Integrand size = 17, antiderivative size = 93 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}+\frac {\left (\frac {3 a}{d^2}-\frac {5 c}{e^2}\right ) x}{8 \left (d+e x^2\right )}+\frac {3 \left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \] Output:
1/4*(a+c*d^2/e^2)*x/d/(e*x^2+d)^2+(3*a/d^2-5*c/e^2)*x/(8*e*x^2+8*d)+3/8*(a *e^2+c*d^2)*arctan(e^(1/2)*x/d^(1/2))/d^(5/2)/e^(5/2)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {a e^2 x \left (5 d+3 e x^2\right )-c d^2 x \left (3 d+5 e x^2\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {3 \left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \] Input:
Integrate[(a + c*x^4)/(d + e*x^2)^3,x]
Output:
(a*e^2*x*(5*d + 3*e*x^2) - c*d^2*x*(3*d + 5*e*x^2))/(8*d^2*e^2*(d + e*x^2) ^2) + (3*(c*d^2 + a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*e^(5/2))
Time = 0.39 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1472, 25, 25, 27, 298, 218}
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.
\(\displaystyle \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1472 |
\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\int -\frac {4 c d x^2+\left (3 a-\frac {c d^2}{e^2}\right ) e}{e \left (e x^2+d\right )^2}dx}{4 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int -\frac {\frac {c d^2}{e}-4 c x^2 d-3 a e}{e \left (e x^2+d\right )^2}dx}{4 d}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\int \frac {\frac {c d^2}{e}-4 c x^2 d-3 a e}{e \left (e x^2+d\right )^2}dx}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\int \frac {\frac {c d^2}{e}-4 c x^2 d-3 a e}{\left (e x^2+d\right )^2}dx}{4 d e}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\frac {x \left (\frac {5 c d}{e}-\frac {3 a e}{d}\right )}{2 \left (d+e x^2\right )}-\frac {3 \left (a e^2+c d^2\right ) \int \frac {1}{e x^2+d}dx}{2 d e}}{4 d e}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x \left (a+\frac {c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\frac {x \left (\frac {5 c d}{e}-\frac {3 a e}{d}\right )}{2 \left (d+e x^2\right )}-\frac {3 \left (a e^2+c d^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{3/2}}}{4 d e}\) |
Input:
Int[(a + c*x^4)/(d + e*x^2)^3,x]
Output:
((a + (c*d^2)/e^2)*x)/(4*d*(d + e*x^2)^2) - ((((5*c*d)/e - (3*a*e)/d)*x)/( 2*(d + e*x^2)) - (3*(c*d^2 + a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2 )*e^(3/2)))/(4*d*e)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom ialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, Simp[(-R)*x*((d + e*x^2) ^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1 )*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {\frac {\left (3 a \,e^{2}-5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {3 \left (a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 d^{2} e^{2} \sqrt {d e}}\) | \(92\) |
risch | \(\frac {\frac {\left (3 a \,e^{2}-5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-d e}\right ) a}{16 \sqrt {-d e}\, d^{2}}-\frac {3 \ln \left (e x +\sqrt {-d e}\right ) c}{16 \sqrt {-d e}\, e^{2}}+\frac {3 \ln \left (-e x +\sqrt {-d e}\right ) a}{16 \sqrt {-d e}\, d^{2}}+\frac {3 \ln \left (-e x +\sqrt {-d e}\right ) c}{16 \sqrt {-d e}\, e^{2}}\) | \(153\) |
Input:
int((c*x^4+a)/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
(1/8*(3*a*e^2-5*c*d^2)/d^2/e*x^3+1/8*(5*a*e^2-3*c*d^2)/d/e^2*x)/(e*x^2+d)^ 2+3/8*(a*e^2+c*d^2)/d^2/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))
Time = 0.08 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.29 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} x^{3} + 3 \, {\left (c d^{4} + a d^{2} e^{2} + {\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \, {\left (c d^{3} e + a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} x}{16 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac {{\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} x^{3} - 3 \, {\left (c d^{4} + a d^{2} e^{2} + {\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \, {\left (c d^{3} e + a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{4} e - 5 \, a d^{2} e^{3}\right )} x}{8 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \] Input:
integrate((c*x^4+a)/(e*x^2+d)^3,x, algorithm="fricas")
Output:
[-1/16*(2*(5*c*d^3*e^2 - 3*a*d*e^4)*x^3 + 3*(c*d^4 + a*d^2*e^2 + (c*d^2*e^ 2 + a*e^4)*x^4 + 2*(c*d^3*e + a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt (-d*e)*x - d)/(e*x^2 + d)) + 2*(3*c*d^4*e - 5*a*d^2*e^3)*x)/(d^3*e^5*x^4 + 2*d^4*e^4*x^2 + d^5*e^3), -1/8*((5*c*d^3*e^2 - 3*a*d*e^4)*x^3 - 3*(c*d^4 + a*d^2*e^2 + (c*d^2*e^2 + a*e^4)*x^4 + 2*(c*d^3*e + a*d*e^3)*x^2)*sqrt(d* e)*arctan(sqrt(d*e)*x/d) + (3*c*d^4*e - 5*a*d^2*e^3)*x)/(d^3*e^5*x^4 + 2*d ^4*e^4*x^2 + d^5*e^3)]
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (90) = 180\).
Time = 0.36 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.35 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log {\left (- \frac {3 d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log {\left (\frac {3 d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a e^{3} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \] Input:
integrate((c*x**4+a)/(e*x**2+d)**3,x)
Output:
-3*sqrt(-1/(d**5*e**5))*(a*e**2 + c*d**2)*log(-3*d**3*e**2*sqrt(-1/(d**5*e **5))*(a*e**2 + c*d**2)/(3*a*e**2 + 3*c*d**2) + x)/16 + 3*sqrt(-1/(d**5*e* *5))*(a*e**2 + c*d**2)*log(3*d**3*e**2*sqrt(-1/(d**5*e**5))*(a*e**2 + c*d* *2)/(3*a*e**2 + 3*c*d**2) + x)/16 + (x**3*(3*a*e**3 - 5*c*d**2*e) + x*(5*a *d*e**2 - 3*c*d**3))/(8*d**4*e**2 + 16*d**3*e**3*x**2 + 8*d**2*e**4*x**4)
Exception generated. \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c*x^4+a)/(e*x^2+d)^3,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(e>0)', see `assume?` for more de tails)Is e
Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {3 \, {\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{2}} - \frac {5 \, c d^{2} e x^{3} - 3 \, a e^{3} x^{3} + 3 \, c d^{3} x - 5 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{2}} \] Input:
integrate((c*x^4+a)/(e*x^2+d)^3,x, algorithm="giac")
Output:
3/8*(c*d^2 + a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^2*e^2) - 1/8*(5*c*d ^2*e*x^3 - 3*a*e^3*x^3 + 3*c*d^3*x - 5*a*d*e^2*x)/((e*x^2 + d)^2*d^2*e^2)
Time = 17.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,a\,e^2-5\,c\,d^2\right )}{8\,d^2\,e}+\frac {x\,\left (5\,a\,e^2-3\,c\,d^2\right )}{8\,d\,e^2}}{d^2+2\,d\,e\,x^2+e^2\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}} \] Input:
int((a + c*x^4)/(d + e*x^2)^3,x)
Output:
((x^3*(3*a*e^2 - 5*c*d^2))/(8*d^2*e) + (x*(5*a*e^2 - 3*c*d^2))/(8*d*e^2))/ (d^2 + e^2*x^4 + 2*d*e*x^2) + (3*atan((e^(1/2)*x)/d^(1/2))*(a*e^2 + c*d^2) )/(8*d^(5/2)*e^(5/2))
Time = 0.18 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.38 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2} e^{2}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d \,e^{3} x^{2}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{4} x^{4}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{4}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{3} e \,x^{2}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c \,d^{2} e^{2} x^{4}+5 a \,d^{2} e^{3} x +3 a d \,e^{4} x^{3}-3 c \,d^{4} e x -5 c \,d^{3} e^{2} x^{3}}{8 d^{3} e^{3} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int((c*x^4+a)/(e*x^2+d)^3,x)
Output:
(3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2*e**2 + 6*sqrt(e)*s qrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e**3*x**2 + 3*sqrt(e)*sqrt(d)*ata n((e*x)/(sqrt(e)*sqrt(d)))*a*e**4*x**4 + 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqr t(e)*sqrt(d)))*c*d**4 + 6*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c* d**3*e*x**2 + 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c*d**2*e**2* x**4 + 5*a*d**2*e**3*x + 3*a*d*e**4*x**3 - 3*c*d**4*e*x - 5*c*d**3*e**2*x* *3)/(8*d**3*e**3*(d**2 + 2*d*e*x**2 + e**2*x**4))