3.42.98 \(\int \frac {e^{e^3-x} (5+x-e x-x \log (2)) (10+21 x+8 x^2+x^3+e (-x-3 x^2-x^3)+(-x-3 x^2-x^3) \log (2))}{(x^2+x^3) (-5 x-6 x^2-x^3+e (x^2+x^3)+(x^2+x^3) \log (2))} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{e^3-x} \left (1-e+\frac {5}{x}-\log (2)\right )}{x+x^2} \]

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Rubi [C]  time = 1.46, antiderivative size = 116, normalized size of antiderivative = 3.74, number of steps used = 21, number of rules used = 10, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6, 1586, 3, 1593, 1584, 21, 6741, 6742, 2177, 2178} \begin {gather*} -5 e^{e^3} \text {Ei}(-x)+e^{e^3} (4+e+\log (2)) \text {Ei}(-x)+e^{e^3} (1-e-\log (2)) \text {Ei}(-x)+\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}+\frac {e^{e^3-x} (4+e+\log (2))}{x+1}+\frac {e^{e^3-x} (1-e-\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^3 - x)*(5 + x - E*x - x*Log[2])*(10 + 21*x + 8*x^2 + x^3 + E*(-x - 3*x^2 - x^3) + (-x - 3*x^2 - x^3)
*Log[2]))/((x^2 + x^3)*(-5*x - 6*x^2 - x^3 + E*(x^2 + x^3) + (x^2 + x^3)*Log[2])),x]

[Out]

(5*E^(E^3 - x))/x^2 - (5*E^(E^3 - x))/x - 5*E^E^3*ExpIntegralEi[-x] + (E^(E^3 - x)*(1 - E - Log[2]))/x + E^E^3
*ExpIntegralEi[-x]*(1 - E - Log[2]) + (E^(E^3 - x)*(4 + E + Log[2]))/(1 + x) + E^E^3*ExpIntegralEi[-x]*(4 + E
+ Log[2])

Rule 3

Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /;
FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1584

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^3-x} (5+(1-e) x-x \log (2)) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+e \left (-x-3 x^2-x^3\right )+\left (-x-3 x^2-x^3\right ) \log (2)\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+e \left (x^2+x^3\right )+\left (x^2+x^3\right ) \log (2)\right )} \, dx\\ &=\int \frac {e^{e^3-x} (5+x (1-e-\log (2))) \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-5 x-6 x^2-x^3+\left (x^2+x^3\right ) (e+\log (2))\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (-\frac {25}{(1-e-\log (2))^3}+\frac {25 e}{(1-e-\log (2))^3}+\frac {30}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {5}{1-e-\log (2)}+\frac {25 \log (2)}{(1-e-\log (2))^3}-\frac {5 \log (2)}{(1-e-\log (2))^2}+x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{\left (x^2+x^3\right ) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^2 (1+x) \left (x^2 \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )+x \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^3 (1+x) \left (\frac {5}{(1-e-\log (2))^2}-\frac {5 e}{(1-e-\log (2))^2}-\frac {6}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}-\frac {5 \log (2)}{(1-e-\log (2))^2}+\frac {\log (2)}{1-e-\log (2)}+x \left (-\frac {1}{1-e-\log (2)}+\frac {e}{1-e-\log (2)}+\frac {\log (2)}{1-e-\log (2)}\right )\right )} \, dx\\ &=-\int \frac {e^{e^3-x} \left (10+21 x+8 x^2+x^3+\left (-x-3 x^2-x^3\right ) (e+\log (2))\right )}{x^3 (1+x)^2} \, dx\\ &=-\int \frac {e^{e^3-x} \left (10+x^3 (1-e-\log (2))+x (21-e-\log (2))+x^2 (8-3 e-\log (8))\right )}{x^3 (1+x)^2} \, dx\\ &=-\int \left (\frac {10 e^{e^3-x}}{x^3}+\frac {e^{e^3-x} (-4-e-\log (2))}{x}+\frac {e^{e^3-x} (1-e-\log (2))}{x^2}+\frac {e^{e^3-x} (4+e+\log (2))}{(1+x)^2}+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}\right ) \, dx\\ &=-\left (10 \int \frac {e^{e^3-x}}{x^3} \, dx\right )-(-4-e-\log (2)) \int \frac {e^{e^3-x}}{x} \, dx-(1-e-\log (2)) \int \frac {e^{e^3-x}}{x^2} \, dx-(4+e+\log (2)) \int \frac {e^{e^3-x}}{(1+x)^2} \, dx-(4+e+\log (2)) \int \frac {e^{e^3-x}}{1+x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}+\frac {e^{e^3-x} (1-e-\log (2))}{x}+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}-e^{1+e^3} \text {Ei}(-1-x) (4+e+\log (2))+e^{e^3} \text {Ei}(-x) (4+e+\log (2))+5 \int \frac {e^{e^3-x}}{x^2} \, dx-(-4-e-\log (2)) \int \frac {e^{e^3-x}}{1+x} \, dx-(-1+e+\log (2)) \int \frac {e^{e^3-x}}{x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}+\frac {e^{e^3-x} (1-e-\log (2))}{x}+e^{e^3} \text {Ei}(-x) (1-e-\log (2))+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}+e^{e^3} \text {Ei}(-x) (4+e+\log (2))-5 \int \frac {e^{e^3-x}}{x} \, dx\\ &=\frac {5 e^{e^3-x}}{x^2}-\frac {5 e^{e^3-x}}{x}-5 e^{e^3} \text {Ei}(-x)+\frac {e^{e^3-x} (1-e-\log (2))}{x}+e^{e^3} \text {Ei}(-x) (1-e-\log (2))+\frac {e^{e^3-x} (4+e+\log (2))}{1+x}+e^{e^3} \text {Ei}(-x) (4+e+\log (2))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 28, normalized size = 0.90 \begin {gather*} \frac {e^{e^3-x} (5-x (-1+e+\log (2)))}{x^2 (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^3 - x)*(5 + x - E*x - x*Log[2])*(10 + 21*x + 8*x^2 + x^3 + E*(-x - 3*x^2 - x^3) + (-x - 3*x^2
- x^3)*Log[2]))/((x^2 + x^3)*(-5*x - 6*x^2 - x^3 + E*(x^2 + x^3) + (x^2 + x^3)*Log[2])),x]

[Out]

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

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fricas [A]  time = 1.04, size = 32, normalized size = 1.03 \begin {gather*} e^{\left (-x + e^{3} + \log \left (-\frac {x e + x \log \relax (2) - x - 5}{x^{3} + x^{2}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(log((-x*log(2)-x*exp(1)+5+x)/(x^
3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.29, size = 277, normalized size = 8.94 \begin {gather*} -\frac {x e^{\left (-x + e^{3}\right )} \log \relax (2)^{3} + 3 \, x e^{\left (-x + e^{3} + 1\right )} \log \relax (2)^{2} + 7 \, x e^{\left (-x + e^{3}\right )} \log \relax (2)^{2} + 3 \, x e^{\left (-x + e^{3} + 2\right )} \log \relax (2) + 14 \, x e^{\left (-x + e^{3} + 1\right )} \log \relax (2) + 8 \, x e^{\left (-x + e^{3}\right )} \log \relax (2) - 5 \, e^{\left (-x + e^{3}\right )} \log \relax (2)^{2} + x e^{\left (-x + e^{3} + 3\right )} + 7 \, x e^{\left (-x + e^{3} + 2\right )} + 8 \, x e^{\left (-x + e^{3} + 1\right )} - 16 \, x e^{\left (-x + e^{3}\right )} - 10 \, e^{\left (-x + e^{3} + 1\right )} \log \relax (2) - 40 \, e^{\left (-x + e^{3}\right )} \log \relax (2) - 5 \, e^{\left (-x + e^{3} + 2\right )} - 40 \, e^{\left (-x + e^{3} + 1\right )} - 80 \, e^{\left (-x + e^{3}\right )}}{2 \, x^{3} e \log \relax (2) + x^{3} \log \relax (2)^{2} + x^{3} e^{2} + 8 \, x^{3} e + 8 \, x^{3} \log \relax (2) + 2 \, x^{2} e \log \relax (2) + x^{2} \log \relax (2)^{2} + 16 \, x^{3} + x^{2} e^{2} + 8 \, x^{2} e + 8 \, x^{2} \log \relax (2) + 16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(log((-x*log(2)-x*exp(1)+5+x)/(x^
3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="giac")

[Out]

-(x*e^(-x + e^3)*log(2)^3 + 3*x*e^(-x + e^3 + 1)*log(2)^2 + 7*x*e^(-x + e^3)*log(2)^2 + 3*x*e^(-x + e^3 + 2)*l
og(2) + 14*x*e^(-x + e^3 + 1)*log(2) + 8*x*e^(-x + e^3)*log(2) - 5*e^(-x + e^3)*log(2)^2 + x*e^(-x + e^3 + 3)
+ 7*x*e^(-x + e^3 + 2) + 8*x*e^(-x + e^3 + 1) - 16*x*e^(-x + e^3) - 10*e^(-x + e^3 + 1)*log(2) - 40*e^(-x + e^
3)*log(2) - 5*e^(-x + e^3 + 2) - 40*e^(-x + e^3 + 1) - 80*e^(-x + e^3))/(2*x^3*e*log(2) + x^3*log(2)^2 + x^3*e
^2 + 8*x^3*e + 8*x^3*log(2) + 2*x^2*e*log(2) + x^2*log(2)^2 + 16*x^3 + x^2*e^2 + 8*x^2*e + 8*x^2*log(2) + 16*x
^2)

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maple [A]  time = 0.22, size = 31, normalized size = 1.00

method	result	size
risch	\(\frac {\left (-x \ln \relax (2)-x \,{\mathrm e}+5+x \right ) {\mathrm e}^{-x +{\mathrm e}^{3}}}{x^{3}+x^{2}}\)	\(31\)
gosper	\({\mathrm e}^{\ln \left (-\frac {x \,{\mathrm e}+x \ln \relax (2)-x -5}{\left (x +1\right ) x^{2}}\right )-x +{\mathrm e}^{3}}\)	\(32\)
norman	\({\mathrm e}^{\ln \left (\frac {-x \ln \relax (2)-x \,{\mathrm e}+5+x}{x^{3}+x^{2}}\right )-x +{\mathrm e}^{3}}\)	\(32\)
default	error in convert/parfrac: cannot convert to partial fraction form\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-3*x^2-x)*ln(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(ln((-x*ln(2)-x*exp(1)+5+x)/(x^3+x^2))-x
+exp(3))/((x^3+x^2)*ln(2)+(x^3+x^2)*exp(1)-x^3-6*x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.53, size = 37, normalized size = 1.19 \begin {gather*} -\frac {{\left ({\left ({\left (\log \relax (2) - 1\right )} e^{\left (e^{3}\right )} + e^{\left (e^{3} + 1\right )}\right )} x - 5 \, e^{\left (e^{3}\right )}\right )} e^{\left (-x\right )}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-3*x^2-x)*log(2)+(-x^3-3*x^2-x)*exp(1)+x^3+8*x^2+21*x+10)*exp(log((-x*log(2)-x*exp(1)+5+x)/(x^
3+x^2))-x+exp(3))/((x^3+x^2)*log(2)+(x^3+x^2)*exp(1)-x^3-6*x^2-5*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.96, size = 78, normalized size = 2.52 \begin {gather*} \frac {5\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x+1}}{x^3+x^2}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3-x}\,\ln \relax (2)}{x^3+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(3) - x + log((x - x*exp(1) - x*log(2) + 5)/(x^2 + x^3)))*(21*x - exp(1)*(x + 3*x^2 + x^3) - log(
2)*(x + 3*x^2 + x^3) + 8*x^2 + x^3 + 10))/(5*x - exp(1)*(x^2 + x^3) + 6*x^2 + x^3 - log(2)*(x^2 + x^3)),x)

[Out]

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

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sympy [A]  time = 0.27, size = 26, normalized size = 0.84 \begin {gather*} \frac {\left (- e x - x \log {\relax (2 )} + x + 5\right ) e^{- x + e^{3}}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-3*x**2-x)*ln(2)+(-x**3-3*x**2-x)*exp(1)+x**3+8*x**2+21*x+10)*exp(ln((-x*ln(2)-x*exp(1)+5+x)/
(x**3+x**2))-x+exp(3))/((x**3+x**2)*ln(2)+(x**3+x**2)*exp(1)-x**3-6*x**2-5*x),x)

[Out]

(-E*x - x*log(2) + x + 5)*exp(-x + exp(3))/(x**3 + x**2)

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