\(\int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 (-25-16 x^2)}{25 e^4}}{50 x^2+32 x^4} \, dx\) [7281]

Optimal result

Integrand size = 43, antiderivative size = 28 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {(-1+2 x)^2}{2 x}+\frac {-1-\frac {16 x^2}{25}}{e^4} \]

[Out]

1/2*(-1+2*x)^2/x+exp(ln(-16/25*x^2-1)-4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {1607, 1600, 14} \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {16 x^2}{25 e^4}+2 x+\frac {1}{2 x} \]

[In]

Int[(-25 + 84*x^2 + 64*x^4 + (64*x^3*(-25 - 16*x^2))/(25*E^4))/(50*x^2 + 32*x^4),x]

[Out]

1/(2*x) + 2*x - (16*x^2)/(25*E^4)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{x^2 \left (50+32 x^2\right )} \, dx \\ & = \int \frac {-\frac {1}{2}+2 x^2-\frac {32 x^3}{25 e^4}}{x^2} \, dx \\ & = \int \left (2-\frac {1}{2 x^2}-\frac {32 x}{25 e^4}\right ) \, dx \\ & = \frac {1}{2 x}+2 x-\frac {16 x^2}{25 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {1}{2 x}+2 x-\frac {16 x^2}{25 e^4} \]

[In]

Integrate[(-25 + 84*x^2 + 64*x^4 + (64*x^3*(-25 - 16*x^2))/(25*E^4))/(50*x^2 + 32*x^4),x]

[Out]

1/(2*x) + 2*x - (16*x^2)/(25*E^4)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {16 \,{\mathrm e}^{-4} x^{2}}{25}+2 x +\frac {1}{2 x}\)	\(17\)
default	\(2 x +\frac {1}{2 x}+{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4}\)	\(21\)
norman	\(\frac {\frac {1}{2}+2 x^{2}-\frac {16 \,{\mathrm e}^{-4} x^{3}}{25}}{x}\)	\(21\)
parts	\(2 x +\frac {1}{2 x}+{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4}\)	\(21\)
parallelrisch	\(-\frac {-4000000-6553600 x^{6}-3276800 \,{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4} x^{5}-22118400 x^{4}-21120000 x^{2}+8000000 \,{\mathrm e}^{\ln \left (-\frac {16 x^{2}}{25}-1\right )-4} x}{12800 \left (16 x^{2}+25\right )^{2} x}\)	\(62\)

[In]

int((64*x^3*exp(ln(-16/25*x^2-1)-4)+64*x^4+84*x^2-25)/(32*x^4+50*x^2),x,method=_RETURNVERBOSE)

[Out]

-16/25*exp(-4)*x^2+2*x+1/2/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {{\left (32 \, x^{3} - 25 \, {\left (4 \, x^{2} + 1\right )} e^{4}\right )} e^{\left (-4\right )}}{50 \, x} \]

[In]

integrate((64*x^3*exp(log(-16/25*x^2-1)-4)+64*x^4+84*x^2-25)/(32*x^4+50*x^2),x, algorithm="fricas")

[Out]

-1/50*(32*x^3 - 25*(4*x^2 + 1)*e^4)*e^(-4)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {- 32 x^{2} + 100 x e^{4} + \frac {25 e^{4}}{x}}{50 e^{4}} \]

[In]

integrate((64*x**3*exp(ln(-16/25*x**2-1)-4)+64*x**4+84*x**2-25)/(32*x**4+50*x**2),x)

[Out]

(-32*x**2 + 100*x*exp(4) + 25*exp(4)/x)*exp(-4)/50

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {2}{25} \, {\left (8 \, x^{2} - 25 \, x e^{4}\right )} e^{\left (-4\right )} + \frac {1}{2 \, x} \]

[In]

integrate((64*x^3*exp(log(-16/25*x^2-1)-4)+64*x^4+84*x^2-25)/(32*x^4+50*x^2),x, algorithm="maxima")

[Out]

-2/25*(8*x^2 - 25*x*e^4)*e^(-4) + 1/2/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=-\frac {2}{25} \, {\left (8 \, x^{2} e^{4} - 25 \, x e^{8}\right )} e^{\left (-8\right )} + \frac {1}{2 \, x} \]

[In]

integrate((64*x^3*exp(log(-16/25*x^2-1)-4)+64*x^4+84*x^2-25)/(32*x^4+50*x^2),x, algorithm="giac")

[Out]

-2/25*(8*x^2*e^4 - 25*x*e^8)*e^(-8) + 1/2/x

Mupad [B] (verification not implemented)

Time = 11.78 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-25+84 x^2+64 x^4+\frac {64 x^3 \left (-25-16 x^2\right )}{25 e^4}}{50 x^2+32 x^4} \, dx=\frac {-32\,{\mathrm {e}}^{-4}\,x^3+100\,x^2+25}{50\,x} \]

[In]

int((84*x^2 + 64*x^4 + 64*x^3*exp(log(- (16*x^2)/25 - 1) - 4) - 25)/(50*x^2 + 32*x^4),x)

[Out]

(100*x^2 - 32*x^3*exp(-4) + 25)/(50*x)