3.90.85 \(\int \frac {-150+e^x (-28+8 x+4 x^2)+40 \log (4)}{15-30 x+15 x^2} \, dx\)

Optimal. Leaf size=34 \[ \frac {1-(3+x) \left (4+\frac {2 e^x}{5}-\log (4)\right )}{-2 x+\frac {3+x}{2}} \]

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Rubi [A]  time = 0.37, antiderivative size = 39, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {27, 12, 6741, 6742, 2199, 2194, 2177, 2178} \begin {gather*} \frac {4 e^x}{15}-\frac {16 e^x}{15 (1-x)}-\frac {2 (15-\log (256))}{3 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-150 + E^x*(-28 + 8*x + 4*x^2) + 40*Log[4])/(15 - 30*x + 15*x^2),x]

[Out]

(4*E^x)/15 - (16*E^x)/(15*(1 - x)) - (2*(15 - Log[256]))/(3*(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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-150+e^x \left (-28+8 x+4 x^2\right )+40 \log (4)}{15 (-1+x)^2} \, dx\\ &=\frac {1}{15} \int \frac {-150+e^x \left (-28+8 x+4 x^2\right )+40 \log (4)}{(-1+x)^2} \, dx\\ &=\frac {1}{15} \int \frac {e^x \left (-28+8 x+4 x^2\right )-150 \left (1-\frac {8 \log (2)}{15}\right )}{(1-x)^2} \, dx\\ &=\frac {1}{15} \int \frac {2 \left (-14 e^x+4 e^x x+2 e^x x^2-75 \left (1-\frac {8 \log (2)}{15}\right )\right )}{(1-x)^2} \, dx\\ &=\frac {2}{15} \int \frac {-14 e^x+4 e^x x+2 e^x x^2-75 \left (1-\frac {8 \log (2)}{15}\right )}{(1-x)^2} \, dx\\ &=\frac {2}{15} \int \left (\frac {2 e^x \left (-7+2 x+x^2\right )}{(-1+x)^2}+\frac {5 (-15+\log (256))}{(-1+x)^2}\right ) \, dx\\ &=-\frac {2 (15-\log (256))}{3 (1-x)}+\frac {4}{15} \int \frac {e^x \left (-7+2 x+x^2\right )}{(-1+x)^2} \, dx\\ &=-\frac {2 (15-\log (256))}{3 (1-x)}+\frac {4}{15} \int \left (e^x-\frac {4 e^x}{(-1+x)^2}+\frac {4 e^x}{-1+x}\right ) \, dx\\ &=-\frac {2 (15-\log (256))}{3 (1-x)}+\frac {4 \int e^x \, dx}{15}-\frac {16}{15} \int \frac {e^x}{(-1+x)^2} \, dx+\frac {16}{15} \int \frac {e^x}{-1+x} \, dx\\ &=\frac {4 e^x}{15}-\frac {16 e^x}{15 (1-x)}+\frac {16}{15} e \text {Ei}(-1+x)-\frac {2 (15-\log (256))}{3 (1-x)}-\frac {16}{15} \int \frac {e^x}{-1+x} \, dx\\ &=\frac {4 e^x}{15}-\frac {16 e^x}{15 (1-x)}-\frac {2 (15-\log (256))}{3 (1-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 0.68 \begin {gather*} \frac {2 \left (75+2 e^x (3+x)-5 \log (256)\right )}{15 (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-150 + E^x*(-28 + 8*x + 4*x^2) + 40*Log[4])/(15 - 30*x + 15*x^2),x]

[Out]

(2*(75 + 2*E^x*(3 + x) - 5*Log[256]))/(15*(-1 + x))

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fricas [A]  time = 0.47, size = 20, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x + 3\right )} e^{x} - 40 \, \log \relax (2) + 75\right )}}{15 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x-28)*exp(x)+80*log(2)-150)/(15*x^2-30*x+15),x, algorithm="fricas")

[Out]

2/15*(2*(x + 3)*e^x - 40*log(2) + 75)/(x - 1)

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giac [A]  time = 0.19, size = 22, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left (2 \, x e^{x} + 6 \, e^{x} - 40 \, \log \relax (2) + 75\right )}}{15 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x-28)*exp(x)+80*log(2)-150)/(15*x^2-30*x+15),x, algorithm="giac")

[Out]

2/15*(2*x*e^x + 6*e^x - 40*log(2) + 75)/(x - 1)

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maple [A]  time = 0.53, size = 22, normalized size = 0.65

method	result	size
norman	\(\frac {\frac {4 \,{\mathrm e}^{x} x}{15}+\frac {4 \,{\mathrm e}^{x}}{5}+10-\frac {16 \ln \relax (2)}{3}}{x -1}\)	\(22\)
risch	\(\frac {10}{x -1}-\frac {16 \ln \relax (2)}{3 \left (x -1\right )}+\frac {4 \left (3+x \right ) {\mathrm e}^{x}}{15 \left (x -1\right )}\)	\(30\)
default	\(\frac {10}{x -1}+\frac {16 \,{\mathrm e}^{x}}{15 \left (x -1\right )}-\frac {16 \ln \relax (2)}{3 \left (x -1\right )}+\frac {4 \,{\mathrm e}^{x}}{15}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+8*x-28)*exp(x)+80*ln(2)-150)/(15*x^2-30*x+15),x,method=_RETURNVERBOSE)

[Out]

(4/15*exp(x)*x+4/5*exp(x)+10-16/3*ln(2))/(x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, {\left (x^{2} + 2 \, x\right )} e^{x}}{15 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {28 \, e E_{2}\left (-x + 1\right )}{15 \, {\left (x - 1\right )}} - \frac {16 \, \log \relax (2)}{3 \, {\left (x - 1\right )}} + \frac {10}{x - 1} + \frac {2}{15} \, \int \frac {4 \, {\left (2 \, x + 1\right )} e^{x}}{x^{3} - 3 \, x^{2} + 3 \, x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+8*x-28)*exp(x)+80*log(2)-150)/(15*x^2-30*x+15),x, algorithm="maxima")

[Out]

4/15*(x^2 + 2*x)*e^x/(x^2 - 2*x + 1) + 28/15*e*exp_integral_e(2, -x + 1)/(x - 1) - 16/3*log(2)/(x - 1) + 10/(x
- 1) + 2/15*integrate(4*(2*x + 1)*e^x/(x^3 - 3*x^2 + 3*x - 1), x)

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mupad [B]  time = 0.15, size = 27, normalized size = 0.79 \begin {gather*} \frac {12\,{\mathrm {e}}^x-x\,\left (80\,\ln \relax (2)-150\right )+4\,x\,{\mathrm {e}}^x}{15\,x-15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((80*log(2) + exp(x)*(8*x + 4*x^2 - 28) - 150)/(15*x^2 - 30*x + 15),x)

[Out]

(12*exp(x) - x*(80*log(2) - 150) + 4*x*exp(x))/(15*x - 15)

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sympy [A]  time = 0.15, size = 24, normalized size = 0.71 \begin {gather*} \frac {\left (4 x + 12\right ) e^{x}}{15 x - 15} - \frac {-30 + 16 \log {\relax (2 )}}{3 x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+8*x-28)*exp(x)+80*ln(2)-150)/(15*x**2-30*x+15),x)

[Out]

(4*x + 12)*exp(x)/(15*x - 15) - (-30 + 16*log(2))/(3*x - 3)

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