e (b d-4 a e)}{\left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^4}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e-b^2 e+c x^4 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right ) \left (a e^2-b d e+c d^2\right )}\right )\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {1}{4} \left (-\frac {\int \left (\frac {c \left (b^2-4 a c\right ) e^3 x^4+2 c^3 d^3+b^3 e^3-5 a b c e^3-3 c^2 d e (b d-2 a e)}{\left (c d^2-b e d+a e^2\right ) \left (c x^8+b x^4+a\right )}-\frac {\left (b^2-4 a c\right ) e^4}{\left (c d^2-b e d+a e^2\right ) \left (e x^4+d\right )}\right )dx^4}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e-b^2 e+c x^4 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right ) \left (a e^2-b d e+c d^2\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (-\frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right ) \left (6 a c e^2-b^2 e^2-2 b c d e+2 c^2 d^2\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {e^3 \left (b^2-4 a c\right ) \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}+\frac {e^3 \left (b^2-4 a c\right ) \log \left (a+b x^4+c x^8\right )}{2 \left (a e^2-b d e+c d^2\right )}}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {2 a c e-b^2 e+c x^4 (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right ) \left (a e^2-b d e+c d^2\right )}\right )\)

Input:

Int[x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

(-((b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x^4)/((b^2 - 4*a*c)*(c*d^2 - 
 b*d*e + a*e^2)*(a + b*x^4 + c*x^8))) - (-(((2*c*d - b*e)*(2*c^2*d^2 - 2*b 
*c*d*e - b^2*e^2 + 6*a*c*e^2)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(S 
qrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2))) - ((b^2 - 4*a*c)*e^3*Log[d + e* 
x^4])/(c*d^2 - b*d*e + a*e^2) + ((b^2 - 4*a*c)*e^3*Log[a + b*x^4 + c*x^8]) 
/(2*(c*d^2 - b*d*e + a*e^2)))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)))/4
 

Defintions of rubi rules used

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1798

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[(d + e*x)^q*(a + b 
*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && 
EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 5.14 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.50

method	result	size
default	\(-\frac {\frac {\frac {c \left (a b \,e^{3}-2 a c d \,e^{2}-b^{2} d \,e^{2}+3 b c \,d^{2} e -2 c^{2} d^{3}\right ) x^{4}}{4 a c -b^{2}}-\frac {2 a^{2} c \,e^{3}-a \,b^{2} e^{3}-a b c d \,e^{2}+2 a \,c^{2} d^{2} e +b^{3} d \,e^{2}-2 b^{2} c \,d^{2} e +b \,c^{2} d^{3}}{4 a c -b^{2}}}{2 c \,x^{8}+2 b \,x^{4}+2 a}+\frac {\frac {\left (4 a \,c^{2} e^{3}-b^{2} c \,e^{3}\right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{2 c}+\frac {2 \left (5 a b c \,e^{3}-6 a \,c^{2} d \,e^{2}-b^{3} e^{3}+3 b \,c^{2} d^{2} e -2 c^{3} d^{3}-\frac {\left (4 a \,c^{2} e^{3}-b^{2} c \,e^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{8 a c -2 b^{2}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {e^{3} \ln \left (x^{4} e +d \right )}{4 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}\)	\(364\)
risch	\(\text {Expression too large to display}\)	\(18132\)

Input:

int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2/(a*e^2-b*d*e+c*d^2)^2*(1/2*(c*(a*b*e^3-2*a*c*d*e^2-b^2*d*e^2+3*b*c*d^ 
2*e-2*c^2*d^3)/(4*a*c-b^2)*x^4-(2*a^2*c*e^3-a*b^2*e^3-a*b*c*d*e^2+2*a*c^2* 
d^2*e+b^3*d*e^2-2*b^2*c*d^2*e+b*c^2*d^3)/(4*a*c-b^2))/(c*x^8+b*x^4+a)+1/2/ 
(4*a*c-b^2)*(1/2*(4*a*c^2*e^3-b^2*c*e^3)/c*ln(c*x^8+b*x^4+a)+2*(5*a*b*c*e^ 
3-6*a*c^2*d*e^2-b^3*e^3+3*b*c^2*d^2*e-2*c^3*d^3-1/2*(4*a*c^2*e^3-b^2*c*e^3 
)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2))))+1/4*e^3*l 
n(e*x^4+d)/(a*e^2-b*d*e+c*d^2)^2
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**3/(e*x**4+d)/(c*x**8+b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^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. 500 vs. \(2 (230) = 460\).

Time = 4.64 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.06 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {e^{4} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )}} - \frac {e^{3} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} - \frac {{\left (4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 12 \, a c^{2} d e^{2} + b^{3} e^{3} - 6 \, a b c e^{3}\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + 2 \, a c^{2} d^{2} e + b^{3} d e^{2} - a b c d e^{2} - a b^{2} e^{3} + 2 \, a^{2} c e^{3} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2} + 2 \, a c^{2} d e^{2} - a b c e^{3}\right )} x^{4}}{4 \, {\left (c x^{8} + b x^{4} + a\right )} {\left (c d^{2} - b d e + a e^{2}\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \] Input:

integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="giac")
 

Output:

1/4*e^4*log(abs(e*x^4 + d))/(c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3 + 2*a 
*c*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5) - 1/8*e^3*log(c*x^8 + b*x^4 + a)/(c^2* 
d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4) - 
 1/4*(4*c^3*d^3 - 6*b*c^2*d^2*e + 12*a*c^2*d*e^2 + b^3*e^3 - 6*a*b*c*e^3)* 
arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/((b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b 
^3*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2 
*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4)*sq 
rt(-b^2 + 4*a*c)) - 1/4*(b*c^2*d^3 - 2*b^2*c*d^2*e + 2*a*c^2*d^2*e + b^3*d 
*e^2 - a*b*c*d*e^2 - a*b^2*e^3 + 2*a^2*c*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e 
+ b^2*c*d*e^2 + 2*a*c^2*d*e^2 - a*b*c*e^3)*x^4)/((c*x^8 + b*x^4 + a)*(c*d^ 
2 - b*d*e + a*e^2)^2*(b^2 - 4*a*c))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Hanged} \] Input:

int(x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\int \frac {x^{3}}{\left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )^{2}}d x \] Input:

int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)
 

Output:

int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)