3.43.93 \(\int \frac {1}{16} (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)) \, dx\)

Optimal. Leaf size=22 \[ e^3 \left (-\frac {37}{8}-5 x\right ) \left (1+\frac {1}{2} x (-1+\log (x))\right ) \]

________________________________________________________________________________________

Rubi [B]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 2.32, number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 2313} \begin {gather*} \frac {5 e^3 x^2}{4}-\frac {1}{16} e^3 \left (40 x^2+37 x\right ) \log (x)+\frac {5}{4} e^3 (2-x)^2+\frac {37 e^3 x}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(-80 + 40*x) + E^3*(-37 - 80*x)*Log[x])/16,x]

[Out]

(5*E^3*(2 - x)^2)/4 + (37*E^3*x)/16 + (5*E^3*x^2)/4 - (E^3*(37*x + 40*x^2)*Log[x])/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {1}{16} \int \left (e^3 (-80+40 x)+e^3 (-37-80 x) \log (x)\right ) \, dx\\ &=\frac {5}{4} e^3 (2-x)^2+\frac {1}{16} e^3 \int (-37-80 x) \log (x) \, dx\\ &=\frac {5}{4} e^3 (2-x)^2-\frac {1}{16} e^3 \left (37 x+40 x^2\right ) \log (x)-\frac {1}{16} e^3 \int (-37-40 x) \, dx\\ &=\frac {5}{4} e^3 (2-x)^2+\frac {37 e^3 x}{16}+\frac {5 e^3 x^2}{4}-\frac {1}{16} e^3 \left (37 x+40 x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 41, normalized size = 1.86 \begin {gather*} -\frac {43 e^3 x}{16}+\frac {5 e^3 x^2}{2}-\frac {37}{16} e^3 x \log (x)-\frac {5}{2} e^3 x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(-80 + 40*x) + E^3*(-37 - 80*x)*Log[x])/16,x]

[Out]

(-43*E^3*x)/16 + (5*E^3*x^2)/2 - (37*E^3*x*Log[x])/16 - (5*E^3*x^2*Log[x])/2

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 29, normalized size = 1.32 \begin {gather*} -\frac {1}{16} \, {\left (40 \, x^{2} + 37 \, x\right )} e^{3} \log \relax (x) + \frac {1}{16} \, {\left (40 \, x^{2} - 43 \, x\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-80*x-37)*exp(3)*log(x)+1/16*(40*x-80)*exp(3),x, algorithm="fricas")

[Out]

-1/16*(40*x^2 + 37*x)*e^3*log(x) + 1/16*(40*x^2 - 43*x)*e^3

________________________________________________________________________________________

giac [B]  time = 0.14, size = 37, normalized size = 1.68 \begin {gather*} -\frac {1}{16} \, {\left (40 \, x^{2} \log \relax (x) - 20 \, x^{2} + 37 \, x \log \relax (x) - 37 \, x\right )} e^{3} + \frac {5}{4} \, {\left (x^{2} - 4 \, x\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-80*x-37)*exp(3)*log(x)+1/16*(40*x-80)*exp(3),x, algorithm="giac")

[Out]

-1/16*(40*x^2*log(x) - 20*x^2 + 37*x*log(x) - 37*x)*e^3 + 5/4*(x^2 - 4*x)*e^3

________________________________________________________________________________________

maple [A]  time = 0.02, size = 29, normalized size = 1.32

method	result	size
risch	\(\frac {{\mathrm e}^{3} \left (-40 x^{2}-37 x \right ) \ln \relax (x )}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{2}-\frac {43 x \,{\mathrm e}^{3}}{16}\)	\(29\)
norman	\(-\frac {43 x \,{\mathrm e}^{3}}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{2}-\frac {37 x \,{\mathrm e}^{3} \ln \relax (x )}{16}-\frac {5 \,{\mathrm e}^{3} \ln \relax (x ) x^{2}}{2}\)	\(30\)
default	\(\frac {{\mathrm e}^{3} \left (20 x^{2}-80 x \right )}{16}-\frac {5 \,{\mathrm e}^{3} \ln \relax (x ) x^{2}}{2}-\frac {37 x \,{\mathrm e}^{3} \ln \relax (x )}{16}+\frac {5 x^{2} {\mathrm e}^{3}}{4}+\frac {37 x \,{\mathrm e}^{3}}{16}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(-80*x-37)*exp(3)*ln(x)+1/16*(40*x-80)*exp(3),x,method=_RETURNVERBOSE)

[Out]

1/16*exp(3)*(-40*x^2-37*x)*ln(x)+5/2*x^2*exp(3)-43/16*x*exp(3)

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 38, normalized size = 1.73 \begin {gather*} \frac {1}{16} \, {\left (20 \, x^{2} - {\left (40 \, x^{2} + 37 \, x\right )} \log \relax (x) + 37 \, x\right )} e^{3} + \frac {5}{4} \, {\left (x^{2} - 4 \, x\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-80*x-37)*exp(3)*log(x)+1/16*(40*x-80)*exp(3),x, algorithm="maxima")

[Out]

1/16*(20*x^2 - (40*x^2 + 37*x)*log(x) + 37*x)*e^3 + 5/4*(x^2 - 4*x)*e^3

________________________________________________________________________________________

mupad [B]  time = 3.26, size = 19, normalized size = 0.86 \begin {gather*} -\frac {x\,{\mathrm {e}}^3\,\left (37\,\ln \relax (x)-40\,x+40\,x\,\ln \relax (x)+43\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(40*x - 80))/16 - (exp(3)*log(x)*(80*x + 37))/16,x)

[Out]

-(x*exp(3)*(37*log(x) - 40*x + 40*x*log(x) + 43))/16

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 41, normalized size = 1.86 \begin {gather*} \frac {5 x^{2} e^{3}}{2} - \frac {43 x e^{3}}{16} + \left (- \frac {5 x^{2} e^{3}}{2} - \frac {37 x e^{3}}{16}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-80*x-37)*exp(3)*ln(x)+1/16*(40*x-80)*exp(3),x)

[Out]

5*x**2*exp(3)/2 - 43*x*exp(3)/16 + (-5*x**2*exp(3)/2 - 37*x*exp(3)/16)*log(x)

________________________________________________________________________________________