\(\int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} (6-2 e^6-3 x)^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx\) [6875]

Optimal result

Integrand size = 48, antiderivative size = 22 \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=5 \left (3-e^6-\frac {3 x}{2}\right )^{2 e^{e^2}} \]

[Out]

5*exp(exp(exp(2))*ln(-exp(6)-3/2*x+3))^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 21, 32} \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=5\ 4^{-e^{e^2}} \left (2 \left (3-e^6\right )-3 x\right )^{2 e^{e^2}} \]

[In]

Int[(15*2^(1 - 2*E^E^2)*E^E^2*(6 - 2*E^6 - 3*x)^(2*E^E^2))/(-6 + 2*E^6 + 3*x),x]

[Out]

(5*(2*(3 - E^6) - 3*x)^(2*E^E^2))/4^E^E^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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \left (15\ 2^{1-2 e^{e^2}} e^{e^2}\right ) \int \frac {\left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx \\ & = -\left (\left (15\ 2^{1-2 e^{e^2}} e^{e^2}\right ) \int \left (6-2 e^6-3 x\right )^{-1+2 e^{e^2}} \, dx\right ) \\ & = 5\ 4^{-e^{e^2}} \left (2 \left (3-e^6\right )-3 x\right )^{2 e^{e^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=5 \left (3-e^6-\frac {3 x}{2}\right )^{2 e^{e^2}} \]

[In]

Integrate[(15*2^(1 - 2*E^E^2)*E^E^2*(6 - 2*E^6 - 3*x)^(2*E^E^2))/(-6 + 2*E^6 + 3*x),x]

[Out]

5*(3 - E^6 - (3*x)/2)^(2*E^E^2)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
risch	\(5 \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )^{2 \,{\mathrm e}^{{\mathrm e}^{2}}}\)	\(18\)
gosper	\(5 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )}\)	\(20\)
derivativedivides	\(5 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )}\)	\(20\)
default	\(5 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )}\)	\(20\)
norman	\(5 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )}\)	\(20\)
parallelrisch	\(5 \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (-{\mathrm e}^{6}-\frac {3 x}{2}+3\right )}\)	\(20\)

[In]

int(30*exp(exp(2))*exp(exp(exp(2))*ln(-exp(6)-3/2*x+3))^2/(2*exp(6)+3*x-6),x,method=_RETURNVERBOSE)

[Out]

5*((-exp(6)-3/2*x+3)^exp(exp(2)))^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=5 \, {\left (-\frac {3}{2} \, x - e^{6} + 3\right )}^{2 \, e^{\left (e^{2}\right )}} \]

[In]

integrate(30*exp(exp(2))*exp(exp(exp(2))*log(-exp(6)-3/2*x+3))^2/(2*exp(6)+3*x-6),x, algorithm="fricas")

[Out]

5*(-3/2*x - e^6 + 3)^(2*e^(e^2))

Sympy [F(-2)]

Exception generated. \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(30*exp(exp(2))*exp(exp(exp(2))*ln(-exp(6)-3/2*x+3))**2/(2*exp(6)+3*x-6),x)

[Out]

Exception raised: TypeError >> Invalid NaN comparison

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=\frac {5 \, {\left (-3 \, x - 2 \, e^{6} + 6\right )}^{2 \, e^{\left (e^{2}\right )}}}{2^{2 \, e^{\left (e^{2}\right )}}} \]

[In]

integrate(30*exp(exp(2))*exp(exp(exp(2))*log(-exp(6)-3/2*x+3))^2/(2*exp(6)+3*x-6),x, algorithm="maxima")

[Out]

5*(-3*x - 2*e^6 + 6)^(2*e^(e^2))/2^(2*e^(e^2))

Giac [F]

\[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=\int { \frac {30 \, {\left (-\frac {3}{2} \, x - e^{6} + 3\right )}^{2 \, e^{\left (e^{2}\right )}} e^{\left (e^{2}\right )}}{3 \, x + 2 \, e^{6} - 6} \,d x } \]

[In]

integrate(30*exp(exp(2))*exp(exp(exp(2))*log(-exp(6)-3/2*x+3))^2/(2*exp(6)+3*x-6),x, algorithm="giac")

[Out]

integrate(30*(-3/2*x - e^6 + 3)^(2*e^(e^2))*e^(e^2)/(3*x + 2*e^6 - 6), x)

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {15\ 2^{1-2 e^{e^2}} e^{e^2} \left (6-2 e^6-3 x\right )^{2 e^{e^2}}}{-6+2 e^6+3 x} \, dx=5\,{\left (3-{\mathrm {e}}^6-\frac {3\,x}{2}\right )}^{2\,{\mathrm {e}}^{{\mathrm {e}}^2}} \]

[In]

int((30*exp(exp(2))*(3 - exp(6) - (3*x)/2)^(2*exp(exp(2))))/(3*x + 2*exp(6) - 6),x)

[Out]

5*(3 - exp(6) - (3*x)/2)^(2*exp(exp(2)))