\(\int \frac {x^{11} (d+e x^4)}{(a+b x^4+c x^8)^2} \, dx\) [89]

Optimal result

Integrand size = 25, antiderivative size = 148 \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\frac {x^4 \left (a (2 c d-b e)+\left (b c d-b^2 e+2 a c e\right ) x^4\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}+\frac {\left (b^3 e+2 a c (2 c d-3 b e)\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {e \log \left (a+b x^4+c x^8\right )}{8 c^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.08 \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\frac {-\frac {2 \left (2 a^2 c e+b^2 (c d-b e) x^4+a \left (-b^2 e-2 c^2 d x^4+b c \left (d+3 e x^4\right )\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}+\frac {2 \left (b^3 e+2 a c (2 c d-3 b e)\right ) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+e \log \left (a+b x^4+c x^8\right )}{8 c^2} \] Input:

Integrate[(x^11*(d + e*x^4))/(a + b*x^4 + c*x^8)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1802, 1233, 25, 1142, 1083, 219, 1103}

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.

\(\displaystyle \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx\)

\(\Big \downarrow \) 1802

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

\(\Big \downarrow \) 1233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^11*(d + e*x^4))/(a + b*x^4 + c*x^8)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 1083

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

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1233

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

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]]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {\frac {\left (3 a b c e -2 a \,c^{2} d -b^{3} e +b^{2} c d \right ) x^{4}}{c^{2} \left (4 a c -b^{2}\right )}+\frac {a \left (2 a c e -b^{2} e +c b d \right )}{c^{2} \left (4 a c -b^{2}\right )}}{4 c \,x^{8}+4 b \,x^{4}+4 a}+\frac {\frac {\left (4 a c e -b^{2} e \right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{2 c}+\frac {2 \left (-a b e +2 a c d -\frac {\left (4 a c e -b^{2} e \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 \left (4 a c -b^{2}\right ) c}\)	\(210\)
risch	\(\text {Expression too large to display}\)	\(1346\)

Input:

int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (138) = 276\).

Time = 1.41 (sec) , antiderivative size = 844, normalized size of antiderivative = 5.70 \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^11*(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="fricas")
 

Output:

[-1/8*(2*((b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*d - (b^5 - 7*a*b^3*c + 12*a^2* 
b*c^2)*e)*x^4 - ((4*a*c^3*d + (b^3*c - 6*a*b*c^2)*e)*x^8 + 4*a^2*c^2*d + ( 
4*a*b*c^2*d + (b^4 - 6*a*b^2*c)*e)*x^4 + (a*b^3 - 6*a^2*b*c)*e)*sqrt(b^2 - 
 4*a*c)*log((2*c^2*x^8 + 2*b*c*x^4 + b^2 - 2*a*c + (2*c*x^4 + b)*sqrt(b^2 
- 4*a*c))/(c*x^8 + b*x^4 + a)) + 2*(a*b^3*c - 4*a^2*b*c^2)*d - 2*(a*b^4 - 
6*a^2*b^2*c + 8*a^3*c^2)*e - ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e*x^8 + ( 
b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e*x^4 + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2) 
*e)*log(c*x^8 + b*x^4 + a))/((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^8 + a* 
b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c 
^4)*x^4), -1/8*(2*((b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*d - (b^5 - 7*a*b^3*c 
+ 12*a^2*b*c^2)*e)*x^4 - 2*((4*a*c^3*d + (b^3*c - 6*a*b*c^2)*e)*x^8 + 4*a^ 
2*c^2*d + (4*a*b*c^2*d + (b^4 - 6*a*b^2*c)*e)*x^4 + (a*b^3 - 6*a^2*b*c)*e) 
*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c) 
) + 2*(a*b^3*c - 4*a^2*b*c^2)*d - 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e - 
((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e*x^8 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^ 
2)*e*x^4 + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e)*log(c*x^8 + b*x^4 + a))/( 
(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^8 + a*b^4*c^2 - 8*a^2*b^2*c^3 + 16* 
a^3*c^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**11*(e*x**4+d)/(c*x**8+b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^11*(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 [A] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=-\frac {{\left (4 \, a c^{2} d + b^{3} e - 6 \, a b c e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {e \log \left (c x^{8} + b x^{4} + a\right )}{8 \, c^{2}} - \frac {b^{2} c e x^{8} - 4 \, a c^{2} e x^{8} + 2 \, b^{2} c d x^{4} - 4 \, a c^{2} d x^{4} - b^{3} e x^{4} + 2 \, a b c e x^{4} + 2 \, a b c d - a b^{2} e}{8 \, {\left (c x^{8} + b x^{4} + a\right )} {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} \] Input:

integrate(x^11*(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 37.25 (sec) , antiderivative size = 12336, normalized size of antiderivative = 83.35 \[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\text {Too large to display} \] Input:

int((x^11*(d + e*x^4))/(a + b*x^4 + c*x^8)^2,x)
                                                                                    
                                                                                    
 

Output:

((a*(2*a*c*e - b^2*e + b*c*d))/(4*c^2*(4*a*c - b^2)) - (x^4*(b^3*e + 2*a*c 
^2*d - b^2*c*d - 3*a*b*c*e))/(4*c^2*(4*a*c - b^2)))/(a + b*x^4 + c*x^8) - 
(log(a + b*x^4 + c*x^8)*(4*b^6*e - 256*a^3*c^3*e + 192*a^2*b^2*c^2*e - 48* 
a*b^4*c*e))/(2*(1024*a^3*c^5 - 16*b^6*c^2 + 192*a*b^4*c^3 - 768*a^2*b^2*c^ 
4)) - (atan(((a*c - b^2)*(512*a^3*c^7*(4*a*c - b^2)^6 - 8*b^6*c^4*(4*a*c - 
 b^2)^6 + 96*a*b^4*c^5*(4*a*c - b^2)^6 - 384*a^2*b^2*c^6*(4*a*c - b^2)^6)* 
((((((((1024*a^2*b^3*c^8*d - 6656*a^2*b^4*c^7*e + 14336*a^3*b^2*c^8*e - 40 
96*a^3*b*c^9*d + 768*a*b^6*c^6*e)/(16*a^2*c^6 + b^4*c^4 - 8*a*b^2*c^5) + ( 
(1024*a*b^6*c^8 - 8192*a^2*b^4*c^9 + 16384*a^3*b^2*c^10)*(4*b^6*e - 256*a^ 
3*c^3*e + 192*a^2*b^2*c^2*e - 48*a*b^4*c*e))/(2*(16*a^2*c^6 + b^4*c^4 - 8* 
a*b^2*c^5)*(1024*a^3*c^5 - 16*b^6*c^2 + 192*a*b^4*c^3 - 768*a^2*b^2*c^4))) 
*(b^3*e + 4*a*c^2*d - 6*a*b*c*e))/(8*c^2*(4*a*c - b^2)^(3/2)) + ((b^3*e + 
4*a*c^2*d - 6*a*b*c*e)*(1024*a*b^6*c^8 - 8192*a^2*b^4*c^9 + 16384*a^3*b^2* 
c^10)*(4*b^6*e - 256*a^3*c^3*e + 192*a^2*b^2*c^2*e - 48*a*b^4*c*e))/(16*c^ 
2*(4*a*c - b^2)^(3/2)*(16*a^2*c^6 + b^4*c^4 - 8*a*b^2*c^5)*(1024*a^3*c^5 - 
 16*b^6*c^2 + 192*a*b^4*c^3 - 768*a^2*b^2*c^4)))*(4*b^6*e - 256*a^3*c^3*e 
+ 192*a^2*b^2*c^2*e - 48*a*b^4*c*e))/(2*(1024*a^3*c^5 - 16*b^6*c^2 + 192*a 
*b^4*c^3 - 768*a^2*b^2*c^4)) + (((256*a^3*c^8*d^2 + 208*a*b^6*c^4*e^2 - 19 
20*a^2*b^4*c^5*e^2 + 4416*a^3*b^2*c^6*e^2 - 2304*a^3*b*c^7*d*e + 512*a^2*b 
^3*c^6*d*e)/(16*a^2*c^6 + b^4*c^4 - 8*a*b^2*c^5) + (((1024*a^2*b^3*c^8*...
 

Reduce [F]

\[ \int \frac {x^{11} \left (d+e x^4\right )}{\left (a+b x^4+c x^8\right )^2} \, dx=\int \frac {x^{11} \left (e \,x^{4}+d \right )}{\left (c \,x^{8}+b \,x^{4}+a \right )^{2}}d x \] Input:

int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)
 

Output:

int(x^11*(e*x^4+d)/(c*x^8+b*x^4+a)^2,x)