3.39.58 \(\int \frac {e^{\frac {1}{15} (35 e^{\frac {e^6}{4}}+3 x)} (20+4 x-4 x^2)}{5-10 x+5 x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {4 e^{\frac {7 e^{\frac {e^6}{4}}}{3}+\frac {x}{5}} x}{1-x} \]

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Rubi [A]  time = 0.20, antiderivative size = 54, normalized size of antiderivative = 1.74, number of steps used = 8, number of rules used = 6, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {27, 12, 2199, 2194, 2177, 2178} \begin {gather*} \frac {4 e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )}}{1-x}-4 e^{\frac {1}{15} \left (3 x+35 e^{\frac {e^6}{4}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((35*E^(E^6/4) + 3*x)/15)*(20 + 4*x - 4*x^2))/(5 - 10*x + 5*x^2),x]

[Out]

-4*E^((35*E^(E^6/4) + 3*x)/15) + (4*E^((35*E^(E^6/4) + 3*x)/15))/(1 - x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{5 (-1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \left (20+4 x-4 x^2\right )}{(-1+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}+\frac {20 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{(-1+x)^2}-\frac {4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{-1+x}\right ) \, dx\\ &=-\left (\frac {4}{5} \int e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )} \, dx\right )-\frac {4}{5} \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{-1+x} \, dx+4 \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{(-1+x)^2} \, dx\\ &=-4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}+\frac {4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{1-x}-\frac {4}{5} e^{\frac {1}{5}+\frac {7 e^{\frac {e^6}{4}}}{3}} \text {Ei}\left (\frac {1}{5} (-1+x)\right )+\frac {4}{5} \int \frac {e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{-1+x} \, dx\\ &=-4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}+\frac {4 e^{\frac {1}{15} \left (35 e^{\frac {e^6}{4}}+3 x\right )}}{1-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 0.94 \begin {gather*} -\frac {4 e^{\frac {7 e^{\frac {e^6}{4}}}{3}+\frac {x}{5}} x}{-1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((35*E^(E^6/4) + 3*x)/15)*(20 + 4*x - 4*x^2))/(5 - 10*x + 5*x^2),x]

[Out]

(-4*E^((7*E^(E^6/4))/3 + x/5)*x)/(-1 + x)

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fricas [A]  time = 1.17, size = 20, normalized size = 0.65 \begin {gather*} -\frac {4 \, x e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x, algorithm="fricas")

[Out]

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1)

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giac [A]  time = 0.16, size = 20, normalized size = 0.65 \begin {gather*} -\frac {4 \, x e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x, algorithm="giac")

[Out]

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1)

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maple [A]  time = 0.13, size = 21, normalized size = 0.68

method	result	size
risch	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{x -1}\)	\(21\)
gosper	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{x -1}\)	\(23\)
norman	\(-\frac {4 x \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{x -1}\)	\(23\)
derivativedivides	\(\frac {12 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-3 x +3}-\frac {4 \,{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{5}+\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{5 \left (-3 x +3\right )}-180 \left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )-4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}-\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (1225 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}+210 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+9\right )}{15 \left (-3 x +3\right )}+4 \left (\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}}{9}+\frac {28 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{5}+\frac {11}{25}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )-420 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{225}\right )-4900 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{225}\right )+4200 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{-675 x +675}-\left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )\right )\)	\(476\)
default	\(\frac {12 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-3 x +3}-\frac {4 \,{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{5}+\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{5 \left (-3 x +3\right )}-180 \left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )-4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}-\frac {4 \,{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (1225 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}+210 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+9\right )}{15 \left (-3 x +3\right )}+4 \left (\frac {49 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}}}{9}+\frac {28 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{5}+\frac {11}{25}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )-420 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{225}\right )-4900 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{2}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}}}{-45 x +45}-\frac {{\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )}{225}\right )+4200 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}} \left (\frac {{\mathrm e}^{\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}+\frac {x}{5}} \left (35 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}+3\right )}{-675 x +675}-\left (\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{675}+\frac {2}{375}\right ) {\mathrm e}^{\frac {1}{5}+\frac {7 \,{\mathrm e}^{\frac {{\mathrm e}^{6}}{4}}}{3}} \expIntegralEi \left (1, -\frac {x}{5}+\frac {1}{5}\right )\right )\)	\(476\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x,method=_RETURNVERBOSE)

[Out]

-4*x*exp(7/3*exp(1/4*exp(6))+1/5*x)/(x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {4 \, x e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x - 1} - \frac {4 \, e^{\left (\frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )} + \frac {1}{5}\right )} E_{2}\left (-\frac {1}{5} \, x + \frac {1}{5}\right )}{x - 1} - 4 \, \int \frac {e^{\left (\frac {1}{5} \, x + \frac {7}{3} \, e^{\left (\frac {1}{4} \, e^{6}\right )}\right )}}{x^{2} - 2 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+4*x+20)*exp(7/3*exp(1/4*exp(3)^2)+1/5*x)/(5*x^2-10*x+5),x, algorithm="maxima")

[Out]

-4*x*e^(1/5*x + 7/3*e^(1/4*e^6))/(x - 1) - 4*e^(7/3*e^(1/4*e^6) + 1/5)*exp_integral_e(2, -1/5*x + 1/5)/(x - 1)
- 4*integrate(e^(1/5*x + 7/3*e^(1/4*e^6))/(x^2 - 2*x + 1), x)

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mupad [B]  time = 2.51, size = 22, normalized size = 0.71 \begin {gather*} -\frac {20\,x\,{\mathrm {e}}^{x/5}\,{\mathrm {e}}^{\frac {7\,{\mathrm {e}}^{\frac {{\mathrm {e}}^6}{4}}}{3}}}{5\,x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/5 + (7*exp(exp(6)/4))/3)*(4*x - 4*x^2 + 20))/(5*x^2 - 10*x + 5),x)

[Out]

-(20*x*exp(x/5)*exp((7*exp(exp(6)/4))/3))/(5*x - 5)

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sympy [A]  time = 0.11, size = 22, normalized size = 0.71 \begin {gather*} - \frac {4 x e^{\frac {x}{5} + \frac {7 e^{\frac {e^{6}}{4}}}{3}}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**2+4*x+20)*exp(7/3*exp(1/4*exp(3)**2)+1/5*x)/(5*x**2-10*x+5),x)

[Out]

-4*x*exp(x/5 + 7*exp(exp(6)/4)/3)/(x - 1)

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