3.71.1 \(\int \frac {1}{3} (-15-3 e^2-30 x-9 x^2+(150 x+45 x^2+e^9 (10 x+3 x^2)) \log (5)) \, dx\)

Optimal. Leaf size=30 \[ x \left (-5-e^2+(5+x) \left (-x+\left (5+\frac {e^9}{3}\right ) x \log (5)\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.90, number of steps used = 4, number of rules used = 1, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {12} \begin {gather*} -x^3+\frac {1}{3} e^9 x^3 \log (5)+5 x^3 \log (5)-5 x^2+\frac {5}{3} e^9 x^2 \log (5)+25 x^2 \log (5)-\left (5+e^2\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 - 3*E^2 - 30*x - 9*x^2 + (150*x + 45*x^2 + E^9*(10*x + 3*x^2))*Log[5])/3,x]

[Out]

-((5 + E^2)*x) - 5*x^2 - x^3 + 25*x^2*Log[5] + (5*E^9*x^2*Log[5])/3 + 5*x^3*Log[5] + (E^9*x^3*Log[5])/3

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (-15-3 e^2-30 x-9 x^2+\left (150 x+45 x^2+e^9 \left (10 x+3 x^2\right )\right ) \log (5)\right ) \, dx\\ &=-\left (\left (5+e^2\right ) x\right )-5 x^2-x^3+\frac {1}{3} \log (5) \int \left (150 x+45 x^2+e^9 \left (10 x+3 x^2\right )\right ) \, dx\\ &=-\left (\left (5+e^2\right ) x\right )-5 x^2-x^3+25 x^2 \log (5)+5 x^3 \log (5)+\frac {1}{3} \left (e^9 \log (5)\right ) \int \left (10 x+3 x^2\right ) \, dx\\ &=-\left (\left (5+e^2\right ) x\right )-5 x^2-x^3+25 x^2 \log (5)+\frac {5}{3} e^9 x^2 \log (5)+5 x^3 \log (5)+\frac {1}{3} e^9 x^3 \log (5)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 55, normalized size = 1.83 \begin {gather*} -5 x-e^2 x+\frac {5}{3} e^9 x^2 \log (5)+\frac {1}{3} e^9 x^3 \log (5)+5 x^2 (-1+5 \log (5))+x^3 (-1+5 \log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 - 3*E^2 - 30*x - 9*x^2 + (150*x + 45*x^2 + E^9*(10*x + 3*x^2))*Log[5])/3,x]

[Out]

-5*x - E^2*x + (5*E^9*x^2*Log[5])/3 + (E^9*x^3*Log[5])/3 + 5*x^2*(-1 + 5*Log[5]) + x^3*(-1 + 5*Log[5])

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 46, normalized size = 1.53 \begin {gather*} -x^{3} - 5 \, x^{2} - x e^{2} + \frac {1}{3} \, {\left (15 \, x^{3} + 75 \, x^{2} + {\left (x^{3} + 5 \, x^{2}\right )} e^{9}\right )} \log \relax (5) - 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+10*x)*exp(9)+45*x^2+150*x)*log(5)-exp(2)-3*x^2-10*x-5,x, algorithm="fricas")

[Out]

-x^3 - 5*x^2 - x*e^2 + 1/3*(15*x^3 + 75*x^2 + (x^3 + 5*x^2)*e^9)*log(5) - 5*x

________________________________________________________________________________________

giac [A]  time = 0.19, size = 46, normalized size = 1.53 \begin {gather*} -x^{3} - 5 \, x^{2} - x e^{2} + \frac {1}{3} \, {\left (15 \, x^{3} + 75 \, x^{2} + {\left (x^{3} + 5 \, x^{2}\right )} e^{9}\right )} \log \relax (5) - 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+10*x)*exp(9)+45*x^2+150*x)*log(5)-exp(2)-3*x^2-10*x-5,x, algorithm="giac")

[Out]

-x^3 - 5*x^2 - x*e^2 + 1/3*(15*x^3 + 75*x^2 + (x^3 + 5*x^2)*e^9)*log(5) - 5*x

________________________________________________________________________________________

maple [A]  time = 0.02, size = 42, normalized size = 1.40

method	result	size
norman	\(\left (-{\mathrm e}^{2}-5\right ) x +\left (\frac {\ln \relax (5) {\mathrm e}^{9}}{3}+5 \ln \relax (5)-1\right ) x^{3}+\left (\frac {5 \ln \relax (5) {\mathrm e}^{9}}{3}+25 \ln \relax (5)-5\right ) x^{2}\)	\(42\)
gosper	\(-\frac {x \left (-\ln \relax (5) {\mathrm e}^{9} x^{2}-5 \ln \relax (5) {\mathrm e}^{9} x -15 x^{2} \ln \relax (5)-75 x \ln \relax (5)+3 x^{2}+3 \,{\mathrm e}^{2}+15 x +15\right )}{3}\)	\(46\)
default	\(\frac {\ln \relax (5) \left ({\mathrm e}^{9} \left (x^{3}+5 x^{2}\right )+15 x^{3}+75 x^{2}\right )}{3}-{\mathrm e}^{2} x -x^{3}-5 x^{2}-5 x\)	\(47\)
risch	\(\frac {\ln \relax (5) x^{3} {\mathrm e}^{9}}{3}+5 x^{3} \ln \relax (5)+\frac {5 \ln \relax (5) {\mathrm e}^{9} x^{2}}{3}+25 x^{2} \ln \relax (5)-{\mathrm e}^{2} x -x^{3}-5 x^{2}-5 x\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((3*x^2+10*x)*exp(9)+45*x^2+150*x)*ln(5)-exp(2)-3*x^2-10*x-5,x,method=_RETURNVERBOSE)

[Out]

(-exp(2)-5)*x+(1/3*ln(5)*exp(9)+5*ln(5)-1)*x^3+(5/3*ln(5)*exp(9)+25*ln(5)-5)*x^2

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 46, normalized size = 1.53 \begin {gather*} -x^{3} - 5 \, x^{2} - x e^{2} + \frac {1}{3} \, {\left (15 \, x^{3} + 75 \, x^{2} + {\left (x^{3} + 5 \, x^{2}\right )} e^{9}\right )} \log \relax (5) - 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^2+10*x)*exp(9)+45*x^2+150*x)*log(5)-exp(2)-3*x^2-10*x-5,x, algorithm="maxima")

[Out]

-x^3 - 5*x^2 - x*e^2 + 1/3*(15*x^3 + 75*x^2 + (x^3 + 5*x^2)*e^9)*log(5) - 5*x

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 40, normalized size = 1.33 \begin {gather*} \left (\frac {\ln \relax (5)\,\left (3\,{\mathrm {e}}^9+45\right )}{9}-1\right )\,x^3+\left (\frac {\ln \relax (5)\,\left (10\,{\mathrm {e}}^9+150\right )}{6}-5\right )\,x^2+\left (-{\mathrm {e}}^2-5\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)*(150*x + exp(9)*(10*x + 3*x^2) + 45*x^2))/3 - exp(2) - 10*x - 3*x^2 - 5,x)

[Out]

x^3*((log(5)*(3*exp(9) + 45))/9 - 1) + x^2*((log(5)*(10*exp(9) + 150))/6 - 5) - x*(exp(2) + 5)

________________________________________________________________________________________

sympy [B]  time = 0.07, size = 46, normalized size = 1.53 \begin {gather*} x^{3} \left (-1 + 5 \log {\relax (5 )} + \frac {e^{9} \log {\relax (5 )}}{3}\right ) + x^{2} \left (-5 + 25 \log {\relax (5 )} + \frac {5 e^{9} \log {\relax (5 )}}{3}\right ) + x \left (- e^{2} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x**2+10*x)*exp(9)+45*x**2+150*x)*ln(5)-exp(2)-3*x**2-10*x-5,x)

[Out]

x**3*(-1 + 5*log(5) + exp(9)*log(5)/3) + x**2*(-5 + 25*log(5) + 5*exp(9)*log(5)/3) + x*(-exp(2) - 5)

________________________________________________________________________________________