3.57.28 \(\int \frac {x+e^5 (-9-23 x-7 x^2)+e^5 (-6 x-2 x^2) \log (x)}{e^5 x} \, dx\)

Optimal. Leaf size=25 \[ -16+x+\frac {x}{e^5}-\log ^2(2)-(3+x)^2 (3+\log (x)) \]

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Rubi [A]  time = 0.05, antiderivative size = 42, normalized size of antiderivative = 1.68, number of steps used = 7, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 14, 2313, 9} \begin {gather*} -\frac {7 x^2}{2}-\left (x^2+6 x\right ) \log (x)-\left (23-\frac {1}{e^5}\right ) x+\frac {1}{2} (x+6)^2-9 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + E^5*(-9 - 23*x - 7*x^2) + E^5*(-6*x - 2*x^2)*Log[x])/(E^5*x),x]

[Out]

-((23 - E^(-5))*x) - (7*x^2)/2 + (6 + x)^2/2 - 9*Log[x] - (6*x + x^2)*Log[x]

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, 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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {x+e^5 \left (-9-23 x-7 x^2\right )+e^5 \left (-6 x-2 x^2\right ) \log (x)}{x} \, dx}{e^5}\\ &=\frac {\int \left (\frac {-9 e^5+\left (1-23 e^5\right ) x-7 e^5 x^2}{x}-2 e^5 (3+x) \log (x)\right ) \, dx}{e^5}\\ &=-(2 \int (3+x) \log (x) \, dx)+\frac {\int \frac {-9 e^5+\left (1-23 e^5\right ) x-7 e^5 x^2}{x} \, dx}{e^5}\\ &=-\left (\left (6 x+x^2\right ) \log (x)\right )+2 \int \frac {6+x}{2} \, dx+\frac {\int \left (1-23 e^5-\frac {9 e^5}{x}-7 e^5 x\right ) \, dx}{e^5}\\ &=-\left (\left (23-\frac {1}{e^5}\right ) x\right )-\frac {7 x^2}{2}+\frac {1}{2} (6+x)^2-9 \log (x)-\left (6 x+x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.20 \begin {gather*} -17 x+\frac {x}{e^5}-3 x^2-9 \log (x)-6 x \log (x)-x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + E^5*(-9 - 23*x - 7*x^2) + E^5*(-6*x - 2*x^2)*Log[x])/(E^5*x),x]

[Out]

-17*x + x/E^5 - 3*x^2 - 9*Log[x] - 6*x*Log[x] - x^2*Log[x]

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fricas [A]  time = 0.56, size = 33, normalized size = 1.32 \begin {gather*} -{\left ({\left (x^{2} + 6 \, x + 9\right )} e^{5} \log \relax (x) + {\left (3 \, x^{2} + 17 \, x\right )} e^{5} - x\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-6*x)*exp(5)*log(x)+(-7*x^2-23*x-9)*exp(5)+x)/x/exp(5),x, algorithm="fricas")

[Out]

-((x^2 + 6*x + 9)*e^5*log(x) + (3*x^2 + 17*x)*e^5 - x)*e^(-5)

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giac [A]  time = 0.22, size = 41, normalized size = 1.64 \begin {gather*} -{\left (x^{2} e^{5} \log \relax (x) + 3 \, x^{2} e^{5} + 6 \, x e^{5} \log \relax (x) + 17 \, x e^{5} + 9 \, e^{5} \log \relax (x) - x\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-6*x)*exp(5)*log(x)+(-7*x^2-23*x-9)*exp(5)+x)/x/exp(5),x, algorithm="giac")

[Out]

-(x^2*e^5*log(x) + 3*x^2*e^5 + 6*x*e^5*log(x) + 17*x*e^5 + 9*e^5*log(x) - x)*e^(-5)

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maple [A]  time = 0.07, size = 26, normalized size = 1.04

method	result	size
risch	\(-\left (x +6\right ) x \ln \relax (x )-3 x^{2}-17 x +{\mathrm e}^{-5} x -9 \ln \relax (x )\)	\(26\)
norman	\(-9 \ln \relax (x )-3 x^{2}-6 x \ln \relax (x )-x^{2} \ln \relax (x )-{\mathrm e}^{-5} \left (-1+17 \,{\mathrm e}^{5}\right ) x\)	\(36\)
default	\({\mathrm e}^{-5} \left (-2 \,{\mathrm e}^{5} \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )-6 \,{\mathrm e}^{5} \left (x \ln \relax (x )-x \right )-\frac {7 x^{2} {\mathrm e}^{5}}{2}-23 x \,{\mathrm e}^{5}-9 \,{\mathrm e}^{5} \ln \relax (x )+x \right )\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-6*x)*exp(5)*ln(x)+(-7*x^2-23*x-9)*exp(5)+x)/x/exp(5),x,method=_RETURNVERBOSE)

[Out]

-(x+6)*x*ln(x)-3*x^2-17*x+exp(-5)*x-9*ln(x)

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maxima [B]  time = 0.43, size = 54, normalized size = 2.16 \begin {gather*} -\frac {1}{2} \, {\left (7 \, x^{2} e^{5} + {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{5} + 12 \, {\left (x \log \relax (x) - x\right )} e^{5} + 46 \, x e^{5} + 18 \, e^{5} \log \relax (x) - 2 \, x\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-6*x)*exp(5)*log(x)+(-7*x^2-23*x-9)*exp(5)+x)/x/exp(5),x, algorithm="maxima")

[Out]

-1/2*(7*x^2*e^5 + (2*x^2*log(x) - x^2)*e^5 + 12*(x*log(x) - x)*e^5 + 46*x*e^5 + 18*e^5*log(x) - 2*x)*e^(-5)

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mupad [B]  time = 3.59, size = 29, normalized size = 1.16 \begin {gather*} x\,{\mathrm {e}}^{-5}-9\,\ln \relax (x)-x^2\,\ln \relax (x)-17\,x-6\,x\,\ln \relax (x)-3\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5)*(exp(5)*(23*x + 7*x^2 + 9) - x + exp(5)*log(x)*(6*x + 2*x^2)))/x,x)

[Out]

x*exp(-5) - 9*log(x) - x^2*log(x) - 17*x - 6*x*log(x) - 3*x^2

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sympy [A]  time = 0.15, size = 41, normalized size = 1.64 \begin {gather*} \left (- x^{2} - 6 x\right ) \log {\relax (x )} + \frac {- 3 x^{2} e^{5} - x \left (-1 + 17 e^{5}\right ) - 9 e^{5} \log {\relax (x )}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-6*x)*exp(5)*ln(x)+(-7*x**2-23*x-9)*exp(5)+x)/x/exp(5),x)

[Out]

(-x**2 - 6*x)*log(x) + (-3*x**2*exp(5) - x*(-1 + 17*exp(5)) - 9*exp(5)*log(x))*exp(-5)

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