3.61.39 \(\int \frac {(-15 x+11 x^2-2 x^3+e^x (3 x-x^2))^x (-15+22 x-6 x^2+e^x (3+x-x^2)+(-15+e^x (3-x)+11 x-2 x^2) \log (-15 x+11 x^2-2 x^3+e^x (3 x-x^2)))}{240-176 x+32 x^2+e^x (-48+16 x)+(-120+e^x (24-8 x)+88 x-16 x^2) (-15 x+11 x^2-2 x^3+e^x (3 x-x^2))^x+(15+e^x (-3+x)-11 x+2 x^2) (-15 x+11 x^2-2 x^3+e^x (3 x-x^2))^{2 x}} \, dx\)

Optimal. Leaf size=23 \[ 1+\frac {1}{-4+\left ((3-x) x \left (-5+e^x+2 x\right )\right )^x} \]

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Rubi [A]  time = 1.19, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 219, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6688, 6708, 32} \begin {gather*} -\frac {1}{4-\left (-\left ((x-3) x \left (2 x+e^x-5\right )\right )\right )^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^x*(-15 + 22*x - 6*x^2 + E^x*(3 + x - x^2) + (-15 + E^x*(3 - x)
+ 11*x - 2*x^2)*Log[-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2)]))/(240 - 176*x + 32*x^2 + E^x*(-48 + 16*x) + (-
120 + E^x*(24 - 8*x) + 88*x - 16*x^2)*(-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^x + (15 + E^x*(-3 + x) - 11*x
+ 2*x^2)*(-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^(2*x)),x]

[Out]

-(4 - (-((-3 + x)*x*(-5 + E^x + 2*x)))^x)^(-1)

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x \left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )^{-1+x} \left (15-22 x+6 x^2+e^x \left (-3-x+x^2\right )+(-3+x) \left (-5+e^x+2 x\right ) \log \left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )\right )}{\left (4-\left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )^x\right )^2} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{(4+x)^2} \, dx,x,-\left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )^x\right )\\ &=-\frac {1}{4-\left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )^x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 20, normalized size = 0.87 \begin {gather*} \frac {1}{-4+\left (-\left ((-3+x) x \left (-5+e^x+2 x\right )\right )\right )^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^x*(-15 + 22*x - 6*x^2 + E^x*(3 + x - x^2) + (-15 + E^x*(
3 - x) + 11*x - 2*x^2)*Log[-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2)]))/(240 - 176*x + 32*x^2 + E^x*(-48 + 16*x
) + (-120 + E^x*(24 - 8*x) + 88*x - 16*x^2)*(-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^x + (15 + E^x*(-3 + x)
- 11*x + 2*x^2)*(-15*x + 11*x^2 - 2*x^3 + E^x*(3*x - x^2))^(2*x)),x]

[Out]

(-4 + (-((-3 + x)*x*(-5 + E^x + 2*x)))^x)^(-1)

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fricas [A]  time = 0.65, size = 31, normalized size = 1.35 \begin {gather*} \frac {1}{{\left (-2 \, x^{3} + 11 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x} - 15 \, x\right )}^{x} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*exp(x)-2*x^2+11*x-15)*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x)+(-x^2+x+3)*exp(x)-6*x^2+22*x-
15)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))/(((x-3)*exp(x)+2*x^2-11*x+15)*exp(x*log((-x^2+3*x)*exp(x)-
2*x^3+11*x^2-15*x))^2+((-8*x+24)*exp(x)-16*x^2+88*x-120)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))+(16*x
-48)*exp(x)+32*x^2-176*x+240),x, algorithm="fricas")

[Out]

1/((-2*x^3 + 11*x^2 - (x^2 - 3*x)*e^x - 15*x)^x - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (6 \, x^{2} + {\left (x^{2} - x - 3\right )} e^{x} + {\left (2 \, x^{2} + {\left (x - 3\right )} e^{x} - 11 \, x + 15\right )} \log \left (-2 \, x^{3} + 11 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x} - 15 \, x\right ) - 22 \, x + 15\right )} {\left (-2 \, x^{3} + 11 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x} - 15 \, x\right )}^{x}}{{\left (2 \, x^{2} + {\left (x - 3\right )} e^{x} - 11 \, x + 15\right )} {\left (-2 \, x^{3} + 11 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x} - 15 \, x\right )}^{2 \, x} - 8 \, {\left (2 \, x^{2} + {\left (x - 3\right )} e^{x} - 11 \, x + 15\right )} {\left (-2 \, x^{3} + 11 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x} - 15 \, x\right )}^{x} + 32 \, x^{2} + 16 \, {\left (x - 3\right )} e^{x} - 176 \, x + 240}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*exp(x)-2*x^2+11*x-15)*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x)+(-x^2+x+3)*exp(x)-6*x^2+22*x-
15)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))/(((x-3)*exp(x)+2*x^2-11*x+15)*exp(x*log((-x^2+3*x)*exp(x)-
2*x^3+11*x^2-15*x))^2+((-8*x+24)*exp(x)-16*x^2+88*x-120)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))+(16*x
-48)*exp(x)+32*x^2-176*x+240),x, algorithm="giac")

[Out]

integrate(-(6*x^2 + (x^2 - x - 3)*e^x + (2*x^2 + (x - 3)*e^x - 11*x + 15)*log(-2*x^3 + 11*x^2 - (x^2 - 3*x)*e^
x - 15*x) - 22*x + 15)*(-2*x^3 + 11*x^2 - (x^2 - 3*x)*e^x - 15*x)^x/((2*x^2 + (x - 3)*e^x - 11*x + 15)*(-2*x^3
+ 11*x^2 - (x^2 - 3*x)*e^x - 15*x)^(2*x) - 8*(2*x^2 + (x - 3)*e^x - 11*x + 15)*(-2*x^3 + 11*x^2 - (x^2 - 3*x)
*e^x - 15*x)^x + 32*x^2 + 16*(x - 3)*e^x - 176*x + 240), x)

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maple [C]  time = 0.16, size = 228, normalized size = 9.91

method	result	size
risch	\(\frac {1}{{\mathrm e}^{\frac {x \left (i \pi \mathrm {csgn}\left (i x \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )^{3}+i \pi \mathrm {csgn}\left (i x \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \mathrm {csgn}\left (i x \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )-i \pi \,\mathrm {csgn}\left (i x \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )-2 i \mathrm {csgn}\left (i x \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )^{2} \pi +2 i \pi +2 \ln \relax (x )+2 \ln \left (x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {11}{2}\right ) x -\frac {3 \,{\mathrm e}^{x}}{2}+\frac {15}{2}\right )\right )}{2}}-4}\)	\(228\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3-x)*exp(x)-2*x^2+11*x-15)*ln((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x)+(-x^2+x+3)*exp(x)-6*x^2+22*x-15)*exp
(x*ln((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))/(((x-3)*exp(x)+2*x^2-11*x+15)*exp(x*ln((-x^2+3*x)*exp(x)-2*x^3+11*
x^2-15*x))^2+((-8*x+24)*exp(x)-16*x^2+88*x-120)*exp(x*ln((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))+(16*x-48)*exp(x
)+32*x^2-176*x+240),x,method=_RETURNVERBOSE)

[Out]

1/(exp(1/2*x*(I*Pi*csgn(I*x*(x^2+(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2))^3+I*Pi*csgn(I*x*(x^2+(1/2*exp(x)-11/2)*
x-3/2*exp(x)+15/2))^2*csgn(I*x)+I*Pi*csgn(I*x*(x^2+(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2))^2*csgn(I*(x^2+(1/2*ex
p(x)-11/2)*x-3/2*exp(x)+15/2))-I*Pi*csgn(I*x*(x^2+(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2))*csgn(I*x)*csgn(I*(x^2+
(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2))-2*I*csgn(I*x*(x^2+(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2))^2*Pi+2*I*Pi+2*ln
(x)+2*ln(x^2+(1/2*exp(x)-11/2)*x-3/2*exp(x)+15/2)))-4)

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maxima [A]  time = 0.50, size = 28, normalized size = 1.22 \begin {gather*} \frac {1}{e^{\left (x \log \left (x - 3\right ) + x \log \relax (x) + x \log \left (-2 \, x - e^{x} + 5\right )\right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*exp(x)-2*x^2+11*x-15)*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x)+(-x^2+x+3)*exp(x)-6*x^2+22*x-
15)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))/(((x-3)*exp(x)+2*x^2-11*x+15)*exp(x*log((-x^2+3*x)*exp(x)-
2*x^3+11*x^2-15*x))^2+((-8*x+24)*exp(x)-16*x^2+88*x-120)*exp(x*log((-x^2+3*x)*exp(x)-2*x^3+11*x^2-15*x))+(16*x
-48)*exp(x)+32*x^2-176*x+240),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.95, size = 32, normalized size = 1.39 \begin {gather*} \frac {1}{{\left (3\,x\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x-15\,x+11\,x^2-2\,x^3\right )}^x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x*log(exp(x)*(3*x - x^2) - 15*x + 11*x^2 - 2*x^3))*(log(exp(x)*(3*x - x^2) - 15*x + 11*x^2 - 2*x^3)*
(exp(x)*(x - 3) - 11*x + 2*x^2 + 15) - exp(x)*(x - x^2 + 3) - 22*x + 6*x^2 + 15))/(exp(x)*(16*x - 48) - exp(x*
log(exp(x)*(3*x - x^2) - 15*x + 11*x^2 - 2*x^3))*(exp(x)*(8*x - 24) - 88*x + 16*x^2 + 120) - 176*x + 32*x^2 +
exp(2*x*log(exp(x)*(3*x - x^2) - 15*x + 11*x^2 - 2*x^3))*(exp(x)*(x - 3) - 11*x + 2*x^2 + 15) + 240),x)

[Out]

1/((3*x*exp(x) - x^2*exp(x) - 15*x + 11*x^2 - 2*x^3)^x - 4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3-x)*exp(x)-2*x**2+11*x-15)*ln((-x**2+3*x)*exp(x)-2*x**3+11*x**2-15*x)+(-x**2+x+3)*exp(x)-6*x**2+
22*x-15)*exp(x*ln((-x**2+3*x)*exp(x)-2*x**3+11*x**2-15*x))/(((x-3)*exp(x)+2*x**2-11*x+15)*exp(x*ln((-x**2+3*x)
*exp(x)-2*x**3+11*x**2-15*x))**2+((-8*x+24)*exp(x)-16*x**2+88*x-120)*exp(x*ln((-x**2+3*x)*exp(x)-2*x**3+11*x**
2-15*x))+(16*x-48)*exp(x)+32*x**2-176*x+240),x)

[Out]

Timed out

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