3.28.57 \(\int \frac {\log (5)+e^{7 x} (-2+14 x+21 x^2) \log (5)}{16 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {\left (-1+x+e^{7 x} (2+3 x)\right ) \log (5)}{16 x} \]

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Rubi [A]  time = 0.10, antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 14, 2199, 2194, 2177, 2178} \begin {gather*} \frac {3}{16} e^{7 x} \log (5)+\frac {e^{7 x} \log (5)}{8 x}-\frac {\log (5)}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[5] + E^(7*x)*(-2 + 14*x + 21*x^2)*Log[5])/(16*x^2),x]

[Out]

(3*E^(7*x)*Log[5])/16 - Log[5]/(16*x) + (E^(7*x)*Log[5])/(8*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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} &=\frac {1}{16} \int \frac {\log (5)+e^{7 x} \left (-2+14 x+21 x^2\right ) \log (5)}{x^2} \, dx\\ &=\frac {1}{16} \int \left (\frac {\log (5)}{x^2}+\frac {e^{7 x} \left (-2+14 x+21 x^2\right ) \log (5)}{x^2}\right ) \, dx\\ &=-\frac {\log (5)}{16 x}+\frac {1}{16} \log (5) \int \frac {e^{7 x} \left (-2+14 x+21 x^2\right )}{x^2} \, dx\\ &=-\frac {\log (5)}{16 x}+\frac {1}{16} \log (5) \int \left (21 e^{7 x}-\frac {2 e^{7 x}}{x^2}+\frac {14 e^{7 x}}{x}\right ) \, dx\\ &=-\frac {\log (5)}{16 x}-\frac {1}{8} \log (5) \int \frac {e^{7 x}}{x^2} \, dx+\frac {1}{8} (7 \log (5)) \int \frac {e^{7 x}}{x} \, dx+\frac {1}{16} (21 \log (5)) \int e^{7 x} \, dx\\ &=\frac {3}{16} e^{7 x} \log (5)-\frac {\log (5)}{16 x}+\frac {e^{7 x} \log (5)}{8 x}+\frac {7}{8} \text {Ei}(7 x) \log (5)-\frac {1}{8} (7 \log (5)) \int \frac {e^{7 x}}{x} \, dx\\ &=\frac {3}{16} e^{7 x} \log (5)-\frac {\log (5)}{16 x}+\frac {e^{7 x} \log (5)}{8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} \frac {\left (-1+e^{7 x} (2+3 x)\right ) \log (5)}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[5] + E^(7*x)*(-2 + 14*x + 21*x^2)*Log[5])/(16*x^2),x]

[Out]

((-1 + E^(7*x)*(2 + 3*x))*Log[5])/(16*x)

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fricas [A]  time = 0.84, size = 22, normalized size = 0.96 \begin {gather*} \frac {{\left (3 \, x + 2\right )} e^{\left (7 \, x\right )} \log \relax (5) - \log \relax (5)}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((21*x^2+14*x-2)*log(5)*exp(7*x)+log(5))/x^2,x, algorithm="fricas")

[Out]

1/16*((3*x + 2)*e^(7*x)*log(5) - log(5))/x

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giac [A]  time = 0.21, size = 27, normalized size = 1.17 \begin {gather*} \frac {3 \, x e^{\left (7 \, x\right )} \log \relax (5) + 2 \, e^{\left (7 \, x\right )} \log \relax (5) - \log \relax (5)}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((21*x^2+14*x-2)*log(5)*exp(7*x)+log(5))/x^2,x, algorithm="giac")

[Out]

1/16*(3*x*e^(7*x)*log(5) + 2*e^(7*x)*log(5) - log(5))/x

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maple [A]  time = 0.06, size = 25, normalized size = 1.09

method	result	size
risch	\(-\frac {\ln \relax (5)}{16 x}+\frac {\ln \relax (5) \left (3 x +2\right ) {\mathrm e}^{7 x}}{16 x}\)	\(25\)
norman	\(\frac {\frac {\ln \relax (5) {\mathrm e}^{7 x}}{8}+\frac {3 x \ln \relax (5) {\mathrm e}^{7 x}}{16}-\frac {\ln \relax (5)}{16}}{x}\)	\(27\)
derivativedivides	\(-\frac {\ln \relax (5)}{16 x}+\frac {3 \ln \relax (5) {\mathrm e}^{7 x}}{16}+\frac {\ln \relax (5) {\mathrm e}^{7 x}}{8 x}\)	\(28\)
default	\(-\frac {\ln \relax (5)}{16 x}+\frac {3 \ln \relax (5) {\mathrm e}^{7 x}}{16}+\frac {\ln \relax (5) {\mathrm e}^{7 x}}{8 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*((21*x^2+14*x-2)*ln(5)*exp(7*x)+ln(5))/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/16*ln(5)/x+1/16*ln(5)*(3*x+2)/x*exp(7*x)

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maxima [C]  time = 0.46, size = 33, normalized size = 1.43 \begin {gather*} \frac {7}{8} \, {\rm Ei}\left (7 \, x\right ) \log \relax (5) + \frac {3}{16} \, e^{\left (7 \, x\right )} \log \relax (5) - \frac {7}{8} \, \Gamma \left (-1, -7 \, x\right ) \log \relax (5) - \frac {\log \relax (5)}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((21*x^2+14*x-2)*log(5)*exp(7*x)+log(5))/x^2,x, algorithm="maxima")

[Out]

7/8*Ei(7*x)*log(5) + 3/16*e^(7*x)*log(5) - 7/8*gamma(-1, -7*x)*log(5) - 1/16*log(5)/x

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mupad [B]  time = 1.79, size = 27, normalized size = 1.17 \begin {gather*} \frac {{\mathrm {e}}^{7\,x}\,\ln \left (125\right )}{16}-\frac {\frac {\ln \relax (5)}{16}-\frac {{\mathrm {e}}^{7\,x}\,\ln \left (25\right )}{16}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)/16 + (exp(7*x)*log(5)*(14*x + 21*x^2 - 2))/16)/x^2,x)

[Out]

(exp(7*x)*log(125))/16 - (log(5)/16 - (exp(7*x)*log(25))/16)/x

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sympy [A]  time = 0.12, size = 26, normalized size = 1.13 \begin {gather*} \frac {\left (3 x \log {\relax (5 )} + 2 \log {\relax (5 )}\right ) e^{7 x}}{16 x} - \frac {\log {\relax (5 )}}{16 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((21*x**2+14*x-2)*ln(5)*exp(7*x)+ln(5))/x**2,x)

[Out]

(3*x*log(5) + 2*log(5))*exp(7*x)/(16*x) - log(5)/(16*x)

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