\(\int \frac {e^{2 x} (-275+3050 x+700 x^2)}{25+140 x+196 x^2} \, dx\) [198]

Optimal result

Integrand size = 28, antiderivative size = 21 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=25-\frac {5 e^{2 x} (4+x)}{-1-\frac {14 x}{5}} \]

[Out]

25-5/(-1-14/5*x)*exp(x)^2*(4+x)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 2230, 2225, 2208, 2209} \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (14 x+5)} \]

[In]

Int[(E^(2*x)*(-275 + 3050*x + 700*x^2))/(25 + 140*x + 196*x^2),x]

[Out]

(25*E^(2*x))/14 + (1275*E^(2*x))/(14*(5 + 14*x))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

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 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{(5+14 x)^2} \, dx \\ & = \int \left (\frac {25 e^{2 x}}{7}-\frac {1275 e^{2 x}}{(5+14 x)^2}+\frac {1275 e^{2 x}}{7 (5+14 x)}\right ) \, dx \\ & = \frac {25}{7} \int e^{2 x} \, dx+\frac {1275}{7} \int \frac {e^{2 x}}{5+14 x} \, dx-1275 \int \frac {e^{2 x}}{(5+14 x)^2} \, dx \\ & = \frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (5+14 x)}+\frac {1275 \operatorname {ExpIntegralEi}\left (\frac {1}{7} (5+14 x)\right )}{98 e^{5/7}}-\frac {1275}{7} \int \frac {e^{2 x}}{5+14 x} \, dx \\ & = \frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (5+14 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {25 e^{2 x} (4+x)}{5+14 x} \]

[In]

Integrate[(E^(2*x)*(-275 + 3050*x + 700*x^2))/(25 + 140*x + 196*x^2),x]

[Out]

(25*E^(2*x)*(4 + x))/(5 + 14*x)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {25 \left (4+x \right ) {\mathrm e}^{2 x}}{14 x +5}\)	\(17\)
risch	\(\frac {25 \left (4+x \right ) {\mathrm e}^{2 x}}{14 x +5}\)	\(17\)
default	\(\frac {1275 \,{\mathrm e}^{2 x}}{196 \left (x +\frac {5}{14}\right )}+\frac {25 \,{\mathrm e}^{2 x}}{14}\)	\(19\)
norman	\(\frac {100 \,{\mathrm e}^{2 x}+25 x \,{\mathrm e}^{2 x}}{14 x +5}\)	\(23\)
parallelrisch	\(\frac {350 x \,{\mathrm e}^{2 x}+1400 \,{\mathrm e}^{2 x}}{196 x +70}\)	\(24\)

[In]

int((700*x^2+3050*x-275)*exp(x)^2/(196*x^2+140*x+25),x,method=_RETURNVERBOSE)

[Out]

25*(4+x)*exp(x)^2/(14*x+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {25 \, {\left (x + 4\right )} e^{\left (2 \, x\right )}}{14 \, x + 5} \]

[In]

integrate((700*x^2+3050*x-275)*exp(x)^2/(196*x^2+140*x+25),x, algorithm="fricas")

[Out]

25*(x + 4)*e^(2*x)/(14*x + 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {\left (25 x + 100\right ) e^{2 x}}{14 x + 5} \]

[In]

integrate((700*x**2+3050*x-275)*exp(x)**2/(196*x**2+140*x+25),x)

[Out]

(25*x + 100)*exp(2*x)/(14*x + 5)

Maxima [F]

\[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\int { \frac {25 \, {\left (28 \, x^{2} + 122 \, x - 11\right )} e^{\left (2 \, x\right )}}{196 \, x^{2} + 140 \, x + 25} \,d x } \]

[In]

integrate((700*x^2+3050*x-275)*exp(x)^2/(196*x^2+140*x+25),x, algorithm="maxima")

[Out]

25*(14*x^2 + 61*x)*e^(2*x)/(196*x^2 + 140*x + 25) + 275/14*e^(-5/7)*exp_integral_e(2, -2*x - 5/7)/(14*x + 5) +
25*integrate((714*x - 305)*e^(2*x)/(2744*x^3 + 2940*x^2 + 1050*x + 125), x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {25 \, {\left (x e^{\left (2 \, x\right )} + 4 \, e^{\left (2 \, x\right )}\right )}}{14 \, x + 5} \]

[In]

integrate((700*x^2+3050*x-275)*exp(x)^2/(196*x^2+140*x+25),x, algorithm="giac")

[Out]

25*(x*e^(2*x) + 4*e^(2*x))/(14*x + 5)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 x} \left (-275+3050 x+700 x^2\right )}{25+140 x+196 x^2} \, dx=\frac {25\,{\mathrm {e}}^{2\,x}\,\left (x+4\right )}{14\,x+5} \]

[In]

int((exp(2*x)*(3050*x + 700*x^2 - 275))/(140*x + 196*x^2 + 25),x)

[Out]

(25*exp(2*x)*(x + 4))/(14*x + 5)