3.57.83 \(\int \frac {-e^3 x^2-x^3+(3 e^3 x^2+4 x^3) \log (\frac {-1+e^4}{12 x})}{\log (4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {x^3 \left (e^3+x\right ) \log \left (\frac {-1+e^4}{12 x}\right )}{\log (4)} \]

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Rubi [A]  time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {12, 1593, 43, 2334} \begin {gather*} \frac {\left (x^4+e^3 x^3\right ) \log \left (-\frac {1-e^4}{12 x}\right )}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^3*x^2) - x^3 + (3*E^3*x^2 + 4*x^3)*Log[(-1 + E^4)/(12*x)])/Log[4],x]

[Out]

((E^3*x^3 + x^4)*Log[-1/12*(1 - E^4)/x])/Log[4]

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-e^3 x^2-x^3+\left (3 e^3 x^2+4 x^3\right ) \log \left (\frac {-1+e^4}{12 x}\right )\right ) \, dx}{\log (4)}\\ &=-\frac {e^3 x^3}{3 \log (4)}-\frac {x^4}{4 \log (4)}+\frac {\int \left (3 e^3 x^2+4 x^3\right ) \log \left (\frac {-1+e^4}{12 x}\right ) \, dx}{\log (4)}\\ &=-\frac {e^3 x^3}{3 \log (4)}-\frac {x^4}{4 \log (4)}+\frac {\int x^2 \left (3 e^3+4 x\right ) \log \left (\frac {-1+e^4}{12 x}\right ) \, dx}{\log (4)}\\ &=-\frac {e^3 x^3}{3 \log (4)}-\frac {x^4}{4 \log (4)}+\frac {\left (e^3 x^3+x^4\right ) \log \left (-\frac {1-e^4}{12 x}\right )}{\log (4)}+\frac {\int x^2 \left (e^3+x\right ) \, dx}{\log (4)}\\ &=-\frac {e^3 x^3}{3 \log (4)}-\frac {x^4}{4 \log (4)}+\frac {\left (e^3 x^3+x^4\right ) \log \left (-\frac {1-e^4}{12 x}\right )}{\log (4)}+\frac {\int \left (e^3 x^2+x^3\right ) \, dx}{\log (4)}\\ &=\frac {\left (e^3 x^3+x^4\right ) \log \left (-\frac {1-e^4}{12 x}\right )}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.73 \begin {gather*} \frac {e^3 x^3 \log \left (-\frac {1-e^4}{12 x}\right )+x^4 \log \left (\frac {-1+e^4}{12 x}\right )}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^3*x^2) - x^3 + (3*E^3*x^2 + 4*x^3)*Log[(-1 + E^4)/(12*x)])/Log[4],x]

[Out]

(E^3*x^3*Log[-1/12*(1 - E^4)/x] + x^4*Log[(-1 + E^4)/(12*x)])/Log[4]

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fricas [A]  time = 0.63, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{4} + x^{3} e^{3}\right )} \log \left (\frac {e^{4} - 1}{12 \, x}\right )}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x^2*exp(3)+4*x^3)*log(1/12*(exp(4)-1)/x)-x^2*exp(3)-x^3)/log(2),x, algorithm="fricas")

[Out]

1/2*(x^4 + x^3*e^3)*log(1/12*(e^4 - 1)/x)/log(2)

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giac [B]  time = 0.19, size = 209, normalized size = 8.04 \begin {gather*} -\frac {3 \, x^{4} + 4 \, x^{3} e^{3} - \frac {\frac {12 \, x^{4} {\left (\frac {{\left (e^{4} - 1\right )} e^{19}}{x} - \frac {4 \, {\left (e^{4} - 1\right )} e^{15}}{x} + \frac {6 \, {\left (e^{4} - 1\right )} e^{11}}{x} - \frac {4 \, {\left (e^{4} - 1\right )} e^{7}}{x} + \frac {{\left (e^{4} - 1\right )} e^{3}}{x} + e^{20} - 5 \, e^{16} + 10 \, e^{12} - 10 \, e^{8} + 5 \, e^{4} - 1\right )} \log \left (\frac {e^{4} - 1}{12 \, x}\right )}{{\left (e^{4} - 1\right )}^{4}} + \frac {x^{4} {\left (\frac {4 \, {\left (e^{4} - 1\right )} e^{19}}{x} - \frac {16 \, {\left (e^{4} - 1\right )} e^{15}}{x} + \frac {24 \, {\left (e^{4} - 1\right )} e^{11}}{x} - \frac {16 \, {\left (e^{4} - 1\right )} e^{7}}{x} + \frac {4 \, {\left (e^{4} - 1\right )} e^{3}}{x} + 3 \, e^{20} - 15 \, e^{16} + 30 \, e^{12} - 30 \, e^{8} + 15 \, e^{4} - 3\right )}}{{\left (e^{4} - 1\right )}^{4}}}{e^{4} - 1}}{24 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x^2*exp(3)+4*x^3)*log(1/12*(exp(4)-1)/x)-x^2*exp(3)-x^3)/log(2),x, algorithm="giac")

[Out]

-1/24*(3*x^4 + 4*x^3*e^3 - (12*x^4*((e^4 - 1)*e^19/x - 4*(e^4 - 1)*e^15/x + 6*(e^4 - 1)*e^11/x - 4*(e^4 - 1)*e
^7/x + (e^4 - 1)*e^3/x + e^20 - 5*e^16 + 10*e^12 - 10*e^8 + 5*e^4 - 1)*log(1/12*(e^4 - 1)/x)/(e^4 - 1)^4 + x^4
*(4*(e^4 - 1)*e^19/x - 16*(e^4 - 1)*e^15/x + 24*(e^4 - 1)*e^11/x - 16*(e^4 - 1)*e^7/x + 4*(e^4 - 1)*e^3/x + 3*
e^20 - 15*e^16 + 30*e^12 - 30*e^8 + 15*e^4 - 3)/(e^4 - 1)^4)/(e^4 - 1))/log(2)

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maple [A]  time = 0.13, size = 27, normalized size = 1.04

method	result	size
risch	\(\frac {\left (x^{3} {\mathrm e}^{3}+x^{4}\right ) \ln \left (\frac {{\mathrm e}^{4}-1}{12 x}\right )}{2 \ln \relax (2)}\)	\(27\)
norman	\(\frac {x^{4} \ln \left (\frac {{\mathrm e}^{4}-1}{12 x}\right )}{2 \ln \relax (2)}+\frac {{\mathrm e}^{3} x^{3} \ln \left (\frac {{\mathrm e}^{4}-1}{12 x}\right )}{2 \ln \relax (2)}\)	\(42\)
derivativedivides	\(-\frac {\left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right ) \left (\frac {{\mathrm e}^{3} {\mathrm e}^{8} \left (-\frac {x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{3 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}-\frac {x^{3}}{9 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}\right )}{48}+\frac {{\mathrm e}^{12} \left (-\frac {x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{4 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {x^{4}}{16 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}\right )}{432}+\frac {{\mathrm e}^{3} {\mathrm e}^{8} x^{3}}{432 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{12} x^{4}}{6912 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{8} \left (-\frac {x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{4 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {x^{4}}{16 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}\right )}{144}-\frac {{\mathrm e}^{8} x^{4}}{2304 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{3} {\mathrm e}^{4} \left (-\frac {x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{3 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}-\frac {x^{3}}{9 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}\right )}{24}-\frac {{\mathrm e}^{3} {\mathrm e}^{4} x^{3}}{216 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{4} \left (-\frac {x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{4 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {x^{4}}{16 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}\right )}{144}+\frac {{\mathrm e}^{4} x^{4}}{2304 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}+\frac {{\mathrm e}^{3} \left (-\frac {x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{3 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}-\frac {x^{3}}{9 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}\right )}{48}+\frac {{\mathrm e}^{3} x^{3}}{432 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{1728 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}\right )}{2 \ln \relax (2)}\)	\(400\)
default	\(\frac {-\frac {{\mathrm e}^{4} x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{5184 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{3} x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{1728 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{12} {\mathrm e}^{3} x^{3}}{5184 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{16} x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{20736 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{12} x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{5184 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}+\frac {{\mathrm e}^{8} x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{3456 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}+\frac {{\mathrm e}^{3} {\mathrm e}^{4} x^{3}}{1728 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}-\frac {{\mathrm e}^{3} {\mathrm e}^{8} x^{3}}{1728 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}-\frac {{\mathrm e}^{8} {\mathrm e}^{3} x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{576 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{4} {\mathrm e}^{3} x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{576 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{12} {\mathrm e}^{3} x^{3} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{1728 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {x^{4}}{82944 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{4} x^{4}}{20736 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{3} x^{3}}{5184 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{3}}+\frac {{\mathrm e}^{16} x^{4}}{82944 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}+\frac {x^{4} \ln \left (\frac {\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}}{x}\right )}{20736 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {{\mathrm e}^{12} x^{4}}{20736 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}+\frac {{\mathrm e}^{8} x^{4}}{13824 \left (\frac {{\mathrm e}^{4}}{12}-\frac {1}{12}\right )^{4}}-\frac {x^{4}}{4}-\frac {x^{3} {\mathrm e}^{3}}{3}}{2 \ln \relax (2)}\)	\(417\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((3*x^2*exp(3)+4*x^3)*ln(1/12*(exp(4)-1)/x)-x^2*exp(3)-x^3)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/2/ln(2)*(x^3*exp(3)+x^4)*ln(1/12*(exp(4)-1)/x)

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maxima [A]  time = 0.36, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{4} + x^{3} e^{3}\right )} \log \left (\frac {e^{4} - 1}{12 \, x}\right )}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x^2*exp(3)+4*x^3)*log(1/12*(exp(4)-1)/x)-x^2*exp(3)-x^3)/log(2),x, algorithm="maxima")

[Out]

1/2*(x^4 + x^3*e^3)*log(1/12*(e^4 - 1)/x)/log(2)

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mupad [B]  time = 3.65, size = 25, normalized size = 0.96 \begin {gather*} \frac {x^3\,\left (\ln \left (\frac {1}{x}\right )+\ln \left (\frac {{\mathrm {e}}^4}{12}-\frac {1}{12}\right )\right )\,\left (x+{\mathrm {e}}^3\right )}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^2*exp(3))/2 - (log((exp(4)/12 - 1/12)/x)*(3*x^2*exp(3) + 4*x^3))/2 + x^3/2)/log(2),x)

[Out]

(x^3*(log(1/x) + log(exp(4)/12 - 1/12))*(x + exp(3)))/(2*log(2))

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sympy [A]  time = 0.15, size = 26, normalized size = 1.00 \begin {gather*} \frac {\left (x^{4} + x^{3} e^{3}\right ) \log {\left (\frac {- \frac {1}{12} + \frac {e^{4}}{12}}{x} \right )}}{2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3*x**2*exp(3)+4*x**3)*ln(1/12*(exp(4)-1)/x)-x**2*exp(3)-x**3)/ln(2),x)

[Out]

(x**4 + x**3*exp(3))*log((-1/12 + exp(4)/12)/x)/(2*log(2))

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