3.2.65 \(\int \frac {7+10 x+2 e^x x+6 x^2+(10 x+12 x^2+e^x (2 x+2 x^2)) \log (x)}{12 x} \, dx\) [165]

Optimal. Leaf size=26 \[ \frac {1}{2} x \left (3+\frac {1}{3} \left (-4+e^x+\frac {7}{2 x}\right )+x\right ) \log (x) \]

[Out]

1/2*(x+5/3+7/6/x+1/3*exp(x))*x*ln(x)

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Rubi [A]
time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 10, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 2326, 2350} \begin {gather*} \frac {1}{2} x^2 \log (x)+\frac {1}{6} e^x x \log (x)+\frac {5}{6} x \log (x)+\frac {7 \log (x)}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(7 + 10*x + 2*E^x*x + 6*x^2 + (10*x + 12*x^2 + E^x*(2*x + 2*x^2))*Log[x])/(12*x),x]

[Out]

(7*Log[x])/12 + (5*x*Log[x])/6 + (E^x*x*Log[x])/6 + (x^2*Log[x])/2

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 2326

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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, 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 {1}{12} \int \frac {7+10 x+2 e^x x+6 x^2+\left (10 x+12 x^2+e^x \left (2 x+2 x^2\right )\right ) \log (x)}{x} \, dx\\ &=\frac {1}{12} \int \left (2 e^x (1+\log (x)+x \log (x))+\frac {7+10 x+6 x^2+10 x \log (x)+12 x^2 \log (x)}{x}\right ) \, dx\\ &=\frac {1}{12} \int \frac {7+10 x+6 x^2+10 x \log (x)+12 x^2 \log (x)}{x} \, dx+\frac {1}{6} \int e^x (1+\log (x)+x \log (x)) \, dx\\ &=\frac {1}{6} e^x x \log (x)+\frac {1}{12} \int \left (\frac {7+10 x+6 x^2}{x}+2 (5+6 x) \log (x)\right ) \, dx\\ &=\frac {1}{6} e^x x \log (x)+\frac {1}{12} \int \frac {7+10 x+6 x^2}{x} \, dx+\frac {1}{6} \int (5+6 x) \log (x) \, dx\\ &=\frac {1}{6} e^x x \log (x)+\frac {1}{6} \left (5 x+3 x^2\right ) \log (x)+\frac {1}{12} \int \left (10+\frac {7}{x}+6 x\right ) \, dx-\frac {1}{6} \int (5+3 x) \, dx\\ &=\frac {7 \log (x)}{12}+\frac {1}{6} e^x x \log (x)+\frac {1}{6} \left (5 x+3 x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 21, normalized size = 0.81 \begin {gather*} \frac {1}{12} \left (7+2 \left (5+e^x\right ) x+6 x^2\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(7 + 10*x + 2*E^x*x + 6*x^2 + (10*x + 12*x^2 + E^x*(2*x + 2*x^2))*Log[x])/(12*x),x]

[Out]

((7 + 2*(5 + E^x)*x + 6*x^2)*Log[x])/12

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Maple [A]
time = 0.04, size = 25, normalized size = 0.96

method	result	size
risch	\(\frac {\left (6 x^{2}+2 \,{\mathrm e}^{x} x +10 x \right ) \ln \left (x \right )}{12}+\frac {7 \ln \left (x \right )}{12}\)	\(24\)
default	\(\frac {x \,{\mathrm e}^{x} \ln \left (x \right )}{6}+\frac {x^{2} \ln \left (x \right )}{2}+\frac {5 x \ln \left (x \right )}{6}+\frac {7 \ln \left (x \right )}{12}\)	\(25\)
norman	\(\frac {x \,{\mathrm e}^{x} \ln \left (x \right )}{6}+\frac {x^{2} \ln \left (x \right )}{2}+\frac {5 x \ln \left (x \right )}{6}+\frac {7 \ln \left (x \right )}{12}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*(((2*x^2+2*x)*exp(x)+12*x^2+10*x)*ln(x)+2*exp(x)*x+6*x^2+10*x+7)/x,x,method=_RETURNVERBOSE)

[Out]

1/6*x*exp(x)*ln(x)+1/2*x^2*ln(x)+5/6*x*ln(x)+7/12*ln(x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(((2*x^2+2*x)*exp(x)+12*x^2+10*x)*log(x)+2*exp(x)*x+6*x^2+10*x+7)/x,x, algorithm="maxima")

[Out]

1/2*x^2*log(x) + 1/6*(x - 1)*e^x*log(x) + 5/6*x*log(x) + 1/6*e^x*log(x) - 1/6*Ei(x) + 1/6*e^x - 1/6*integrate(
(x - 1)*e^x/x, x) + 7/12*log(x)

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Fricas [A]
time = 0.33, size = 19, normalized size = 0.73 \begin {gather*} \frac {1}{12} \, {\left (6 \, x^{2} + 2 \, x e^{x} + 10 \, x + 7\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(((2*x^2+2*x)*exp(x)+12*x^2+10*x)*log(x)+2*exp(x)*x+6*x^2+10*x+7)/x,x, algorithm="fricas")

[Out]

1/12*(6*x^2 + 2*x*e^x + 10*x + 7)*log(x)

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Sympy [A]
time = 0.12, size = 29, normalized size = 1.12 \begin {gather*} \frac {x e^{x} \log {\left (x \right )}}{6} + \left (\frac {x^{2}}{2} + \frac {5 x}{6}\right ) \log {\left (x \right )} + \frac {7 \log {\left (x \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(((2*x**2+2*x)*exp(x)+12*x**2+10*x)*ln(x)+2*exp(x)*x+6*x**2+10*x+7)/x,x)

[Out]

x*exp(x)*log(x)/6 + (x**2/2 + 5*x/6)*log(x) + 7*log(x)/12

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Giac [A]
time = 0.40, size = 24, normalized size = 0.92 \begin {gather*} \frac {1}{2} \, x^{2} \log \left (x\right ) + \frac {1}{6} \, x e^{x} \log \left (x\right ) + \frac {5}{6} \, x \log \left (x\right ) + \frac {7}{12} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(((2*x^2+2*x)*exp(x)+12*x^2+10*x)*log(x)+2*exp(x)*x+6*x^2+10*x+7)/x,x, algorithm="giac")

[Out]

1/2*x^2*log(x) + 1/6*x*e^x*log(x) + 5/6*x*log(x) + 7/12*log(x)

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Mupad [B]
time = 0.35, size = 19, normalized size = 0.73 \begin {gather*} \frac {\ln \left (x\right )\,\left (10\,x+2\,x\,{\mathrm {e}}^x+6\,x^2+7\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x)/6 + (log(x)*(10*x + exp(x)*(2*x + 2*x^2) + 12*x^2))/12 + (x*exp(x))/6 + x^2/2 + 7/12)/x,x)

[Out]

(log(x)*(10*x + 2*x*exp(x) + 6*x^2 + 7))/12

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