\(\int \frac {x^7}{(d+e x^4) (a+b x^4+c x^8)^2} \, dx\) [106]

Optimal result

Integrand size = 27, antiderivative size = 254 \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {a (2 c d-b e)+c (b d-2 a e) x^4}{4 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x^4+c x^8\right )}+\frac {\left (4 b^2 c d^2 e-b^3 d e^2-4 a c e \left (c d^2-a e^2\right )-2 b c d \left (c d^2+a e^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 \left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^2}-\frac {d e^2 \log \left (d+e x^4\right )}{4 \left (c d^2-b d e+a e^2\right )^2}+\frac {d e^2 \log \left (a+b x^4+c x^8\right )}{8 \left (c d^2-b d e+a e^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00 \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {1}{8} \left (\frac {2 \left (-b c d x^4+a \left (-2 c d+b e+2 c e x^4\right )\right )}{\left (b^2-4 a c\right ) \left (-c d^2+e (b d-a e)\right ) \left (a+b x^4+c x^8\right )}-\frac {2 \left (-4 b^2 c d^2 e+b^3 d e^2+4 a c e \left (c d^2-a e^2\right )+2 b c d \left (c d^2+a e^2\right )\right ) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2} \left (c d^2+e (-b d+a e)\right )^2}-\frac {2 d e^2 \log \left (d+e x^4\right )}{\left (c d^2+e (-b d+a e)\right )^2}+\frac {d e^2 \log \left (a+b x^4+c x^8\right )}{\left (c d^2+e (-b d+a e)\right )^2}\right ) \] Input:

Integrate[x^7/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1802, 1235, 25, 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.

\(\displaystyle \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx\)

\(\Big \downarrow \) 1802

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

\(\Big \downarrow \) 1235

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^7/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

((a*(2*c*d - b*e) + c*(b*d - 2*a*e)*x^4)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a 
*e^2)*(a + b*x^4 + c*x^8)) + (((4*b^2*c*d^2*e - b^3*d*e^2 - 4*a*c*e*(c*d^2 
 - a*e^2) - 2*b*c*d*(c*d^2 + a*e^2))*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a* 
c]])/(Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)) - ((b^2 - 4*a*c)*d*e^2*Lo 
g[d + e*x^4])/(c*d^2 - b*d*e + a*e^2) + ((b^2 - 4*a*c)*d*e^2*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 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*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 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] 
)
 

rule 1802

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1 
)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

Maple [A] (verified)

Time = 5.31 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.40

method	result	size
default	\(\frac {\frac {\frac {c \left (2 a^{2} e^{3}-3 a b d \,e^{2}+2 a c \,d^{2} e +b^{2} d^{2} e -b c \,d^{3}\right ) x^{4}}{4 a c -b^{2}}+\frac {a \left (a b \,e^{3}-2 a c d \,e^{2}-b^{2} d \,e^{2}+3 b c \,d^{2} e -2 c^{2} d^{3}\right )}{4 a c -b^{2}}}{2 c \,x^{8}+2 b \,x^{4}+2 a}+\frac {\frac {\left (4 a \,c^{2} d \,e^{2}-b^{2} c d \,e^{2}\right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{2 c}+\frac {2 \left (2 a^{2} c \,e^{3}+a b c d \,e^{2}-2 a \,c^{2} d^{2} e -b^{3} d \,e^{2}+2 b^{2} c \,d^{2} e -b \,c^{2} d^{3}-\frac {\left (4 a \,c^{2} d \,e^{2}-b^{2} c d \,e^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{8 a c -2 b^{2}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {d \,e^{2} \ln \left (x^{4} e +d \right )}{4 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}\)	\(356\)
risch	\(\text {Expression too large to display}\)	\(2068\)

Input:

int(x^7/(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*(2*a^2*e^3-3*a*b*d*e^2+2*a*c*d^2*e+b^2*d 
^2*e-b*c*d^3)/(4*a*c-b^2)*x^4+a*(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))/(c*x^8+b*x^4+a)+1/2/(4*a*c-b^2)*(1/2*(4*a*c^2*d*e 
^2-b^2*c*d*e^2)/c*ln(c*x^8+b*x^4+a)+2*(2*a^2*c*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^3-1/2*(4*a*c^2*d*e^2-b^2*c*d*e^2)*b/c)/ 
(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2))))-1/4*d*e^2*ln(e*x 
^4+d)/(a*e^2-b*d*e+c*d^2)^2
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x^7/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**7/(e*x**4+d)/(c*x**8+b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^7/(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. 498 vs. \(2 (242) = 484\).

Time = 4.43 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.96 \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=-\frac {d e^{3} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )}} + \frac {d e^{2} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} + \frac {{\left (2 \, b c^{2} d^{3} - 4 \, b^{2} c d^{2} e + 4 \, a c^{2} d^{2} e + b^{3} d e^{2} + 2 \, a b c d e^{2} - 4 \, a^{2} c e^{3}\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {2 \, a c^{2} d^{3} - 3 \, a b c d^{2} e + a b^{2} d e^{2} + 2 \, a^{2} c d e^{2} - a^{2} b e^{3} + {\left (b c^{2} d^{3} - b^{2} c d^{2} e - 2 \, a c^{2} d^{2} e + 3 \, a b c d e^{2} - 2 \, a^{2} c e^{3}\right )} x^{4}}{4 \, {\left (c x^{8} + b x^{4} + a\right )} {\left (c d^{2} - b d e + a e^{2}\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \] Input:

integrate(x^7/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="giac")
 

Output:

-1/4*d*e^3*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*d*e^2*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*(2*b*c^2*d^3 - 4*b^2*c*d^2*e + 4*a*c^2*d^2*e + b^3*d*e^2 + 2*a*b 
*c*d*e^2 - 4*a^2*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)*sqrt(-b^2 + 4*a*c)) + 1/4*(2*a*c^2*d^3 - 3*a*b*c*d^2 
*e + a*b^2*d*e^2 + 2*a^2*c*d*e^2 - a^2*b*e^3 + (b*c^2*d^3 - b^2*c*d^2*e - 
2*a*c^2*d^2*e + 3*a*b*c*d*e^2 - 2*a^2*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. \[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Hanged} \] Input:

int(x^7/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {x^7}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\int \frac {x^{7}}{\left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )^{2}}d x \] Input:

int(x^7/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)
 

Output:

int(x^7/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)