\(\int \frac {630-172 x+65 x^2+5 x^4+e^x (405+90 x^2+5 x^4)}{405+90 x^2+5 x^4} \, dx\) [608]

Optimal result

Integrand size = 46, antiderivative size = 31 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=e^x+x-\frac {\frac {4}{5}+(3-x)^2+x+2 x^2}{9+x^2} \]

[Out]

exp(x)+x-((-x+3)^2+2*x^2+x+4/5)/(x^2+9)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {28, 6874, 2225, 205, 209, 267, 294, 327} \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=\frac {x}{2 \left (x^2+9\right )}+\frac {86}{5 \left (x^2+9\right )}-\frac {x^3}{2 \left (x^2+9\right )}+\frac {3 x}{2}+e^x \]

[In]

Int[(630 - 172*x + 65*x^2 + 5*x^4 + E^x*(405 + 90*x^2 + 5*x^4))/(405 + 90*x^2 + 5*x^4),x]

[Out]

E^x + (3*x)/2 + 86/(5*(9 + x^2)) + x/(2*(9 + x^2)) - x^3/(2*(9 + x^2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = 5 \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{\left (45+5 x^2\right )^2} \, dx \\ & = 5 \int \left (\frac {e^x}{5}+\frac {126}{5 \left (9+x^2\right )^2}-\frac {172 x}{25 \left (9+x^2\right )^2}+\frac {13 x^2}{5 \left (9+x^2\right )^2}+\frac {x^4}{5 \left (9+x^2\right )^2}\right ) \, dx \\ & = 13 \int \frac {x^2}{\left (9+x^2\right )^2} \, dx-\frac {172}{5} \int \frac {x}{\left (9+x^2\right )^2} \, dx+126 \int \frac {1}{\left (9+x^2\right )^2} \, dx+\int e^x \, dx+\int \frac {x^4}{\left (9+x^2\right )^2} \, dx \\ & = e^x+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )}+\frac {3}{2} \int \frac {x^2}{9+x^2} \, dx+\frac {13}{2} \int \frac {1}{9+x^2} \, dx+7 \int \frac {1}{9+x^2} \, dx \\ & = e^x+\frac {3 x}{2}+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )}+\frac {9}{2} \arctan \left (\frac {x}{3}\right )-\frac {27}{2} \int \frac {1}{9+x^2} \, dx \\ & = e^x+\frac {3 x}{2}+\frac {86}{5 \left (9+x^2\right )}+\frac {x}{2 \left (9+x^2\right )}-\frac {x^3}{2 \left (9+x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=\frac {1}{5} \left (5 e^x+5 x+\frac {86+25 x}{9+x^2}\right ) \]

[In]

Integrate[(630 - 172*x + 65*x^2 + 5*x^4 + E^x*(405 + 90*x^2 + 5*x^4))/(405 + 90*x^2 + 5*x^4),x]

[Out]

(5*E^x + 5*x + (86 + 25*x)/(9 + x^2))/5

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58

method	result	size
risch	\(x +\frac {5 x +\frac {86}{5}}{x^{2}+9}+{\mathrm e}^{x}\)	\(18\)
parts	\(x -\frac {-25 x -86}{5 \left (x^{2}+9\right )}+{\mathrm e}^{x}\)	\(19\)
default	\(\frac {5 x}{x^{2}+9}+\frac {86}{5 \left (x^{2}+9\right )}+x +{\mathrm e}^{x}\)	\(24\)
norman	\(\frac {x^{3}+{\mathrm e}^{x} x^{2}+14 x +9 \,{\mathrm e}^{x}+\frac {86}{5}}{x^{2}+9}\)	\(27\)
parallelrisch	\(\frac {5 \,{\mathrm e}^{x} x^{2}+5 x^{3}+45 \,{\mathrm e}^{x}+70 x +86}{5 x^{2}+45}\)	\(31\)

[In]

int(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x,method=_RETURNVERBOSE)

[Out]

x+(5*x+86/5)/(x^2+9)+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=\frac {5 \, x^{3} + 5 \, {\left (x^{2} + 9\right )} e^{x} + 70 \, x + 86}{5 \, {\left (x^{2} + 9\right )}} \]

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="fricas")

[Out]

1/5*(5*x^3 + 5*(x^2 + 9)*e^x + 70*x + 86)/(x^2 + 9)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=x + \frac {25 x + 86}{5 x^{2} + 45} + e^{x} \]

[In]

integrate(((5*x**4+90*x**2+405)*exp(x)+5*x**4+65*x**2-172*x+630)/(5*x**4+90*x**2+405),x)

[Out]

x + (25*x + 86)/(5*x**2 + 45) + exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=x + \frac {5 \, x}{x^{2} + 9} + \frac {86}{5 \, {\left (x^{2} + 9\right )}} + e^{x} \]

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="maxima")

[Out]

x + 5*x/(x^2 + 9) + 86/5/(x^2 + 9) + e^x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=\frac {5 \, x^{3} + 5 \, x^{2} e^{x} + 70 \, x + 45 \, e^{x} + 86}{5 \, {\left (x^{2} + 9\right )}} \]

[In]

integrate(((5*x^4+90*x^2+405)*exp(x)+5*x^4+65*x^2-172*x+630)/(5*x^4+90*x^2+405),x, algorithm="giac")

[Out]

1/5*(5*x^3 + 5*x^2*e^x + 70*x + 45*e^x + 86)/(x^2 + 9)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {630-172 x+65 x^2+5 x^4+e^x \left (405+90 x^2+5 x^4\right )}{405+90 x^2+5 x^4} \, dx=x+{\mathrm {e}}^x+\frac {25\,x+86}{5\,x^2+45} \]

[In]

int((exp(x)*(90*x^2 + 5*x^4 + 405) - 172*x + 65*x^2 + 5*x^4 + 630)/(90*x^2 + 5*x^4 + 405),x)

[Out]

x + exp(x) + (25*x + 86)/(5*x^2 + 45)