3.101.29 \(\int \frac {25+25 x+e^{1-x^6} (4+4 x-24 x^7)-24 e^{1-x^6} x^6 \log (x)}{4 x} \, dx\)

Optimal. Leaf size=20 \[ 1+\left (\frac {25}{4}+e^{1-x^6}\right ) (x+\log (x)) \]

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Rubi [A]  time = 0.10, antiderivative size = 35, normalized size of antiderivative = 1.75, number of steps used = 6, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 43, 2288} \begin {gather*} \frac {e^{1-x^6} \left (x^7+x^6 \log (x)\right )}{x^6}+\frac {25 x}{4}+\frac {25 \log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 25*x + E^(1 - x^6)*(4 + 4*x - 24*x^7) - 24*E^(1 - x^6)*x^6*Log[x])/(4*x),x]

[Out]

(25*x)/4 + (25*Log[x])/4 + (E^(1 - x^6)*(x^7 + x^6*Log[x]))/x^6

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {25+25 x+e^{1-x^6} \left (4+4 x-24 x^7\right )-24 e^{1-x^6} x^6 \log (x)}{x} \, dx\\ &=\frac {1}{4} \int \left (\frac {25 (1+x)}{x}-\frac {4 e^{1-x^6} \left (-1-x+6 x^7+6 x^6 \log (x)\right )}{x}\right ) \, dx\\ &=\frac {25}{4} \int \frac {1+x}{x} \, dx-\int \frac {e^{1-x^6} \left (-1-x+6 x^7+6 x^6 \log (x)\right )}{x} \, dx\\ &=\frac {e^{1-x^6} \left (x^7+x^6 \log (x)\right )}{x^6}+\frac {25}{4} \int \left (1+\frac {1}{x}\right ) \, dx\\ &=\frac {25 x}{4}+\frac {25 \log (x)}{4}+\frac {e^{1-x^6} \left (x^7+x^6 \log (x)\right )}{x^6}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 26, normalized size = 1.30 \begin {gather*} \frac {1}{4} e^{-x^6} \left (4 e+25 e^{x^6}\right ) (x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 25*x + E^(1 - x^6)*(4 + 4*x - 24*x^7) - 24*E^(1 - x^6)*x^6*Log[x])/(4*x),x]

[Out]

((4*E + 25*E^x^6)*(x + Log[x]))/(4*E^x^6)

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fricas [A]  time = 1.35, size = 30, normalized size = 1.50 \begin {gather*} x e^{\left (-x^{6} + 1\right )} + \frac {1}{4} \, {\left (4 \, e^{\left (-x^{6} + 1\right )} + 25\right )} \log \relax (x) + \frac {25}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*x^6*exp(-x^6+1)*log(x)+(-24*x^7+4*x+4)*exp(-x^6+1)+25*x+25)/x,x, algorithm="fricas")

[Out]

x*e^(-x^6 + 1) + 1/4*(4*e^(-x^6 + 1) + 25)*log(x) + 25/4*x

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giac [A]  time = 0.17, size = 29, normalized size = 1.45 \begin {gather*} x e^{\left (-x^{6} + 1\right )} + e^{\left (-x^{6} + 1\right )} \log \relax (x) + \frac {25}{4} \, x + \frac {25}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*x^6*exp(-x^6+1)*log(x)+(-24*x^7+4*x+4)*exp(-x^6+1)+25*x+25)/x,x, algorithm="giac")

[Out]

x*e^(-x^6 + 1) + e^(-x^6 + 1)*log(x) + 25/4*x + 25/4*log(x)

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maple [B]  time = 0.04, size = 60, normalized size = 3.00




method result size



risch \({\mathrm e}^{-\left (x -1\right ) \left (x +1\right ) \left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )} \ln \relax (x )+\frac {25 x}{4}+\frac {25 \ln \relax (x )}{4}+{\mathrm e}^{-\left (x -1\right ) \left (x +1\right ) \left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )} x\) \(60\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-24*x^6*exp(-x^6+1)*ln(x)+(-24*x^7+4*x+4)*exp(-x^6+1)+25*x+25)/x,x,method=_RETURNVERBOSE)

[Out]

exp(-(x-1)*(x+1)*(x^2+x+1)*(x^2-x+1))*ln(x)+25/4*x+25/4*ln(x)+exp(-(x-1)*(x+1)*(x^2+x+1)*(x^2-x+1))*x

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maxima [C]  time = 0.43, size = 50, normalized size = 2.50 \begin {gather*} \frac {x^{7} e \Gamma \left (\frac {7}{6}, x^{6}\right )}{{\left (x^{6}\right )}^{\frac {7}{6}}} - \frac {x e \Gamma \left (\frac {1}{6}, x^{6}\right )}{6 \, {\left (x^{6}\right )}^{\frac {1}{6}}} + e^{\left (-x^{6} + 1\right )} \log \relax (x) + \frac {25}{4} \, x + \frac {25}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*x^6*exp(-x^6+1)*log(x)+(-24*x^7+4*x+4)*exp(-x^6+1)+25*x+25)/x,x, algorithm="maxima")

[Out]

x^7*e*gamma(7/6, x^6)/(x^6)^(7/6) - 1/6*x*e*gamma(1/6, x^6)/(x^6)^(1/6) + e^(-x^6 + 1)*log(x) + 25/4*x + 25/4*
log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\frac {25\,x}{4}+\frac {{\mathrm {e}}^{1-x^6}\,\left (-24\,x^7+4\,x+4\right )}{4}-6\,x^6\,{\mathrm {e}}^{1-x^6}\,\ln \relax (x)+\frac {25}{4}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x)/4 + (exp(1 - x^6)*(4*x - 24*x^7 + 4))/4 - 6*x^6*exp(1 - x^6)*log(x) + 25/4)/x,x)

[Out]

int(((25*x)/4 + (exp(1 - x^6)*(4*x - 24*x^7 + 4))/4 - 6*x^6*exp(1 - x^6)*log(x) + 25/4)/x, x)

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sympy [A]  time = 0.39, size = 22, normalized size = 1.10 \begin {gather*} \frac {25 x}{4} + \left (x + \log {\relax (x )}\right ) e^{1 - x^{6}} + \frac {25 \log {\relax (x )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*x**6*exp(-x**6+1)*ln(x)+(-24*x**7+4*x+4)*exp(-x**6+1)+25*x+25)/x,x)

[Out]

25*x/4 + (x + log(x))*exp(1 - x**6) + 25*log(x)/4

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