3.90.85 \(\int \frac {-150+e^x (-28+8 x+4 x^2)+40 \log (4)}{15-30 x+15 x^2} \, dx\) [8985]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.38, 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, 6873, 6874, 2230, 2225, 2208, 2209} \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 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]

Rule 6873

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

Rule 6874

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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 23, normalized size = 0.68 \begin {gather*} \frac {150+4 e^x (3+x)-40 \log (4)}{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]

(150 + 4*E^x*(3 + x) - 40*Log[4])/(15*(-1 + x))

________________________________________________________________________________________

Maple [A]
time = 2.95, size = 31, normalized size = 0.91

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

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]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 20, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x + 3\right )} e^{x} - 40 \, \log \left (2\right ) + 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)

________________________________________________________________________________________

Sympy [A]
time = 0.06, size = 24, normalized size = 0.71 \begin {gather*} \frac {\left (4 x + 12\right ) e^{x}}{15 x - 15} - \frac {-30 + 16 \log {\left (2 \right )}}{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)

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 22, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left (2 \, x e^{x} + 6 \, e^{x} - 40 \, \log \left (2\right ) + 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)

________________________________________________________________________________________

Mupad [B]
time = 0.15, size = 27, normalized size = 0.79 \begin {gather*} \frac {12\,{\mathrm {e}}^x-x\,\left (80\,\ln \left (2\right )-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)

________________________________________________________________________________________