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

Optimal. Leaf size=18 \[ 2+\frac {e^{71 x} x^2 \log (256)}{-5+x} \]

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Rubi [A]  time = 0.14, antiderivative size = 35, normalized size of antiderivative = 1.94, number of steps used = 11, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 27, 1594, 2199, 2194, 2177, 2178, 2176} \begin {gather*} e^{71 x} x \log (256)+5 e^{71 x} \log (256)-\frac {25 e^{71 x} \log (256)}{5-x} \end {gather*}

Antiderivative was successfully verified.

[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 1594

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 2176

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] &&  !$UseGamma === True

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} &=\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} \text {Ei}(-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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 18, normalized size = 1.00 \begin {gather*} e^{71 x} \left (5+\frac {25}{-5+x}+x\right ) \log (256) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.73, size = 16, normalized size = 0.89 \begin {gather*} \frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \relax (2)}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 16, normalized size = 0.89 \begin {gather*} \frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \relax (2)}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.25, size = 17, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-5)*exp(71*x)

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maxima [A]  time = 0.39, size = 16, normalized size = 0.89 \begin {gather*} \frac {8 \, x^{2} e^{\left (71 \, x\right )} \log \relax (2)}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.13, size = 16, normalized size = 0.89 \begin {gather*} \frac {8\,x^2\,{\mathrm {e}}^{71\,x}\,\ln \relax (2)}{x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.83 \begin {gather*} \frac {8 x^{2} e^{71 x} \log {\relax (2 )}}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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