\(\int \frac {1}{3} (42+72 x-18 x^2+e^5 (12 x^2+16 x^3) \log (5)) \, dx\) [4692]

Optimal result

Integrand size = 31, antiderivative size = 24 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=x (2+2 x) \left (7-x+\frac {2}{3} e^5 x^2 \log (5)\right ) \]

[Out]

x*(2/3*x^2*exp(5)*ln(5)-x+7)*(2+2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12} \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=\frac {4}{3} e^5 x^4 \log (5)-2 x^3+\frac {4}{3} e^5 x^3 \log (5)+12 x^2+14 x \]

[In]

Int[(42 + 72*x - 18*x^2 + E^5*(12*x^2 + 16*x^3)*Log[5])/3,x]

[Out]

14*x + 12*x^2 - 2*x^3 + (4*E^5*x^3*Log[5])/3 + (4*E^5*x^4*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{align*} \text {integral}& = \frac {1}{3} \int \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx \\ & = 14 x+12 x^2-2 x^3+\frac {1}{3} \left (e^5 \log (5)\right ) \int \left (12 x^2+16 x^3\right ) \, dx \\ & = 14 x+12 x^2-2 x^3+\frac {4}{3} e^5 x^3 \log (5)+\frac {4}{3} e^5 x^4 \log (5) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=14 x+12 x^2-2 x^3+\frac {4}{3} e^5 x^3 \log (5)+\frac {4}{3} e^5 x^4 \log (5) \]

[In]

Integrate[(42 + 72*x - 18*x^2 + E^5*(12*x^2 + 16*x^3)*Log[5])/3,x]

[Out]

14*x + 12*x^2 - 2*x^3 + (4*E^5*x^3*Log[5])/3 + (4*E^5*x^4*Log[5])/3

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29

method	result	size
norman	\(\left (\frac {4 \,{\mathrm e}^{5} \ln \left (5\right )}{3}-2\right ) x^{3}+14 x +12 x^{2}+\frac {4 x^{4} {\mathrm e}^{5} \ln \left (5\right )}{3}\)	\(31\)
gosper	\(\frac {2 x \left (2 \,{\mathrm e}^{5} \ln \left (5\right ) x^{3}+2 x^{2} {\mathrm e}^{5} \ln \left (5\right )-3 x^{2}+18 x +21\right )}{3}\)	\(32\)
default	\(\frac {4 x^{4} {\mathrm e}^{5} \ln \left (5\right )}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (5\right ) x^{3}}{3}-2 x^{3}+12 x^{2}+14 x\)	\(33\)
risch	\(\frac {4 x^{4} {\mathrm e}^{5} \ln \left (5\right )}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (5\right ) x^{3}}{3}-2 x^{3}+12 x^{2}+14 x\)	\(33\)
parallelrisch	\(\frac {4 x^{4} {\mathrm e}^{5} \ln \left (5\right )}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (5\right ) x^{3}}{3}-2 x^{3}+12 x^{2}+14 x\)	\(33\)
parts	\(\frac {4 x^{4} {\mathrm e}^{5} \ln \left (5\right )}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (5\right ) x^{3}}{3}-2 x^{3}+12 x^{2}+14 x\)	\(33\)

[In]

int(1/3*(16*x^3+12*x^2)*exp(5)*ln(5)-6*x^2+24*x+14,x,method=_RETURNVERBOSE)

[Out]

(4/3*exp(5)*ln(5)-2)*x^3+14*x+12*x^2+4/3*x^4*exp(5)*ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=-2 \, x^{3} + \frac {4}{3} \, {\left (x^{4} + x^{3}\right )} e^{5} \log \left (5\right ) + 12 \, x^{2} + 14 \, x \]

[In]

integrate(1/3*(16*x^3+12*x^2)*exp(5)*log(5)-6*x^2+24*x+14,x, algorithm="fricas")

[Out]

-2*x^3 + 4/3*(x^4 + x^3)*e^5*log(5) + 12*x^2 + 14*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=\frac {4 x^{4} e^{5} \log {\left (5 \right )}}{3} + x^{3} \left (-2 + \frac {4 e^{5} \log {\left (5 \right )}}{3}\right ) + 12 x^{2} + 14 x \]

[In]

integrate(1/3*(16*x**3+12*x**2)*exp(5)*ln(5)-6*x**2+24*x+14,x)

[Out]

4*x**4*exp(5)*log(5)/3 + x**3*(-2 + 4*exp(5)*log(5)/3) + 12*x**2 + 14*x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=-2 \, x^{3} + \frac {4}{3} \, {\left (x^{4} + x^{3}\right )} e^{5} \log \left (5\right ) + 12 \, x^{2} + 14 \, x \]

[In]

integrate(1/3*(16*x^3+12*x^2)*exp(5)*log(5)-6*x^2+24*x+14,x, algorithm="maxima")

[Out]

-2*x^3 + 4/3*(x^4 + x^3)*e^5*log(5) + 12*x^2 + 14*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=-2 \, x^{3} + \frac {4}{3} \, {\left (x^{4} + x^{3}\right )} e^{5} \log \left (5\right ) + 12 \, x^{2} + 14 \, x \]

[In]

integrate(1/3*(16*x^3+12*x^2)*exp(5)*log(5)-6*x^2+24*x+14,x, algorithm="giac")

[Out]

-2*x^3 + 4/3*(x^4 + x^3)*e^5*log(5) + 12*x^2 + 14*x

Mupad [B] (verification not implemented)

Time = 11.55 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {1}{3} \left (42+72 x-18 x^2+e^5 \left (12 x^2+16 x^3\right ) \log (5)\right ) \, dx=\frac {4\,{\mathrm {e}}^5\,\ln \left (5\right )\,x^4}{3}+\left (\frac {4\,{\mathrm {e}}^5\,\ln \left (5\right )}{3}-2\right )\,x^3+12\,x^2+14\,x \]

[In]

int(24*x - 6*x^2 + (exp(5)*log(5)*(12*x^2 + 16*x^3))/3 + 14,x)

[Out]

14*x + x^3*((4*exp(5)*log(5))/3 - 2) + 12*x^2 + (4*x^4*exp(5)*log(5))/3