3.37.86 \(\int \frac {-6 x^4-2 e^{10} x^4+e^{12+4 x} (-4+4 x)+e^{6+2 x} (-14 x^2+12 x^3+e^{10} (-4 x^2+4 x^3))+(2 x^4+e^{6+2 x} (4 x^2-4 x^3)) \log (x)}{e^8 x^5} \, dx\)

Optimal. Leaf size=26 \[ \frac {\left (3+e^{10}+\frac {e^{6+2 x}}{x^2}-\log (x)\right )^2}{e^8} \]

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Rubi [A]  time = 0.19, antiderivative size = 52, normalized size of antiderivative = 2.00, number of steps used = 7, number of rules used = 6, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6, 12, 14, 2197, 2301, 2288} \begin {gather*} \frac {e^{4 x+4}}{x^4}+\frac {2 e^{2 x-2} \left (\left (3+e^{10}\right ) x-x \log (x)\right )}{x^3}+\frac {\left (-\log (x)+e^{10}+3\right )^2}{e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x^4 - 2*E^10*x^4 + E^(12 + 4*x)*(-4 + 4*x) + E^(6 + 2*x)*(-14*x^2 + 12*x^3 + E^10*(-4*x^2 + 4*x^3)) +
(2*x^4 + E^(6 + 2*x)*(4*x^2 - 4*x^3))*Log[x])/(E^8*x^5),x]

[Out]

E^(4 + 4*x)/x^4 + (3 + E^10 - Log[x])^2/E^8 + (2*E^(-2 + 2*x)*((3 + E^10)*x - x*Log[x]))/x^3

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-6-2 e^{10}\right ) x^4+e^{12+4 x} (-4+4 x)+e^{6+2 x} \left (-14 x^2+12 x^3+e^{10} \left (-4 x^2+4 x^3\right )\right )+\left (2 x^4+e^{6+2 x} \left (4 x^2-4 x^3\right )\right ) \log (x)}{e^8 x^5} \, dx\\ &=\frac {\int \frac {\left (-6-2 e^{10}\right ) x^4+e^{12+4 x} (-4+4 x)+e^{6+2 x} \left (-14 x^2+12 x^3+e^{10} \left (-4 x^2+4 x^3\right )\right )+\left (2 x^4+e^{6+2 x} \left (4 x^2-4 x^3\right )\right ) \log (x)}{x^5} \, dx}{e^8}\\ &=\frac {\int \left (\frac {4 e^{12+4 x} (-1+x)}{x^5}+\frac {2 \left (-3 \left (1+\frac {e^{10}}{3}\right )+\log (x)\right )}{x}+\frac {2 e^{6+2 x} \left (-7 \left (1+\frac {2 e^{10}}{7}\right )+6 \left (1+\frac {e^{10}}{3}\right ) x+2 \log (x)-2 x \log (x)\right )}{x^3}\right ) \, dx}{e^8}\\ &=\frac {2 \int \frac {-3 \left (1+\frac {e^{10}}{3}\right )+\log (x)}{x} \, dx}{e^8}+\frac {2 \int \frac {e^{6+2 x} \left (-7 \left (1+\frac {2 e^{10}}{7}\right )+6 \left (1+\frac {e^{10}}{3}\right ) x+2 \log (x)-2 x \log (x)\right )}{x^3} \, dx}{e^8}+\frac {4 \int \frac {e^{12+4 x} (-1+x)}{x^5} \, dx}{e^8}\\ &=\frac {e^{4+4 x}}{x^4}+\frac {\left (3+e^{10}-\log (x)\right )^2}{e^8}+\frac {2 e^{-2+2 x} \left (\left (3+e^{10}\right ) x-x \log (x)\right )}{x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 36, normalized size = 1.38 \begin {gather*} \frac {\left (e^{6+2 x}+3 x^2+e^{10} x^2-x^2 \log (x)\right )^2}{e^8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x^4 - 2*E^10*x^4 + E^(12 + 4*x)*(-4 + 4*x) + E^(6 + 2*x)*(-14*x^2 + 12*x^3 + E^10*(-4*x^2 + 4*x^
3)) + (2*x^4 + E^(6 + 2*x)*(4*x^2 - 4*x^3))*Log[x])/(E^8*x^5),x]

[Out]

(E^(6 + 2*x) + 3*x^2 + E^10*x^2 - x^2*Log[x])^2/(E^8*x^4)

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fricas [B]  time = 1.01, size = 67, normalized size = 2.58 \begin {gather*} \frac {{\left (x^{4} \log \relax (x)^{2} + 2 \, {\left (x^{2} e^{10} + 3 \, x^{2}\right )} e^{\left (2 \, x + 6\right )} - 2 \, {\left (x^{4} e^{10} + 3 \, x^{4} + x^{2} e^{\left (2 \, x + 6\right )}\right )} \log \relax (x) + e^{\left (4 \, x + 12\right )}\right )} e^{\left (-8\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+4*x^2)*exp(3+x)^2+2*x^4)*log(x)+(4*x-4)*exp(3+x)^4+((4*x^3-4*x^2)*exp(5)^2+12*x^3-14*x^2)*
exp(3+x)^2-2*x^4*exp(5)^2-6*x^4)/x^5/exp(4)^2,x, algorithm="fricas")

[Out]

(x^4*log(x)^2 + 2*(x^2*e^10 + 3*x^2)*e^(2*x + 6) - 2*(x^4*e^10 + 3*x^4 + x^2*e^(2*x + 6))*log(x) + e^(4*x + 12
))*e^(-8)/x^4

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giac [B]  time = 0.15, size = 76, normalized size = 2.92 \begin {gather*} -\frac {{\left (2 \, x^{4} e^{10} \log \relax (x) - x^{4} \log \relax (x)^{2} + 6 \, x^{4} \log \relax (x) + 2 \, x^{2} e^{\left (2 \, x + 6\right )} \log \relax (x) - 2 \, x^{2} e^{\left (2 \, x + 16\right )} - 6 \, x^{2} e^{\left (2 \, x + 6\right )} - e^{\left (4 \, x + 12\right )}\right )} e^{\left (-8\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+4*x^2)*exp(3+x)^2+2*x^4)*log(x)+(4*x-4)*exp(3+x)^4+((4*x^3-4*x^2)*exp(5)^2+12*x^3-14*x^2)*
exp(3+x)^2-2*x^4*exp(5)^2-6*x^4)/x^5/exp(4)^2,x, algorithm="giac")

[Out]

-(2*x^4*e^10*log(x) - x^4*log(x)^2 + 6*x^4*log(x) + 2*x^2*e^(2*x + 6)*log(x) - 2*x^2*e^(2*x + 16) - 6*x^2*e^(2
*x + 6) - e^(4*x + 12))*e^(-8)/x^4

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maple [B]  time = 0.12, size = 59, normalized size = 2.27

method	result	size
default	\({\mathrm e}^{-8} \left (\frac {\left (2 \,{\mathrm e}^{10}+6\right ) {\mathrm e}^{2 x +6}-2 \ln \relax (x ) {\mathrm e}^{2 x +6}}{x^{2}}-2 \ln \relax (x ) {\mathrm e}^{10}-6 \ln \relax (x )+\ln \relax (x )^{2}+\frac {{\mathrm e}^{4 x +12}}{x^{4}}\right )\)	\(59\)
risch	\({\mathrm e}^{-8} \ln \relax (x )^{2}-\frac {2 \ln \relax (x ) {\mathrm e}^{2 x -2}}{x^{2}}-\frac {{\mathrm e}^{-8} \left (2 \,{\mathrm e}^{10} \ln \relax (x ) x^{4}-2 x^{2} {\mathrm e}^{2 x +16}+6 x^{4} \ln \relax (x )-6 \,{\mathrm e}^{2 x +6} x^{2}-{\mathrm e}^{4 x +12}\right )}{x^{4}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3+4*x^2)*exp(3+x)^2+2*x^4)*ln(x)+(4*x-4)*exp(3+x)^4+((4*x^3-4*x^2)*exp(5)^2+12*x^3-14*x^2)*exp(3+x
)^2-2*x^4*exp(5)^2-6*x^4)/x^5/exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

1/exp(4)^2*(((2*exp(10)+6)*exp(2*x+6)-2*ln(x)*exp(2*x+6))/x^2-2*ln(x)*exp(10)-6*ln(x)+ln(x)^2+exp(4*x+12)/x^4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\left (8 \, e^{16} \Gamma \left (-1, -2 \, x\right ) + 24 \, e^{6} \Gamma \left (-1, -2 \, x\right ) + 16 \, e^{16} \Gamma \left (-2, -2 \, x\right ) + 56 \, e^{6} \Gamma \left (-2, -2 \, x\right ) + 256 \, e^{12} \Gamma \left (-3, -4 \, x\right ) + 1024 \, e^{12} \Gamma \left (-4, -4 \, x\right ) - 2 \, e^{10} \log \relax (x) + \log \relax (x)^{2} - \frac {2 \, e^{\left (2 \, x + 6\right )} \log \relax (x)}{x^{2}} + 2 \, \int \frac {e^{\left (2 \, x + 6\right )}}{x^{3}}\,{d x} - 6 \, \log \relax (x)\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+4*x^2)*exp(3+x)^2+2*x^4)*log(x)+(4*x-4)*exp(3+x)^4+((4*x^3-4*x^2)*exp(5)^2+12*x^3-14*x^2)*
exp(3+x)^2-2*x^4*exp(5)^2-6*x^4)/x^5/exp(4)^2,x, algorithm="maxima")

[Out]

(8*e^16*gamma(-1, -2*x) + 24*e^6*gamma(-1, -2*x) + 16*e^16*gamma(-2, -2*x) + 56*e^6*gamma(-2, -2*x) + 256*e^12
*gamma(-3, -4*x) + 1024*e^12*gamma(-4, -4*x) - 2*e^10*log(x) + log(x)^2 - 2*e^(2*x + 6)*log(x)/x^2 + 2*integra
te(e^(2*x + 6)/x^3, x) - 6*log(x))*e^(-8)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-8}\,\left ({\mathrm {e}}^{2\,x+6}\,\left ({\mathrm {e}}^{10}\,\left (4\,x^2-4\,x^3\right )+14\,x^2-12\,x^3\right )-{\mathrm {e}}^{4\,x+12}\,\left (4\,x-4\right )+2\,x^4\,{\mathrm {e}}^{10}-\ln \relax (x)\,\left ({\mathrm {e}}^{2\,x+6}\,\left (4\,x^2-4\,x^3\right )+2\,x^4\right )+6\,x^4\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(exp(2*x + 6)*(exp(10)*(4*x^2 - 4*x^3) + 14*x^2 - 12*x^3) - exp(4*x + 12)*(4*x - 4) + 2*x^4*exp(
10) - log(x)*(exp(2*x + 6)*(4*x^2 - 4*x^3) + 2*x^4) + 6*x^4))/x^5,x)

[Out]

int(-(exp(-8)*(exp(2*x + 6)*(exp(10)*(4*x^2 - 4*x^3) + 14*x^2 - 12*x^3) - exp(4*x + 12)*(4*x - 4) + 2*x^4*exp(
10) - log(x)*(exp(2*x + 6)*(4*x^2 - 4*x^3) + 2*x^4) + 6*x^4))/x^5, x)

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sympy [B]  time = 0.43, size = 78, normalized size = 3.00 \begin {gather*} \frac {\log {\relax (x )}^{2}}{e^{8}} + \frac {\left (- 2 e^{10} - 6\right ) \log {\relax (x )}}{e^{8}} + \frac {x^{2} e^{8} e^{4 x + 12} + \left (- 2 x^{4} e^{8} \log {\relax (x )} + 6 x^{4} e^{8} + 2 x^{4} e^{18}\right ) e^{2 x + 6}}{x^{6} e^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**3+4*x**2)*exp(3+x)**2+2*x**4)*ln(x)+(4*x-4)*exp(3+x)**4+((4*x**3-4*x**2)*exp(5)**2+12*x**3-
14*x**2)*exp(3+x)**2-2*x**4*exp(5)**2-6*x**4)/x**5/exp(4)**2,x)

[Out]

exp(-8)*log(x)**2 + (-2*exp(10) - 6)*exp(-8)*log(x) + (x**2*exp(8)*exp(4*x + 12) + (-2*x**4*exp(8)*log(x) + 6*
x**4*exp(8) + 2*x**4*exp(18))*exp(2*x + 6))*exp(-16)/x**6

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