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

Optimal. Leaf size=21 \[ 25-\frac {5 e^{2 x} (4+x)}{-1-\frac {14 x}{5}} \]

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Rubi [A]  time = 0.12, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 2199, 2194, 2177, 2178} \begin {gather*} \frac {25 e^{2 x}}{14}+\frac {1275 e^{2 x}}{14 (14 x+5)} \end {gather*}

Antiderivative was successfully verified.

[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 2177

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] &&  !$UseGamma ===
True

Rule 2178

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

Rule 2194

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 2199

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] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \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 \text {Ei}\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 0.81 \begin {gather*} \frac {25 e^{2 x} (4+x)}{5+14 x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.93, size = 16, normalized size = 0.76 \begin {gather*} \frac {25 \, {\left (x + 4\right )} e^{\left (2 \, x\right )}}{14 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 22, normalized size = 1.05 \begin {gather*} \frac {25 \, {\left (x e^{\left (2 \, x\right )} + 4 \, e^{\left (2 \, x\right )}\right )}}{14 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.08, size = 17, normalized size = 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	\(\frac {100 \,{\mathrm e}^{2 x}+25 x \,{\mathrm e}^{2 x}}{14 x +5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {25 \, {\left (14 \, x^{2} + 61 \, x\right )} e^{\left (2 \, x\right )}}{196 \, x^{2} + 140 \, x + 25} + \frac {275 \, e^{\left (-\frac {5}{7}\right )} E_{2}\left (-2 \, x - \frac {5}{7}\right )}{14 \, {\left (14 \, x + 5\right )}} + 25 \, \int \frac {{\left (714 \, x - 305\right )} e^{\left (2 \, x\right )}}{2744 \, x^{3} + 2940 \, x^{2} + 1050 \, x + 125}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.76 \begin {gather*} \frac {25\,{\mathrm {e}}^{2\,x}\,\left (x+4\right )}{14\,x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.67 \begin {gather*} \frac {\left (25 x + 100\right ) e^{2 x}}{14 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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