\(\int \frac {e^{71 x} (-10 x-354 x^2+71 x^3) \log (256)}{25-10 x+x^2} \, dx\) [7604]

Optimal result

Integrand size = 32, antiderivative size = 18 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=2+\frac {e^{71 x} x^2 \log (256)}{-5+x} \]

[Out]

8*x^2*ln(2)/(-5+x)*exp(71*x)+2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.94, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 27, 1608, 2230, 2225, 2208, 2209, 2207} \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=e^{71 x} x \log (256)+5 e^{71 x} \log (256)-\frac {25 e^{71 x} \log (256)}{5-x} \]

[In]

Int[(E^(71*x)*(-10*x - 354*x^2 + 71*x^3)*Log[256])/(25 - 10*x + x^2),x]

[Out]

5*E^(71*x)*Log[256] - (25*E^(71*x)*Log[256])/(5 - x) + E^(71*x)*x*Log[256]

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 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rubi steps \begin{align*} \text {integral}& = \log (256) \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right )}{25-10 x+x^2} \, dx \\ & = \log (256) \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right )}{(-5+x)^2} \, dx \\ & = \log (256) \int \frac {e^{71 x} x \left (-10-354 x+71 x^2\right )}{(-5+x)^2} \, dx \\ & = \log (256) \int \left (356 e^{71 x}-\frac {25 e^{71 x}}{(-5+x)^2}+\frac {1775 e^{71 x}}{-5+x}+71 e^{71 x} x\right ) \, dx \\ & = -\left ((25 \log (256)) \int \frac {e^{71 x}}{(-5+x)^2} \, dx\right )+(71 \log (256)) \int e^{71 x} x \, dx+(356 \log (256)) \int e^{71 x} \, dx+(1775 \log (256)) \int \frac {e^{71 x}}{-5+x} \, dx \\ & = \frac {356}{71} e^{71 x} \log (256)-\frac {25 e^{71 x} \log (256)}{5-x}+e^{71 x} x \log (256)+1775 e^{355} \operatorname {ExpIntegralEi}(-71 (5-x)) \log (256)-\log (256) \int e^{71 x} \, dx-(1775 \log (256)) \int \frac {e^{71 x}}{-5+x} \, dx \\ & = 5 e^{71 x} \log (256)-\frac {25 e^{71 x} \log (256)}{5-x}+e^{71 x} x \log (256) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=e^{71 x} \left (5+\frac {25}{-5+x}+x\right ) \log (256) \]

[In]

Integrate[(E^(71*x)*(-10*x - 354*x^2 + 71*x^3)*Log[256])/(25 - 10*x + x^2),x]

[Out]

E^(71*x)*(5 + 25/(-5 + x) + x)*Log[256]

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
gosper	\(\frac {8 x^{2} \ln \left (2\right ) {\mathrm e}^{71 x}}{-5+x}\)	\(17\)
norman	\(\frac {8 x^{2} \ln \left (2\right ) {\mathrm e}^{71 x}}{-5+x}\)	\(17\)
risch	\(\frac {8 x^{2} \ln \left (2\right ) {\mathrm e}^{71 x}}{-5+x}\)	\(17\)
parallelrisch	\(\frac {8 x^{2} \ln \left (2\right ) {\mathrm e}^{71 x}}{-5+x}\)	\(17\)
default	\(8 \ln \left (2\right ) \left (x \,{\mathrm e}^{71 x}+5 \,{\mathrm e}^{71 x}+\frac {1775 \,{\mathrm e}^{71 x}}{71 x -355}\right )\)	\(31\)
derivativedivides	\(\frac {8 \ln \left (2\right ) \left (71 x \,{\mathrm e}^{71 x}+355 \,{\mathrm e}^{71 x}+\frac {126025 \,{\mathrm e}^{71 x}}{71 x -355}\right )}{71}\)	\(32\)

[In]

int(8*(71*x^3-354*x^2-10*x)*ln(2)*exp(71*x)/(x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

8*x^2*ln(2)/(-5+x)*exp(71*x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=\frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \left (2\right )}{x - 5} \]

[In]

integrate(8*(71*x^3-354*x^2-10*x)*log(2)*exp(71*x)/(x^2-10*x+25),x, algorithm="fricas")

[Out]

8*x^2*e^(71*x)*log(2)/(x - 5)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=\frac {8 x^{2} e^{71 x} \log {\left (2 \right )}}{x - 5} \]

[In]

integrate(8*(71*x**3-354*x**2-10*x)*ln(2)*exp(71*x)/(x**2-10*x+25),x)

[Out]

8*x**2*exp(71*x)*log(2)/(x - 5)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=\frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \left (2\right )}{x - 5} \]

[In]

integrate(8*(71*x^3-354*x^2-10*x)*log(2)*exp(71*x)/(x^2-10*x+25),x, algorithm="maxima")

[Out]

8*x^2*e^(71*x)*log(2)/(x - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=\frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \left (2\right )}{x - 5} \]

[In]

integrate(8*(71*x^3-354*x^2-10*x)*log(2)*exp(71*x)/(x^2-10*x+25),x, algorithm="giac")

[Out]

8*x^2*e^(71*x)*log(2)/(x - 5)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e^{71 x} \left (-10 x-354 x^2+71 x^3\right ) \log (256)}{25-10 x+x^2} \, dx=\frac {8\,x^2\,{\mathrm {e}}^{71\,x}\,\ln \left (2\right )}{x-5} \]

[In]

int(-(8*exp(71*x)*log(2)*(10*x + 354*x^2 - 71*x^3))/(x^2 - 10*x + 25),x)

[Out]

(8*x^2*exp(71*x)*log(2))/(x - 5)