3.56.72 \(\int \frac {-9 x^2+6 x^3-16 x^4+8 x^5-8 x^6+2 x^7-x^8+e^x (3 x^2-5 x^3+x^4+4 x^5-4 x^6+x^7)+(-9 x^2+6 x^3+2 x^4-4 x^5+6 x^6-2 x^7+x^8+e^x (6 x-10 x^2+2 x^3+8 x^4-8 x^5+2 x^6)) \log (\frac {-1+x^2}{x})+e^x (3-5 x+x^2+4 x^3-4 x^4+x^5) \log ^2(\frac {-1+x^2}{x})}{-36 x^4+24 x^5+8 x^6-16 x^7+24 x^8-8 x^9+4 x^{10}+(-72 x^3+48 x^4+16 x^5-32 x^6+48 x^7-16 x^8+8 x^9) \log (\frac {-1+x^2}{x})+(-36 x^2+24 x^3+8 x^4-16 x^5+24 x^6-8 x^7+4 x^8) \log ^2(\frac {-1+x^2}{x})} \, dx\)

Optimal. Leaf size=36 \[ \frac {1}{4} \left (\frac {e^x}{x \left (3-x+x^2\right )}+\frac {x}{x+\log \left (-\frac {1}{x}+x\right )}\right ) \]

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Rubi [C]  time = 7.45, antiderivative size = 523, normalized size of antiderivative = 14.53, number of steps used = 33, number of rules used = 9, integrand size = 317, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6688, 12, 6725, 6742, 2177, 2178, 6728, 6711, 32} \begin {gather*} -\frac {1}{264} \left (11+3 i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x-i \sqrt {11}-1\right )\right )+\frac {1}{264} \left (1+i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x-i \sqrt {11}-1\right )\right )+\frac {i e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x-i \sqrt {11}-1\right )\right )}{12 \sqrt {11}}+\frac {5}{132} e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x-i \sqrt {11}-1\right )\right )-\frac {1}{264} \left (11-3 i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x+i \sqrt {11}-1\right )\right )+\frac {1}{264} \left (1-i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x+i \sqrt {11}-1\right )\right )-\frac {i e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x+i \sqrt {11}-1\right )\right )}{12 \sqrt {11}}+\frac {5}{132} e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (2 x+i \sqrt {11}-1\right )\right )+\frac {\left (1-i \sqrt {11}\right ) e^x}{132 \left (-2 x-i \sqrt {11}+1\right )}+\frac {5 e^x}{66 \left (-2 x-i \sqrt {11}+1\right )}+\frac {\left (1+i \sqrt {11}\right ) e^x}{132 \left (-2 x+i \sqrt {11}+1\right )}+\frac {5 e^x}{66 \left (-2 x+i \sqrt {11}+1\right )}+\frac {e^x}{12 x}-\frac {1}{4 \left (\frac {x}{\log \left (x-\frac {1}{x}\right )}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*x^2 + 6*x^3 - 16*x^4 + 8*x^5 - 8*x^6 + 2*x^7 - x^8 + E^x*(3*x^2 - 5*x^3 + x^4 + 4*x^5 - 4*x^6 + x^7) +
(-9*x^2 + 6*x^3 + 2*x^4 - 4*x^5 + 6*x^6 - 2*x^7 + x^8 + E^x*(6*x - 10*x^2 + 2*x^3 + 8*x^4 - 8*x^5 + 2*x^6))*L
og[(-1 + x^2)/x] + E^x*(3 - 5*x + x^2 + 4*x^3 - 4*x^4 + x^5)*Log[(-1 + x^2)/x]^2)/(-36*x^4 + 24*x^5 + 8*x^6 -
16*x^7 + 24*x^8 - 8*x^9 + 4*x^10 + (-72*x^3 + 48*x^4 + 16*x^5 - 32*x^6 + 48*x^7 - 16*x^8 + 8*x^9)*Log[(-1 + x^
2)/x] + (-36*x^2 + 24*x^3 + 8*x^4 - 16*x^5 + 24*x^6 - 8*x^7 + 4*x^8)*Log[(-1 + x^2)/x]^2),x]

[Out]

(5*E^x)/(66*(1 - I*Sqrt[11] - 2*x)) + ((1 - I*Sqrt[11])*E^x)/(132*(1 - I*Sqrt[11] - 2*x)) + (5*E^x)/(66*(1 + I
*Sqrt[11] - 2*x)) + ((1 + I*Sqrt[11])*E^x)/(132*(1 + I*Sqrt[11] - 2*x)) + E^x/(12*x) + (5*E^(1/2 + (I/2)*Sqrt[
11])*ExpIntegralEi[(-1 - I*Sqrt[11] + 2*x)/2])/132 + ((I/12)*E^(1/2 + (I/2)*Sqrt[11])*ExpIntegralEi[(-1 - I*Sq
rt[11] + 2*x)/2])/Sqrt[11] + ((1 + I*Sqrt[11])*E^(1/2 + (I/2)*Sqrt[11])*ExpIntegralEi[(-1 - I*Sqrt[11] + 2*x)/
2])/264 - ((11 + (3*I)*Sqrt[11])*E^(1/2 + (I/2)*Sqrt[11])*ExpIntegralEi[(-1 - I*Sqrt[11] + 2*x)/2])/264 + (5*E
^(1/2 - (I/2)*Sqrt[11])*ExpIntegralEi[(-1 + I*Sqrt[11] + 2*x)/2])/132 - ((I/12)*E^(1/2 - (I/2)*Sqrt[11])*ExpIn
tegralEi[(-1 + I*Sqrt[11] + 2*x)/2])/Sqrt[11] + ((1 - I*Sqrt[11])*E^(1/2 - (I/2)*Sqrt[11])*ExpIntegralEi[(-1 +
I*Sqrt[11] + 2*x)/2])/264 - ((11 - (3*I)*Sqrt[11])*E^(1/2 - (I/2)*Sqrt[11])*ExpIntegralEi[(-1 + I*Sqrt[11] +
2*x)/2])/264 - 1/(4*(1 + x/Log[-x^(-1) + x]))

Rule 12

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

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 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

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 {x^2 \left (\left (1+x^2\right ) \left (3-x+x^2\right )^2-e^x \left (3-5 x+x^2+4 x^3-4 x^4+x^5\right )\right )-x \left (-1+x^2\right ) \left (x \left (3-x+x^2\right )^2+2 e^x \left (-3+5 x-4 x^2+x^3\right )\right ) \log \left (-\frac {1}{x}+x\right )-e^x \left (3-5 x+x^2+4 x^3-4 x^4+x^5\right ) \log ^2\left (-\frac {1}{x}+x\right )}{4 x^2 \left (1-x^2\right ) \left (3-x+x^2\right )^2 \left (x+\log \left (-\frac {1}{x}+x\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {x^2 \left (\left (1+x^2\right ) \left (3-x+x^2\right )^2-e^x \left (3-5 x+x^2+4 x^3-4 x^4+x^5\right )\right )-x \left (-1+x^2\right ) \left (x \left (3-x+x^2\right )^2+2 e^x \left (-3+5 x-4 x^2+x^3\right )\right ) \log \left (-\frac {1}{x}+x\right )-e^x \left (3-5 x+x^2+4 x^3-4 x^4+x^5\right ) \log ^2\left (-\frac {1}{x}+x\right )}{x^2 \left (1-x^2\right ) \left (3-x+x^2\right )^2 \left (x+\log \left (-\frac {1}{x}+x\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {e^x \left (-3+5 x-4 x^2+x^3\right )}{x^2 \left (3-x+x^2\right )^2}+\frac {-1-x^2-\log \left (-\frac {1}{x}+x\right )+x^2 \log \left (-\frac {1}{x}+x\right )}{\left (-1+x^2\right ) \left (x+\log \left (-\frac {1}{x}+x\right )\right )^2}\right ) \, dx\\ &=\frac {1}{4} \int \frac {e^x \left (-3+5 x-4 x^2+x^3\right )}{x^2 \left (3-x+x^2\right )^2} \, dx+\frac {1}{4} \int \frac {-1-x^2-\log \left (-\frac {1}{x}+x\right )+x^2 \log \left (-\frac {1}{x}+x\right )}{\left (-1+x^2\right ) \left (x+\log \left (-\frac {1}{x}+x\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {e^x}{3 x^2}+\frac {e^x}{3 x}+\frac {e^x (-5-x)}{3 \left (3-x+x^2\right )^2}+\frac {e^x (2-x)}{3 \left (3-x+x^2\right )}\right ) \, dx+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )\\ &=-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}-\frac {1}{12} \int \frac {e^x}{x^2} \, dx+\frac {1}{12} \int \frac {e^x}{x} \, dx+\frac {1}{12} \int \frac {e^x (-5-x)}{\left (3-x+x^2\right )^2} \, dx+\frac {1}{12} \int \frac {e^x (2-x)}{3-x+x^2} \, dx\\ &=\frac {e^x}{12 x}+\frac {\text {Ei}(x)}{12}-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}-\frac {1}{12} \int \frac {e^x}{x} \, dx+\frac {1}{12} \int \left (\frac {\left (-1-\frac {3 i}{\sqrt {11}}\right ) e^x}{-1-i \sqrt {11}+2 x}+\frac {\left (-1+\frac {3 i}{\sqrt {11}}\right ) e^x}{-1+i \sqrt {11}+2 x}\right ) \, dx+\frac {1}{12} \int \left (-\frac {5 e^x}{\left (3-x+x^2\right )^2}-\frac {e^x x}{\left (3-x+x^2\right )^2}\right ) \, dx\\ &=\frac {e^x}{12 x}-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}-\frac {1}{12} \int \frac {e^x x}{\left (3-x+x^2\right )^2} \, dx-\frac {5}{12} \int \frac {e^x}{\left (3-x+x^2\right )^2} \, dx+\frac {1}{132} \left (-11+3 i \sqrt {11}\right ) \int \frac {e^x}{-1+i \sqrt {11}+2 x} \, dx-\frac {1}{132} \left (11+3 i \sqrt {11}\right ) \int \frac {e^x}{-1-i \sqrt {11}+2 x} \, dx\\ &=\frac {e^x}{12 x}-\frac {1}{264} \left (11+3 i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )-\frac {1}{264} \left (11-3 i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}-\frac {1}{12} \int \left (-\frac {2 \left (1+i \sqrt {11}\right ) e^x}{11 \left (1+i \sqrt {11}-2 x\right )^2}+\frac {2 i e^x}{11 \sqrt {11} \left (1+i \sqrt {11}-2 x\right )}-\frac {2 \left (1-i \sqrt {11}\right ) e^x}{11 \left (-1+i \sqrt {11}+2 x\right )^2}+\frac {2 i e^x}{11 \sqrt {11} \left (-1+i \sqrt {11}+2 x\right )}\right ) \, dx-\frac {5}{12} \int \left (-\frac {4 e^x}{11 \left (1+i \sqrt {11}-2 x\right )^2}+\frac {4 i e^x}{11 \sqrt {11} \left (1+i \sqrt {11}-2 x\right )}-\frac {4 e^x}{11 \left (-1+i \sqrt {11}+2 x\right )^2}+\frac {4 i e^x}{11 \sqrt {11} \left (-1+i \sqrt {11}+2 x\right )}\right ) \, dx\\ &=\frac {e^x}{12 x}-\frac {1}{264} \left (11+3 i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )-\frac {1}{264} \left (11-3 i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}+\frac {5}{33} \int \frac {e^x}{\left (1+i \sqrt {11}-2 x\right )^2} \, dx+\frac {5}{33} \int \frac {e^x}{\left (-1+i \sqrt {11}+2 x\right )^2} \, dx-\frac {i \int \frac {e^x}{1+i \sqrt {11}-2 x} \, dx}{66 \sqrt {11}}-\frac {i \int \frac {e^x}{-1+i \sqrt {11}+2 x} \, dx}{66 \sqrt {11}}-\frac {(5 i) \int \frac {e^x}{1+i \sqrt {11}-2 x} \, dx}{33 \sqrt {11}}-\frac {(5 i) \int \frac {e^x}{-1+i \sqrt {11}+2 x} \, dx}{33 \sqrt {11}}-\frac {1}{66} \left (-1-i \sqrt {11}\right ) \int \frac {e^x}{\left (1+i \sqrt {11}-2 x\right )^2} \, dx-\frac {1}{66} \left (-1+i \sqrt {11}\right ) \int \frac {e^x}{\left (-1+i \sqrt {11}+2 x\right )^2} \, dx\\ &=\frac {5 e^x}{66 \left (1-i \sqrt {11}-2 x\right )}+\frac {\left (1-i \sqrt {11}\right ) e^x}{132 \left (1-i \sqrt {11}-2 x\right )}+\frac {5 e^x}{66 \left (1+i \sqrt {11}-2 x\right )}+\frac {\left (1+i \sqrt {11}\right ) e^x}{132 \left (1+i \sqrt {11}-2 x\right )}+\frac {e^x}{12 x}+\frac {i e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )}{12 \sqrt {11}}-\frac {1}{264} \left (11+3 i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )-\frac {i e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )}{12 \sqrt {11}}-\frac {1}{264} \left (11-3 i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}-\frac {5}{66} \int \frac {e^x}{1+i \sqrt {11}-2 x} \, dx+\frac {5}{66} \int \frac {e^x}{-1+i \sqrt {11}+2 x} \, dx-\frac {1}{132} \left (-1+i \sqrt {11}\right ) \int \frac {e^x}{-1+i \sqrt {11}+2 x} \, dx-\frac {1}{132} \left (1+i \sqrt {11}\right ) \int \frac {e^x}{1+i \sqrt {11}-2 x} \, dx\\ &=\frac {5 e^x}{66 \left (1-i \sqrt {11}-2 x\right )}+\frac {\left (1-i \sqrt {11}\right ) e^x}{132 \left (1-i \sqrt {11}-2 x\right )}+\frac {5 e^x}{66 \left (1+i \sqrt {11}-2 x\right )}+\frac {\left (1+i \sqrt {11}\right ) e^x}{132 \left (1+i \sqrt {11}-2 x\right )}+\frac {e^x}{12 x}+\frac {5}{132} e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )+\frac {i e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )}{12 \sqrt {11}}+\frac {1}{264} \left (1+i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )-\frac {1}{264} \left (11+3 i \sqrt {11}\right ) e^{\frac {1}{2}+\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1-i \sqrt {11}+2 x\right )\right )+\frac {5}{132} e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {i e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )}{12 \sqrt {11}}+\frac {1}{264} \left (1-i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {1}{264} \left (11-3 i \sqrt {11}\right ) e^{\frac {1}{2}-\frac {i \sqrt {11}}{2}} \text {Ei}\left (\frac {1}{2} \left (-1+i \sqrt {11}+2 x\right )\right )-\frac {1}{4 \left (1+\frac {x}{\log \left (-\frac {1}{x}+x\right )}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 37, normalized size = 1.03 \begin {gather*} \frac {1}{4} \left (\frac {e^x}{3 x-x^2+x^3}+\frac {x}{x+\log \left (-\frac {1}{x}+x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*x^2 + 6*x^3 - 16*x^4 + 8*x^5 - 8*x^6 + 2*x^7 - x^8 + E^x*(3*x^2 - 5*x^3 + x^4 + 4*x^5 - 4*x^6 +
x^7) + (-9*x^2 + 6*x^3 + 2*x^4 - 4*x^5 + 6*x^6 - 2*x^7 + x^8 + E^x*(6*x - 10*x^2 + 2*x^3 + 8*x^4 - 8*x^5 + 2*x
^6))*Log[(-1 + x^2)/x] + E^x*(3 - 5*x + x^2 + 4*x^3 - 4*x^4 + x^5)*Log[(-1 + x^2)/x]^2)/(-36*x^4 + 24*x^5 + 8*
x^6 - 16*x^7 + 24*x^8 - 8*x^9 + 4*x^10 + (-72*x^3 + 48*x^4 + 16*x^5 - 32*x^6 + 48*x^7 - 16*x^8 + 8*x^9)*Log[(-
1 + x^2)/x] + (-36*x^2 + 24*x^3 + 8*x^4 - 16*x^5 + 24*x^6 - 8*x^7 + 4*x^8)*Log[(-1 + x^2)/x]^2),x]

[Out]

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

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fricas [B]  time = 0.77, size = 72, normalized size = 2.00 \begin {gather*} \frac {x^{4} - x^{3} + 3 \, x^{2} + x e^{x} + e^{x} \log \left (\frac {x^{2} - 1}{x}\right )}{4 \, {\left (x^{4} - x^{3} + 3 \, x^{2} + {\left (x^{3} - x^{2} + 3 \, x\right )} \log \left (\frac {x^{2} - 1}{x}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-4*x^4+4*x^3+x^2-5*x+3)*exp(x)*log((x^2-1)/x)^2+((2*x^6-8*x^5+8*x^4+2*x^3-10*x^2+6*x)*exp(x)+x^
8-2*x^7+6*x^6-4*x^5+2*x^4+6*x^3-9*x^2)*log((x^2-1)/x)+(x^7-4*x^6+4*x^5+x^4-5*x^3+3*x^2)*exp(x)-x^8+2*x^7-8*x^6
+8*x^5-16*x^4+6*x^3-9*x^2)/((4*x^8-8*x^7+24*x^6-16*x^5+8*x^4+24*x^3-36*x^2)*log((x^2-1)/x)^2+(8*x^9-16*x^8+48*
x^7-32*x^6+16*x^5+48*x^4-72*x^3)*log((x^2-1)/x)+4*x^10-8*x^9+24*x^8-16*x^7+8*x^6+24*x^5-36*x^4),x, algorithm="
fricas")

[Out]

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

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giac [B]  time = 2.80, size = 91, normalized size = 2.53 \begin {gather*} \frac {x^{4} - x^{3} + 3 \, x^{2} + x e^{x} + e^{x} \log \left (\frac {x^{2} - 1}{x}\right )}{4 \, {\left (x^{4} + x^{3} \log \left (\frac {x^{2} - 1}{x}\right ) - x^{3} - x^{2} \log \left (\frac {x^{2} - 1}{x}\right ) + 3 \, x^{2} + 3 \, x \log \left (\frac {x^{2} - 1}{x}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-4*x^4+4*x^3+x^2-5*x+3)*exp(x)*log((x^2-1)/x)^2+((2*x^6-8*x^5+8*x^4+2*x^3-10*x^2+6*x)*exp(x)+x^
8-2*x^7+6*x^6-4*x^5+2*x^4+6*x^3-9*x^2)*log((x^2-1)/x)+(x^7-4*x^6+4*x^5+x^4-5*x^3+3*x^2)*exp(x)-x^8+2*x^7-8*x^6
+8*x^5-16*x^4+6*x^3-9*x^2)/((4*x^8-8*x^7+24*x^6-16*x^5+8*x^4+24*x^3-36*x^2)*log((x^2-1)/x)^2+(8*x^9-16*x^8+48*
x^7-32*x^6+16*x^5+48*x^4-72*x^3)*log((x^2-1)/x)+4*x^10-8*x^9+24*x^8-16*x^7+8*x^6+24*x^5-36*x^4),x, algorithm="
giac")

[Out]

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

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maple [C]  time = 0.20, size = 142, normalized size = 3.94

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{4 x \left (x^{2}-x +3\right )}+\frac {x}{-2 i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x^{2}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )+2 i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )^{2}+2 i \pi \,\mathrm {csgn}\left (i \left (x^{2}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )^{2}-2 i \pi \mathrm {csgn}\left (\frac {i \left (x^{2}-1\right )}{x}\right )^{3}+4 x -4 \ln \relax (x )+4 \ln \left (x^{2}-1\right )}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5-4*x^4+4*x^3+x^2-5*x+3)*exp(x)*ln((x^2-1)/x)^2+((2*x^6-8*x^5+8*x^4+2*x^3-10*x^2+6*x)*exp(x)+x^8-2*x^7
+6*x^6-4*x^5+2*x^4+6*x^3-9*x^2)*ln((x^2-1)/x)+(x^7-4*x^6+4*x^5+x^4-5*x^3+3*x^2)*exp(x)-x^8+2*x^7-8*x^6+8*x^5-1
6*x^4+6*x^3-9*x^2)/((4*x^8-8*x^7+24*x^6-16*x^5+8*x^4+24*x^3-36*x^2)*ln((x^2-1)/x)^2+(8*x^9-16*x^8+48*x^7-32*x^
6+16*x^5+48*x^4-72*x^3)*ln((x^2-1)/x)+4*x^10-8*x^9+24*x^8-16*x^7+8*x^6+24*x^5-36*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.45, size = 105, normalized size = 2.92 \begin {gather*} \frac {x^{4} - x^{3} + 3 \, x^{2} + {\left (x - \log \relax (x)\right )} e^{x} + e^{x} \log \left (x + 1\right ) + e^{x} \log \left (x - 1\right )}{4 \, {\left (x^{4} - x^{3} + 3 \, x^{2} + {\left (x^{3} - x^{2} + 3 \, x\right )} \log \left (x + 1\right ) + {\left (x^{3} - x^{2} + 3 \, x\right )} \log \left (x - 1\right ) - {\left (x^{3} - x^{2} + 3 \, x\right )} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5-4*x^4+4*x^3+x^2-5*x+3)*exp(x)*log((x^2-1)/x)^2+((2*x^6-8*x^5+8*x^4+2*x^3-10*x^2+6*x)*exp(x)+x^
8-2*x^7+6*x^6-4*x^5+2*x^4+6*x^3-9*x^2)*log((x^2-1)/x)+(x^7-4*x^6+4*x^5+x^4-5*x^3+3*x^2)*exp(x)-x^8+2*x^7-8*x^6
+8*x^5-16*x^4+6*x^3-9*x^2)/((4*x^8-8*x^7+24*x^6-16*x^5+8*x^4+24*x^3-36*x^2)*log((x^2-1)/x)^2+(8*x^9-16*x^8+48*
x^7-32*x^6+16*x^5+48*x^4-72*x^3)*log((x^2-1)/x)+4*x^10-8*x^9+24*x^8-16*x^7+8*x^6+24*x^5-36*x^4),x, algorithm="
maxima")

[Out]

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

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mupad [B]  time = 4.09, size = 89, normalized size = 2.47 \begin {gather*} \frac {x^2\,\ln \left (\frac {x^2-1}{x}\right )-x^3\,\ln \left (\frac {x^2-1}{x}\right )+{\mathrm {e}}^x\,\ln \left (\frac {x^2-1}{x}\right )-3\,x\,\ln \left (\frac {x^2-1}{x}\right )+x\,{\mathrm {e}}^x}{4\,x\,\left (x+\ln \left (\frac {x^2-1}{x}\right )\right )\,\left (x^2-x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(3*x^2 - 5*x^3 + x^4 + 4*x^5 - 4*x^6 + x^7) + log((x^2 - 1)/x)*(exp(x)*(6*x - 10*x^2 + 2*x^3 + 8*x
^4 - 8*x^5 + 2*x^6) - 9*x^2 + 6*x^3 + 2*x^4 - 4*x^5 + 6*x^6 - 2*x^7 + x^8) - 9*x^2 + 6*x^3 - 16*x^4 + 8*x^5 -
8*x^6 + 2*x^7 - x^8 + exp(x)*log((x^2 - 1)/x)^2*(x^2 - 5*x + 4*x^3 - 4*x^4 + x^5 + 3))/(log((x^2 - 1)/x)*(48*x
^4 - 72*x^3 + 16*x^5 - 32*x^6 + 48*x^7 - 16*x^8 + 8*x^9) + log((x^2 - 1)/x)^2*(24*x^3 - 36*x^2 + 8*x^4 - 16*x^
5 + 24*x^6 - 8*x^7 + 4*x^8) - 36*x^4 + 24*x^5 + 8*x^6 - 16*x^7 + 24*x^8 - 8*x^9 + 4*x^10),x)

[Out]

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

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sympy [A]  time = 0.61, size = 31, normalized size = 0.86 \begin {gather*} \frac {x}{4 x + 4 \log {\left (\frac {x^{2} - 1}{x} \right )}} + \frac {e^{x}}{4 x^{3} - 4 x^{2} + 12 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**5-4*x**4+4*x**3+x**2-5*x+3)*exp(x)*ln((x**2-1)/x)**2+((2*x**6-8*x**5+8*x**4+2*x**3-10*x**2+6*x)
*exp(x)+x**8-2*x**7+6*x**6-4*x**5+2*x**4+6*x**3-9*x**2)*ln((x**2-1)/x)+(x**7-4*x**6+4*x**5+x**4-5*x**3+3*x**2)
*exp(x)-x**8+2*x**7-8*x**6+8*x**5-16*x**4+6*x**3-9*x**2)/((4*x**8-8*x**7+24*x**6-16*x**5+8*x**4+24*x**3-36*x**
2)*ln((x**2-1)/x)**2+(8*x**9-16*x**8+48*x**7-32*x**6+16*x**5+48*x**4-72*x**3)*ln((x**2-1)/x)+4*x**10-8*x**9+24
*x**8-16*x**7+8*x**6+24*x**5-36*x**4),x)

[Out]

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

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