Integrand size = 27, antiderivative size = 133 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {e \log \left (d+e x^4\right )}{4 \left (c d^2-b d e+a e^2\right )}-\frac {e \log \left (a+b x^4+c x^8\right )}{8 \left (c d^2-b d e+a e^2\right )} \] Output:
-1/4*(-b*e+2*c*d)*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/ 2)/(a*e^2-b*d*e+c*d^2)+e*ln(e*x^4+d)/(4*a*e^2-4*b*d*e+4*c*d^2)-e*ln(c*x^8+ b*x^4+a)/(8*a*e^2-8*b*d*e+8*c*d^2)
Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {(-4 c d+2 b e) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} e \left (-2 \log \left (d+e x^4\right )+\log \left (a+b x^4+c x^8\right )\right )}{8 \sqrt {-b^2+4 a c} \left (-c d^2+e (b d-a e)\right )} \] Input:
Integrate[x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
((-4*c*d + 2*b*e)*ArcTan[(b + 2*c*x^4)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4 *a*c]*e*(-2*Log[d + e*x^4] + Log[a + b*x^4 + c*x^8]))/(8*Sqrt[-b^2 + 4*a*c ]*(-(c*d^2) + e*(b*d - a*e)))
Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1798, 1144, 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.
\(\displaystyle \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^4\) |
\(\Big \downarrow \) 1144 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {-c e x^4+c d-b e}{c x^8+b x^4+a}dx^4}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{4} \left (\frac {\frac {1}{2} (2 c d-b e) \int \frac {1}{c x^8+b x^4+a}dx^4-\frac {1}{2} e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{4} \left (\frac {-\left ((2 c d-b e) \int \frac {1}{-x^8+b^2-4 a c}d\left (2 c x^4+b\right )\right )-\frac {1}{2} e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {1}{2} e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{4} \left (\frac {-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {1}{2} e \log \left (a+b x^4+c x^8\right )}{a e^2-b d e+c d^2}+\frac {e \log \left (d+e x^4\right )}{a e^2-b d e+c d^2}\right )\) |
Input:
Int[x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
((e*Log[d + e*x^4])/(c*d^2 - b*d*e + a*e^2) + (-(((2*c*d - b*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/Sqrt[b^2 - 4*a*c]) - (e*Log[a + b*x^4 + c* x^8])/2)/(c*d^2 - b*d*e + a*e^2))/4
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])
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]
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]
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]
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[e*(Log[RemoveContent[d + e*x, x]]/(c*d^2 - b*d*e + a*e^2)), x] + S imp[1/(c*d^2 - b*d*e + a*e^2) Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
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]
Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\frac {e \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4}+\frac {\left (\frac {e b}{2}-c d \right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {e \ln \left (x^{4} e +d \right )}{4 a \,e^{2}-4 b d e +4 c \,d^{2}}\) | \(112\) |
risch | \(\frac {e \ln \left (x^{4} e +d \right )}{4 a \,e^{2}-4 b d e +4 c \,d^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c \,e^{2}-a \,b^{2} e^{2}-4 a b c d e +4 a \,c^{2} d^{2}+b^{3} d e -b^{2} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (4 a c e -b^{2} e \right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (4 a^{2} c \,e^{3}-a \,b^{2} e^{3}+3 a b c d \,e^{2}-4 a \,c^{2} d^{2} e -b^{3} d \,e^{2}+2 b^{2} c \,d^{2} e -b \,c^{2} d^{3}\right ) \textit {\_R}^{2}+\left (4 a c \,e^{2}-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \textit {\_R} +c e \right ) x^{4}+\left (6 a^{2} c d \,e^{2}-2 a \,b^{2} d \,e^{2}+2 a b c \,d^{2} e -2 a \,c^{2} d^{3}\right ) \textit {\_R}^{2}+3 \textit {\_R} a c d e \right )\right )}{4}\) | \(271\) |
Input:
int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/2/(a*e^2-b*d*e+c*d^2)*(1/4*e*ln(c*x^8+b*x^4+a)+(1/2*e*b-c*d)/(4*a*c-b^2 )^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2)))+1/4*e/(a*e^2-b*d*e+c*d^2)*l n(e*x^4+d)
Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(x**3/(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),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
Time = 1.04 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {e^{2} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac {e \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (c d^{2} - b d e + a e^{2}\right )}} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \] Input:
integrate(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
1/4*e^2*log(abs(e*x^4 + d))/(c*d^2*e - b*d*e^2 + a*e^3) - 1/8*e*log(c*x^8 + b*x^4 + a)/(c*d^2 - b*d*e + a*e^2) + 1/4*(2*c*d - b*e)*arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/((c*d^2 - b*d*e + a*e^2)*sqrt(-b^2 + 4*a*c))
Timed out. \[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Hanged} \] Input:
int(x^3/((d + e*x^4)*(a + b*x^4 + c*x^8)),x)
Output:
\text{Hanged}
\[ \int \frac {x^3}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {x^{3}}{\left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x)