Integrand size = 25, antiderivative size = 72 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 c \sqrt {b^2-4 a c}}+\frac {e \log \left (a+b x^4+c x^8\right )}{8 c} \] Output:
-1/4*(-b*e+2*c*d)*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^( 1/2)+1/8*e*ln(c*x^8+b*x^4+a)/c
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {-\frac {2 (-2 c d+b e) \arctan \left (\frac {b+2 c x^4}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+e \log \left (a+b x^4+c x^8\right )}{8 c} \] Input:
Integrate[(x^3*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
((-2*(-2*c*d + b*e)*ArcTan[(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])/(8*c)
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1798, 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.
\(\displaystyle \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx\) |
\(\Big \downarrow \) 1798 |
\(\displaystyle \frac {1}{4} \int \frac {e x^4+d}{c x^8+b x^4+a}dx^4\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{4} \left (\frac {(2 c d-b e) \int \frac {1}{c x^8+b x^4+a}dx^4}{2 c}+\frac {e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{2 c}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{4} \left (\frac {e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{2 c}-\frac {(2 c d-b e) \int \frac {1}{-x^8+b^2-4 a c}d\left (2 c x^4+b\right )}{c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (\frac {e \int \frac {2 c x^4+b}{c x^8+b x^4+a}dx^4}{2 c}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{4} \left (\frac {e \log \left (a+b x^4+c x^8\right )}{2 c}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )\) |
Input:
Int[(x^3*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
(-(((2*c*d - b*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c])) + (e*Log[a + b*x^4 + c*x^8])/(2*c))/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[(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 = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {e \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{8 c}+\frac {\left (d -\frac {e b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}}\) | \(66\) |
risch | \(\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}+2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) a e}{8 a c -2 b^{2}}-\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}+2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2} e}{8 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d +\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}+2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}}{8 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}-2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) a e}{8 a c -2 b^{2}}-\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}-2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2} e}{8 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c e +8 a \,c^{2} d +b^{3} e -2 b^{2} c d -\sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{4}-2 \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-\left (e b -2 c d \right )^{2} \left (4 a c -b^{2}\right )}}{8 c \left (4 a c -b^{2}\right )}\) | \(671\) |
Input:
int(x^3*(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
1/8*e*ln(c*x^8+b*x^4+a)/c+1/2*(d-1/2*e*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c* x^4+b)/(4*a*c-b^2)^(1/2))
Time = 0.18 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.00 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\left [\frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{8} + b x^{4} + a\right ) - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c d - b e\right )} \log \left (\frac {2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c + {\left (2 \, c x^{4} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right )}{8 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {{\left (b^{2} - 4 \, a c\right )} e \log \left (c x^{8} + b x^{4} + a\right ) - 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {{\left (2 \, c x^{4} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{8 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \] Input:
integrate(x^3*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
[1/8*((b^2 - 4*a*c)*e*log(c*x^8 + b*x^4 + a) - sqrt(b^2 - 4*a*c)*(2*c*d - b*e)*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)))/(b^2*c - 4*a*c^2), 1/8*((b^2 - 4*a*c)*e*log(c *x^8 + b*x^4 + a) - 2*sqrt(-b^2 + 4*a*c)*(2*c*d - b*e)*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)))/(b^2*c - 4*a*c^2)]
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (63) = 126\).
Time = 10.55 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.99 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\left (\frac {e}{8 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{4} + \frac {- 16 a c \left (\frac {e}{8 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + 4 b^{2} \left (\frac {e}{8 c} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} + \left (\frac {e}{8 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{4} + \frac {- 16 a c \left (\frac {e}{8 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) + 2 a e + 4 b^{2} \left (\frac {e}{8 c} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{8 c \left (4 a c - b^{2}\right )}\right ) - b d}{b e - 2 c d} \right )} \] Input:
integrate(x**3*(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
(e/(8*c) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(8*c*(4*a*c - b**2)))*log(x** 4 + (-16*a*c*(e/(8*c) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(8*c*(4*a*c - b* *2))) + 2*a*e + 4*b**2*(e/(8*c) - sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(8*c*( 4*a*c - b**2))) - b*d)/(b*e - 2*c*d)) + (e/(8*c) + sqrt(-4*a*c + b**2)*(b* e - 2*c*d)/(8*c*(4*a*c - b**2)))*log(x**4 + (-16*a*c*(e/(8*c) + sqrt(-4*a* c + b**2)*(b*e - 2*c*d)/(8*c*(4*a*c - b**2))) + 2*a*e + 4*b**2*(e/(8*c) + sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(8*c*(4*a*c - b**2))) - b*d)/(b*e - 2*c* d))
Exception generated. \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Exception raised: ValueError} \] 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 = 1.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {e \log \left (c x^{8} + b x^{4} + a\right )}{8 \, c} + \frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} c} \] Input:
integrate(x^3*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
1/8*e*log(c*x^8 + b*x^4 + a)/c + 1/4*(2*c*d - b*e)*arctan((2*c*x^4 + b)/sq rt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c)
Time = 23.41 (sec) , antiderivative size = 3704, normalized size of antiderivative = 51.44 \[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:
int((x^3*(d + e*x^4))/(a + b*x^4 + c*x^8),x)
Output:
- (log(a + b*x^4 + c*x^8)*(4*b^2*e - 16*a*c*e))/(2*(64*a*c^2 - 16*b^2*c)) - (atan((8*x^4*(((a*c - b^2)*(((((4*b^2*e - 16*a*c*e)*(((b*e - 2*c*d)*(448 *b^3*c^3*e - 384*b^2*c^4*d + (256*b^3*c^4*(4*b^2*e - 16*a*c*e))/(64*a*c^2 - 16*b^2*c)))/(8*c*(4*a*c - b^2)^(1/2)) + (32*b^3*c^3*(4*b^2*e - 16*a*c*e) *(b*e - 2*c*d))/((64*a*c^2 - 16*b^2*c)*(4*a*c - b^2)^(1/2))))/(2*(64*a*c^2 - 16*b^2*c)) + ((b*e - 2*c*d)*(96*b*c^4*d^2 + ((4*b^2*e - 16*a*c*e)*(448* b^3*c^3*e - 384*b^2*c^4*d + (256*b^3*c^4*(4*b^2*e - 16*a*c*e))/(64*a*c^2 - 16*b^2*c)))/(2*(64*a*c^2 - 16*b^2*c)) + 144*b^3*c^2*e^2 - 240*b^2*c^3*d*e ))/(8*c*(4*a*c - b^2)^(1/2)))*(4*b^2*e - 16*a*c*e))/(2*(64*a*c^2 - 16*b^2* c)) - ((((b*e - 2*c*d)*(((b*e - 2*c*d)*(448*b^3*c^3*e - 384*b^2*c^4*d + (2 56*b^3*c^4*(4*b^2*e - 16*a*c*e))/(64*a*c^2 - 16*b^2*c)))/(8*c*(4*a*c - b^2 )^(1/2)) + (32*b^3*c^3*(4*b^2*e - 16*a*c*e)*(b*e - 2*c*d))/((64*a*c^2 - 16 *b^2*c)*(4*a*c - b^2)^(1/2))))/(8*c*(4*a*c - b^2)^(1/2)) + (4*b^3*c^2*(4*b ^2*e - 16*a*c*e)*(b*e - 2*c*d)^2)/((64*a*c^2 - 16*b^2*c)*(4*a*c - b^2)))*( b*e - 2*c*d))/(8*c*(4*a*c - b^2)^(1/2)) + ((b*e - 2*c*d)*(((4*b^2*e - 16*a *c*e)*(96*b*c^4*d^2 + ((4*b^2*e - 16*a*c*e)*(448*b^3*c^3*e - 384*b^2*c^4*d + (256*b^3*c^4*(4*b^2*e - 16*a*c*e))/(64*a*c^2 - 16*b^2*c)))/(2*(64*a*c^2 - 16*b^2*c)) + 144*b^3*c^2*e^2 - 240*b^2*c^3*d*e))/(2*(64*a*c^2 - 16*b^2* c)) - 8*c^4*d^3 + 20*b^3*c*e^3 - 48*b^2*c^2*d*e^2 + 36*b*c^3*d^2*e))/(8*c* (4*a*c - b^2)^(1/2)) - (b^3*c*(4*b^2*e - 16*a*c*e)*(b*e - 2*c*d)^3)/(2*...
\[ \int \frac {x^3 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\int \frac {x^{3} \left (e \,x^{4}+d \right )}{c \,x^{8}+b \,x^{4}+a}d x \] 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)