3.80.71 \(\int \frac {e^{4 x} (3-4 x)+4 e^{3 x} x+25 x^2+25 x^4+(-12 e^{2 x} x^2+e^{3 x} (-8 x+12 x^2)) \log (x)+(12 e^x x^3+e^{2 x} (6 x^2-12 x^3)) \log ^2(x)+(-4 x^4+4 e^x x^4) \log ^3(x)-x^4 \log ^4(x)}{25 x^4} \, dx\)

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

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Rubi [B]  time = 0.31, antiderivative size = 69, normalized size of antiderivative = 2.76, number of steps used = 18, number of rules used = 7, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 14, 2197, 2202, 2288, 2296, 2295} \begin {gather*} -\frac {e^{4 x}}{25 x^3}+\frac {4 e^{3 x} \log (x)}{25 x^2}+x-\frac {1}{x}-\frac {1}{25} x \log ^4(x)+\frac {4}{25} e^x \log ^3(x)-\frac {6 e^{2 x} \log ^2(x)}{25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4*x)*(3 - 4*x) + 4*E^(3*x)*x + 25*x^2 + 25*x^4 + (-12*E^(2*x)*x^2 + E^(3*x)*(-8*x + 12*x^2))*Log[x] +
(12*E^x*x^3 + E^(2*x)*(6*x^2 - 12*x^3))*Log[x]^2 + (-4*x^4 + 4*E^x*x^4)*Log[x]^3 - x^4*Log[x]^4)/(25*x^4),x]

[Out]

-1/25*E^(4*x)/x^3 - x^(-1) + x + (4*E^(3*x)*Log[x])/(25*x^2) - (6*E^(2*x)*Log[x]^2)/(25*x) + (4*E^x*Log[x]^3)/
25 - (x*Log[x]^4)/25

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2202

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(x_)^(m_.)*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) +
(g_.)*(x_))), x_Symbol] :> Simp[(e*x^(m + 1)*F^(c*(a + b*x))*Log[d*x]^(n + 1))/(n + 1), x] /; FreeQ[{F, a, b,
c, d, e, f, g, h, m, n}, x] && EqQ[e*(m + 1) - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n,
-1]

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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {e^{4 x} (3-4 x)+4 e^{3 x} x+25 x^2+25 x^4+\left (-12 e^{2 x} x^2+e^{3 x} \left (-8 x+12 x^2\right )\right ) \log (x)+\left (12 e^x x^3+e^{2 x} \left (6 x^2-12 x^3\right )\right ) \log ^2(x)+\left (-4 x^4+4 e^x x^4\right ) \log ^3(x)-x^4 \log ^4(x)}{x^4} \, dx\\ &=\frac {1}{25} \int \left (-\frac {e^{4 x} (-3+4 x)}{x^4}+\frac {4 e^x \log ^2(x) (3+x \log (x))}{x}-\frac {6 e^{2 x} \log (x) (2-\log (x)+2 x \log (x))}{x^2}+\frac {4 e^{3 x} (1-2 \log (x)+3 x \log (x))}{x^3}+\frac {25+25 x^2-4 x^2 \log ^3(x)-x^2 \log ^4(x)}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{25} \int \frac {e^{4 x} (-3+4 x)}{x^4} \, dx\right )+\frac {1}{25} \int \frac {25+25 x^2-4 x^2 \log ^3(x)-x^2 \log ^4(x)}{x^2} \, dx+\frac {4}{25} \int \frac {e^x \log ^2(x) (3+x \log (x))}{x} \, dx+\frac {4}{25} \int \frac {e^{3 x} (1-2 \log (x)+3 x \log (x))}{x^3} \, dx-\frac {6}{25} \int \frac {e^{2 x} \log (x) (2-\log (x)+2 x \log (x))}{x^2} \, dx\\ &=-\frac {e^{4 x}}{25 x^3}+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {4}{25} e^x \log ^3(x)+\frac {1}{25} \int \left (\frac {25 \left (1+x^2\right )}{x^2}-4 \log ^3(x)-\log ^4(x)\right ) \, dx\\ &=-\frac {e^{4 x}}{25 x^3}+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {4}{25} e^x \log ^3(x)-\frac {1}{25} \int \log ^4(x) \, dx-\frac {4}{25} \int \log ^3(x) \, dx+\int \frac {1+x^2}{x^2} \, dx\\ &=-\frac {e^{4 x}}{25 x^3}+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {4}{25} e^x \log ^3(x)-\frac {4}{25} x \log ^3(x)-\frac {1}{25} x \log ^4(x)+\frac {4}{25} \int \log ^3(x) \, dx+\frac {12}{25} \int \log ^2(x) \, dx+\int \left (1+\frac {1}{x^2}\right ) \, dx\\ &=-\frac {e^{4 x}}{25 x^3}-\frac {1}{x}+x+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {12}{25} x \log ^2(x)+\frac {4}{25} e^x \log ^3(x)-\frac {1}{25} x \log ^4(x)-\frac {12}{25} \int \log ^2(x) \, dx-\frac {24}{25} \int \log (x) \, dx\\ &=-\frac {e^{4 x}}{25 x^3}-\frac {1}{x}+\frac {49 x}{25}+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {24}{25} x \log (x)-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {4}{25} e^x \log ^3(x)-\frac {1}{25} x \log ^4(x)+\frac {24}{25} \int \log (x) \, dx\\ &=-\frac {e^{4 x}}{25 x^3}-\frac {1}{x}+x+\frac {4 e^{3 x} \log (x)}{25 x^2}-\frac {6 e^{2 x} \log ^2(x)}{25 x}+\frac {4}{25} e^x \log ^3(x)-\frac {1}{25} x \log ^4(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 65, normalized size = 2.60 \begin {gather*} \frac {1}{25} \left (-\frac {e^{4 x}}{x^3}-\frac {25}{x}+25 x+\frac {4 e^{3 x} \log (x)}{x^2}-\frac {6 e^{2 x} \log ^2(x)}{x}+4 e^x \log ^3(x)-x \log ^4(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x)*(3 - 4*x) + 4*E^(3*x)*x + 25*x^2 + 25*x^4 + (-12*E^(2*x)*x^2 + E^(3*x)*(-8*x + 12*x^2))*Log
[x] + (12*E^x*x^3 + E^(2*x)*(6*x^2 - 12*x^3))*Log[x]^2 + (-4*x^4 + 4*E^x*x^4)*Log[x]^3 - x^4*Log[x]^4)/(25*x^4
),x]

[Out]

(-(E^(4*x)/x^3) - 25/x + 25*x + (4*E^(3*x)*Log[x])/x^2 - (6*E^(2*x)*Log[x]^2)/x + 4*E^x*Log[x]^3 - x*Log[x]^4)
/25

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fricas [B]  time = 0.84, size = 61, normalized size = 2.44 \begin {gather*} -\frac {x^{4} \log \relax (x)^{4} - 4 \, x^{3} e^{x} \log \relax (x)^{3} + 6 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 25 \, x^{4} - 4 \, x e^{\left (3 \, x\right )} \log \relax (x) + 25 \, x^{2} + e^{\left (4 \, x\right )}}{25 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-x^4*log(x)^4+(4*exp(x)*x^4-4*x^4)*log(x)^3+((-12*x^3+6*x^2)*exp(x)^2+12*exp(x)*x^3)*log(x)^2+
((12*x^2-8*x)*exp(x)^3-12*exp(x)^2*x^2)*log(x)+(3-4*x)*exp(x)^4+4*x*exp(x)^3+25*x^4+25*x^2)/x^4,x, algorithm="
fricas")

[Out]

-1/25*(x^4*log(x)^4 - 4*x^3*e^x*log(x)^3 + 6*x^2*e^(2*x)*log(x)^2 - 25*x^4 - 4*x*e^(3*x)*log(x) + 25*x^2 + e^(
4*x))/x^3

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giac [B]  time = 0.22, size = 61, normalized size = 2.44 \begin {gather*} -\frac {x^{4} \log \relax (x)^{4} - 4 \, x^{3} e^{x} \log \relax (x)^{3} + 6 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 25 \, x^{4} - 4 \, x e^{\left (3 \, x\right )} \log \relax (x) + 25 \, x^{2} + e^{\left (4 \, x\right )}}{25 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-x^4*log(x)^4+(4*exp(x)*x^4-4*x^4)*log(x)^3+((-12*x^3+6*x^2)*exp(x)^2+12*exp(x)*x^3)*log(x)^2+
((12*x^2-8*x)*exp(x)^3-12*exp(x)^2*x^2)*log(x)+(3-4*x)*exp(x)^4+4*x*exp(x)^3+25*x^4+25*x^2)/x^4,x, algorithm="
giac")

[Out]

-1/25*(x^4*log(x)^4 - 4*x^3*e^x*log(x)^3 + 6*x^2*e^(2*x)*log(x)^2 - 25*x^4 - 4*x*e^(3*x)*log(x) + 25*x^2 + e^(
4*x))/x^3

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maple [B]  time = 0.10, size = 56, normalized size = 2.24

method	result	size
default	\(\frac {4 \,{\mathrm e}^{x} \ln \relax (x )^{3}}{25}+\frac {4 \ln \relax (x ) {\mathrm e}^{3 x}}{25 x^{2}}-\frac {6 \ln \relax (x )^{2} {\mathrm e}^{2 x}}{25 x}-\frac {x \ln \relax (x )^{4}}{25}+x -\frac {1}{x}-\frac {{\mathrm e}^{4 x}}{25 x^{3}}\)	\(56\)
risch	\(-\frac {x \ln \relax (x )^{4}}{25}+\frac {4 \,{\mathrm e}^{x} \ln \relax (x )^{3}}{25}-\frac {6 \ln \relax (x )^{2} {\mathrm e}^{2 x}}{25 x}+\frac {4 \ln \relax (x ) {\mathrm e}^{3 x}}{25 x^{2}}+\frac {25 x^{4}-{\mathrm e}^{4 x}-25 x^{2}}{25 x^{3}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-x^4*ln(x)^4+(4*exp(x)*x^4-4*x^4)*ln(x)^3+((-12*x^3+6*x^2)*exp(x)^2+12*exp(x)*x^3)*ln(x)^2+((12*x^2-
8*x)*exp(x)^3-12*exp(x)^2*x^2)*ln(x)+(3-4*x)*exp(x)^4+4*x*exp(x)^3+25*x^4+25*x^2)/x^4,x,method=_RETURNVERBOSE)

[Out]

4/25*exp(x)*ln(x)^3+4/25/x^2*ln(x)*exp(3*x)-6/25*ln(x)^2*exp(2*x)/x-1/25*x*ln(x)^4+x-1/x-1/25*exp(4*x)/x^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x - \frac {x^{3} \log \relax (x)^{4} - 4 \, x^{2} e^{x} \log \relax (x)^{3} + 6 \, x e^{\left (2 \, x\right )} \log \relax (x)^{2} - 4 \, e^{\left (3 \, x\right )} \log \relax (x)}{25 \, x^{2}} - \frac {1}{x} - \frac {36}{25} \, \Gamma \left (-2, -3 \, x\right ) + \frac {64}{25} \, \Gamma \left (-2, -4 \, x\right ) + \frac {192}{25} \, \Gamma \left (-3, -4 \, x\right ) - \frac {4}{25} \, \int \frac {e^{\left (3 \, x\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-x^4*log(x)^4+(4*exp(x)*x^4-4*x^4)*log(x)^3+((-12*x^3+6*x^2)*exp(x)^2+12*exp(x)*x^3)*log(x)^2+
((12*x^2-8*x)*exp(x)^3-12*exp(x)^2*x^2)*log(x)+(3-4*x)*exp(x)^4+4*x*exp(x)^3+25*x^4+25*x^2)/x^4,x, algorithm="
maxima")

[Out]

x - 1/25*(x^3*log(x)^4 - 4*x^2*e^x*log(x)^3 + 6*x*e^(2*x)*log(x)^2 - 4*e^(3*x)*log(x))/x^2 - 1/x - 36/25*gamma
(-2, -3*x) + 64/25*gamma(-2, -4*x) + 192/25*gamma(-3, -4*x) - 4/25*integrate(e^(3*x)/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int -\frac {\frac {4\,x\,{\mathrm {e}}^{3\,x}}{25}-\frac {\ln \relax (x)\,\left ({\mathrm {e}}^{3\,x}\,\left (8\,x-12\,x^2\right )+12\,x^2\,{\mathrm {e}}^{2\,x}\right )}{25}+\frac {{\ln \relax (x)}^3\,\left (4\,x^4\,{\mathrm {e}}^x-4\,x^4\right )}{25}-\frac {x^4\,{\ln \relax (x)}^4}{25}+\frac {{\ln \relax (x)}^2\,\left (12\,x^3\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (6\,x^2-12\,x^3\right )\right )}{25}-\frac {{\mathrm {e}}^{4\,x}\,\left (4\,x-3\right )}{25}+x^2+x^4}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x*exp(3*x))/25 - (log(x)*(exp(3*x)*(8*x - 12*x^2) + 12*x^2*exp(2*x)))/25 + (log(x)^3*(4*x^4*exp(x) - 4
*x^4))/25 - (x^4*log(x)^4)/25 + (log(x)^2*(12*x^3*exp(x) + exp(2*x)*(6*x^2 - 12*x^3)))/25 - (exp(4*x)*(4*x - 3
))/25 + x^2 + x^4)/x^4,x)

[Out]

-int(-((4*x*exp(3*x))/25 - (log(x)*(exp(3*x)*(8*x - 12*x^2) + 12*x^2*exp(2*x)))/25 + (log(x)^3*(4*x^4*exp(x) -
4*x^4))/25 - (x^4*log(x)^4)/25 + (log(x)^2*(12*x^3*exp(x) + exp(2*x)*(6*x^2 - 12*x^3)))/25 - (exp(4*x)*(4*x -
3))/25 + x^2 + x^4)/x^4, x)

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sympy [B]  time = 0.50, size = 70, normalized size = 2.80 \begin {gather*} - \frac {x \log {\relax (x )}^{4}}{25} + x - \frac {1}{x} + \frac {62500 x^{6} e^{x} \log {\relax (x )}^{3} - 93750 x^{5} e^{2 x} \log {\relax (x )}^{2} + 62500 x^{4} e^{3 x} \log {\relax (x )} - 15625 x^{3} e^{4 x}}{390625 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-x**4*ln(x)**4+(4*exp(x)*x**4-4*x**4)*ln(x)**3+((-12*x**3+6*x**2)*exp(x)**2+12*exp(x)*x**3)*ln
(x)**2+((12*x**2-8*x)*exp(x)**3-12*exp(x)**2*x**2)*ln(x)+(3-4*x)*exp(x)**4+4*x*exp(x)**3+25*x**4+25*x**2)/x**4
,x)

[Out]

-x*log(x)**4/25 + x - 1/x + (62500*x**6*exp(x)*log(x)**3 - 93750*x**5*exp(2*x)*log(x)**2 + 62500*x**4*exp(3*x)
*log(x) - 15625*x**3*exp(4*x))/(390625*x**6)

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