Integrand size = 51, antiderivative size = 42 \[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} x^{1+m}}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \] Output:
1/2*arctan(2*d^(1/2)*f^(1/2)*x^(1+m)/(2*f*x^3+e))/d^(1/2)/f^(1/2)
Time = 2.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} x^{1+m}}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}} \] Input:
Integrate[(x^m*(e*(1 + m) + 2*f*(-2 + m)*x^3))/(e^2 + 4*e*f*x^3 + 4*f^2*x^ 6 + 4*d*f*x^(2 + 2*m)),x]
Output:
ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^3)]/(2*Sqrt[d]*Sqrt[f])
Time = 0.61 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2520, 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.
\(\displaystyle \int \frac {x^m \left (e (m+1)+2 f (m-2) x^3\right )}{4 d f x^{2 m+2}+e^2+4 e f x^3+4 f^2 x^6} \, dx\) |
\(\Big \downarrow \) 2520 |
\(\displaystyle -e^2 (2-m) (m+1) \int \frac {1}{\frac {4 d e^2 f x^{2 m+2}}{\left (2 f x^3+e\right )^2}+e^2}d\left (-\frac {x^{m+1}}{(2-m) (m+1) \left (2 f x^3+e\right )}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} x^{m+1}}{e+2 f x^3}\right )}{2 \sqrt {d} \sqrt {f}}\) |
Input:
Int[(x^m*(e*(1 + m) + 2*f*(-2 + m)*x^3))/(e^2 + 4*e*f*x^3 + 4*f^2*x^6 + 4* d*f*x^(2 + 2*m)),x]
Output:
ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^3)]/(2*Sqrt[d]*Sqrt[f])
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[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.) *(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_Symbol] :> Simp[A^2*((m - n + 1)/(m + 1) ) Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m - n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && EqQ[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]
Leaf count of result is larger than twice the leaf count of optimal. \(77\) vs. \(2(32)=64\).
Time = 1.41 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.86
method | result | size |
risch | \(-\frac {\ln \left (x^{m}+\frac {\left (2 f \,x^{3}+e \right ) \sqrt {-d f}}{2 d f x}\right )}{4 \sqrt {-d f}}+\frac {\ln \left (x^{m}-\frac {\left (2 f \,x^{3}+e \right ) \sqrt {-d f}}{2 d f x}\right )}{4 \sqrt {-d f}}\) | \(78\) |
Input:
int(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)) ,x,method=_RETURNVERBOSE)
Output:
-1/4/(-d*f)^(1/2)*ln(x^m+1/2*(2*f*x^3+e)*(-d*f)^(1/2)/d/f/x)+1/4/(-d*f)^(1 /2)*ln(x^m-1/2*(2*f*x^3+e)*(-d*f)^(1/2)/d/f/x)
Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.29 \[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\left [-\frac {\sqrt {-d f} \log \left (-\frac {4 \, f^{2} x^{6} - 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{3} + 4 \, {\left (2 \, f x^{4} + e x\right )} \sqrt {-d f} x^{m} + e^{2}}{4 \, f^{2} x^{6} + 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{3} + e^{2}}\right )}{4 \, d f}, \frac {\sqrt {d f} \arctan \left (\frac {2 \, \sqrt {d f} x x^{m}}{2 \, f x^{3} + e}\right )}{2 \, d f}\right ] \] Input:
integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2 +2*m)),x, algorithm="fricas")
Output:
[-1/4*sqrt(-d*f)*log(-(4*f^2*x^6 - 4*d*f*x^2*x^(2*m) + 4*e*f*x^3 + 4*(2*f* x^4 + e*x)*sqrt(-d*f)*x^m + e^2)/(4*f^2*x^6 + 4*d*f*x^2*x^(2*m) + 4*e*f*x^ 3 + e^2))/(d*f), 1/2*sqrt(d*f)*arctan(2*sqrt(d*f)*x*x^m/(2*f*x^3 + e))/(d* f)]
Timed out. \[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\text {Timed out} \] Input:
integrate(x**m*(e*(1+m)+2*f*(-2+m)*x**3)/(e**2+4*e*f*x**3+4*f**2*x**6+4*d* f*x**(2+2*m)),x)
Output:
Timed out
\[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\int { \frac {{\left (2 \, f {\left (m - 2\right )} x^{3} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2 \, m + 2} + e^{2}} \,d x } \] Input:
integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2 +2*m)),x, algorithm="maxima")
Output:
integrate((2*f*(m - 2)*x^3 + e*(m + 1))*x^m/(4*f^2*x^6 + 4*e*f*x^3 + 4*d*f *x^(2*m + 2) + e^2), x)
\[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\int { \frac {{\left (2 \, f {\left (m - 2\right )} x^{3} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2 \, m + 2} + e^{2}} \,d x } \] Input:
integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2 +2*m)),x, algorithm="giac")
Output:
integrate((2*f*(m - 2)*x^3 + e*(m + 1))*x^m/(4*f^2*x^6 + 4*e*f*x^3 + 4*d*f *x^(2*m + 2) + e^2), x)
Timed out. \[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\int \frac {x^m\,\left (2\,f\,\left (m-2\right )\,x^3+e\,\left (m+1\right )\right )}{e^2+4\,f^2\,x^6+4\,e\,f\,x^3+4\,d\,f\,x^{2\,m+2}} \,d x \] Input:
int((x^m*(e*(m + 1) + 2*f*x^3*(m - 2)))/(e^2 + 4*f^2*x^6 + 4*e*f*x^3 + 4*d *f*x^(2*m + 2)),x)
Output:
int((x^m*(e*(m + 1) + 2*f*x^3*(m - 2)))/(e^2 + 4*f^2*x^6 + 4*e*f*x^3 + 4*d *f*x^(2*m + 2)), x)
\[ \int \frac {x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx=\left (\int \frac {x^{m}}{4 x^{2 m} d f \,x^{2}+e^{2}+4 e f \,x^{3}+4 f^{2} x^{6}}d x \right ) e m +\left (\int \frac {x^{m}}{4 x^{2 m} d f \,x^{2}+e^{2}+4 e f \,x^{3}+4 f^{2} x^{6}}d x \right ) e +2 \left (\int \frac {x^{m} x^{3}}{4 x^{2 m} d f \,x^{2}+e^{2}+4 e f \,x^{3}+4 f^{2} x^{6}}d x \right ) f m -4 \left (\int \frac {x^{m} x^{3}}{4 x^{2 m} d f \,x^{2}+e^{2}+4 e f \,x^{3}+4 f^{2} x^{6}}d x \right ) f \] Input:
int(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)) ,x)
Output:
int(x**m/(4*x**(2*m)*d*f*x**2 + e**2 + 4*e*f*x**3 + 4*f**2*x**6),x)*e*m + int(x**m/(4*x**(2*m)*d*f*x**2 + e**2 + 4*e*f*x**3 + 4*f**2*x**6),x)*e + 2* int((x**m*x**3)/(4*x**(2*m)*d*f*x**2 + e**2 + 4*e*f*x**3 + 4*f**2*x**6),x) *f*m - 4*int((x**m*x**3)/(4*x**(2*m)*d*f*x**2 + e**2 + 4*e*f*x**3 + 4*f**2 *x**6),x)*f