3.15.87 \(\int \frac {-252+180 x+e^{2 x} (-50+150 x-150 x^2+50 x^3)+e^x (-270+630 x-510 x^2+150 x^3)}{-25+75 x-75 x^2+25 x^3} \, dx\)

Optimal. Leaf size=35 \[ \left (e^x+x-\frac {(-1+x)^2-x-\frac {5+x}{5 (1-x)}}{x}\right )^2 \]

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Rubi [A]  time = 0.20, antiderivative size = 43, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 6, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6688, 2194, 37, 2199, 2177, 2178} \begin {gather*} \frac {9 (7-5 x)^2}{25 (1-x)^2}+6 e^x+e^{2 x}+\frac {12 e^x}{5 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-252 + 180*x + E^(2*x)*(-50 + 150*x - 150*x^2 + 50*x^3) + E^x*(-270 + 630*x - 510*x^2 + 150*x^3))/(-25 +
75*x - 75*x^2 + 25*x^3),x]

[Out]

6*E^x + E^(2*x) + (9*(7 - 5*x)^2)/(25*(1 - x)^2) + (12*E^x)/(5*(1 - x))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{2 x}+\frac {36 (-7+5 x)}{25 (-1+x)^3}+\frac {6 e^x \left (9-12 x+5 x^2\right )}{5 (-1+x)^2}\right ) \, dx\\ &=\frac {6}{5} \int \frac {e^x \left (9-12 x+5 x^2\right )}{(-1+x)^2} \, dx+\frac {36}{25} \int \frac {-7+5 x}{(-1+x)^3} \, dx+2 \int e^{2 x} \, dx\\ &=e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {6}{5} \int \left (5 e^x+\frac {2 e^x}{(-1+x)^2}-\frac {2 e^x}{-1+x}\right ) \, dx\\ &=e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12}{5} \int \frac {e^x}{(-1+x)^2} \, dx-\frac {12}{5} \int \frac {e^x}{-1+x} \, dx+6 \int e^x \, dx\\ &=6 e^x+e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12 e^x}{5 (1-x)}-\frac {12}{5} e \text {Ei}(-1+x)+\frac {12}{5} \int \frac {e^x}{-1+x} \, dx\\ &=6 e^x+e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12 e^x}{5 (1-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 43, normalized size = 1.23 \begin {gather*} 6 e^x+e^{2 x}+\frac {12 e^x}{5 (1-x)}+\frac {36}{25 (-1+x)^2}-\frac {36}{5 (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-252 + 180*x + E^(2*x)*(-50 + 150*x - 150*x^2 + 50*x^3) + E^x*(-270 + 630*x - 510*x^2 + 150*x^3))/(
-25 + 75*x - 75*x^2 + 25*x^3),x]

[Out]

6*E^x + E^(2*x) + (12*E^x)/(5*(1 - x)) + 36/(25*(-1 + x)^2) - 36/(5*(-1 + x))

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fricas [A]  time = 0.66, size = 45, normalized size = 1.29 \begin {gather*} \frac {25 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 30 \, {\left (5 \, x^{2} - 12 \, x + 7\right )} e^{x} - 180 \, x + 216}{25 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-150*x^2+150*x-50)*exp(x)^2+(150*x^3-510*x^2+630*x-270)*exp(x)+180*x-252)/(25*x^3-75*x^2+75*
x-25),x, algorithm="fricas")

[Out]

1/25*(25*(x^2 - 2*x + 1)*e^(2*x) + 30*(5*x^2 - 12*x + 7)*e^x - 180*x + 216)/(x^2 - 2*x + 1)

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giac [A]  time = 0.28, size = 55, normalized size = 1.57 \begin {gather*} \frac {25 \, x^{2} e^{\left (2 \, x\right )} + 150 \, x^{2} e^{x} - 50 \, x e^{\left (2 \, x\right )} - 360 \, x e^{x} - 180 \, x + 25 \, e^{\left (2 \, x\right )} + 210 \, e^{x} + 216}{25 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-150*x^2+150*x-50)*exp(x)^2+(150*x^3-510*x^2+630*x-270)*exp(x)+180*x-252)/(25*x^3-75*x^2+75*
x-25),x, algorithm="giac")

[Out]

1/25*(25*x^2*e^(2*x) + 150*x^2*e^x - 50*x*e^(2*x) - 360*x*e^x - 180*x + 25*e^(2*x) + 210*e^x + 216)/(x^2 - 2*x
+ 1)

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maple [A]  time = 0.05, size = 33, normalized size = 0.94

method	result	size
default	\(\frac {36}{25 \left (x -1\right )^{2}}-\frac {36}{5 \left (x -1\right )}-\frac {12 \,{\mathrm e}^{x}}{5 \left (x -1\right )}+6 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}\)	\(33\)
risch	\(\frac {-\frac {36 x}{5}+\frac {216}{25}}{x^{2}-2 x +1}+{\mathrm e}^{2 x}+\frac {6 \left (5 x -7\right ) {\mathrm e}^{x}}{5 \left (x -1\right )}\)	\(36\)
norman	\(\frac {{\mathrm e}^{2 x}-\frac {36 x}{5}+{\mathrm e}^{2 x} x^{2}-2 x \,{\mathrm e}^{2 x}-\frac {72 \,{\mathrm e}^{x} x}{5}+6 \,{\mathrm e}^{x} x^{2}+\frac {42 \,{\mathrm e}^{x}}{5}+\frac {216}{25}}{\left (x -1\right )^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^3-150*x^2+150*x-50)*exp(x)^2+(150*x^3-510*x^2+630*x-270)*exp(x)+180*x-252)/(25*x^3-75*x^2+75*x-25),
x,method=_RETURNVERBOSE)

[Out]

36/25/(x-1)^2-36/5/(x-1)-12/5/(x-1)*exp(x)+6*exp(x)+exp(x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )} e^{\left (2 \, x\right )} + 6 \, {\left (5 \, x^{3} - 17 \, x^{2} + 19 \, x\right )} e^{x}}{5 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} - \frac {18 \, {\left (2 \, x - 1\right )}}{5 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {54 \, e E_{3}\left (-x + 1\right )}{5 \, {\left (x - 1\right )}^{2}} + \frac {126}{25 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {2}{25} \, \int \frac {15 \, {\left (2 \, x + 19\right )} e^{x}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-150*x^2+150*x-50)*exp(x)^2+(150*x^3-510*x^2+630*x-270)*exp(x)+180*x-252)/(25*x^3-75*x^2+75*
x-25),x, algorithm="maxima")

[Out]

1/5*(5*(x^3 - 3*x^2 + 3*x - 1)*e^(2*x) + 6*(5*x^3 - 17*x^2 + 19*x)*e^x)/(x^3 - 3*x^2 + 3*x - 1) - 18/5*(2*x -
1)/(x^2 - 2*x + 1) + 54/5*e*exp_integral_e(3, -x + 1)/(x - 1)^2 + 126/25/(x^2 - 2*x + 1) + 2/25*integrate(15*(
2*x + 19)*e^x/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1), x)

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mupad [B]  time = 0.13, size = 30, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^x+\frac {\frac {12\,{\mathrm {e}}^x}{5}-x\,\left (\frac {12\,{\mathrm {e}}^x}{5}+\frac {36}{5}\right )+\frac {216}{25}}{{\left (x-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((180*x + exp(2*x)*(150*x - 150*x^2 + 50*x^3 - 50) + exp(x)*(630*x - 510*x^2 + 150*x^3 - 270) - 252)/(75*x
- 75*x^2 + 25*x^3 - 25),x)

[Out]

exp(2*x) + 6*exp(x) + ((12*exp(x))/5 - x*((12*exp(x))/5 + 36/5) + 216/25)/(x - 1)^2

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sympy [A]  time = 0.18, size = 37, normalized size = 1.06 \begin {gather*} \frac {216 - 180 x}{25 x^{2} - 50 x + 25} + \frac {\left (5 x - 5\right ) e^{2 x} + \left (30 x - 42\right ) e^{x}}{5 x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**3-150*x**2+150*x-50)*exp(x)**2+(150*x**3-510*x**2+630*x-270)*exp(x)+180*x-252)/(25*x**3-75*x
**2+75*x-25),x)

[Out]

(216 - 180*x)/(25*x**2 - 50*x + 25) + ((5*x - 5)*exp(2*x) + (30*x - 42)*exp(x))/(5*x - 5)

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