3.7.77 \(\int \frac {300 x^2-440 x^3-20 x^4+160 x^5+e^3 (-25 x-85 x^2+35 x^3+40 x^4)+(100-120 x-60 x^2+80 x^3+e^3 (-25+5 x+20 x^2)) \log (\frac {4-e^3-4 x}{-1+x})}{4+e^3 (-1+x)-8 x+4 x^2} \, dx\)

Optimal. Leaf size=26 \[ 5 x (5+2 x) \left (x^2+\log \left (-4+\frac {e^3}{1-x}\right )\right ) \]

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Rubi [B]  time = 1.92, antiderivative size = 101, normalized size of antiderivative = 3.88, number of steps used = 9, number of rules used = 6, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6688, 12, 6742, 1612, 2492, 72} \begin {gather*} 10 x^4+25 x^3+\frac {5}{8} \left (56-18 e^3+e^6\right ) \log \left (-4 x-e^3+4\right )-\frac {5}{8} \left (9-e^3\right )^2 \log \left (-4 x-e^3+4\right )+\frac {5}{8} (4 x+5)^2 \log \left (-\frac {-4 x-e^3+4}{1-x}\right )+\frac {125}{8} \log (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(300*x^2 - 440*x^3 - 20*x^4 + 160*x^5 + E^3*(-25*x - 85*x^2 + 35*x^3 + 40*x^4) + (100 - 120*x - 60*x^2 + 8
0*x^3 + E^3*(-25 + 5*x + 20*x^2))*Log[(4 - E^3 - 4*x)/(-1 + x)])/(4 + E^3*(-1 + x) - 8*x + 4*x^2),x]

[Out]

25*x^3 + 10*x^4 - (5*(9 - E^3)^2*Log[4 - E^3 - 4*x])/8 + (5*(56 - 18*E^3 + E^6)*Log[4 - E^3 - 4*x])/8 + (5*(5
+ 4*x)^2*Log[-((4 - E^3 - 4*x)/(1 - x))])/8 + (125*Log[1 - x])/8

Rule 12

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rule 2492

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*((g_.) + (h_.)*(x_))^
(m_.), x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(h*(m + 1)), x] - Dist[(p*
r*s*(b*c - a*d))/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*
(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0]
&& IGtQ[s, 0] && NeQ[m, -1]

Rule 6688

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

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 {5 \left (-\frac {x \left (4 (-1+x)^2 x (15+8 x)+e^3 \left (-5-17 x+7 x^2+8 x^3\right )\right )}{-4+e^3+4 x}-\left (-5+x+4 x^2\right ) \log \left (-\frac {-4+e^3+4 x}{-1+x}\right )\right )}{1-x} \, dx\\ &=5 \int \frac {-\frac {x \left (4 (-1+x)^2 x (15+8 x)+e^3 \left (-5-17 x+7 x^2+8 x^3\right )\right )}{-4+e^3+4 x}-\left (-5+x+4 x^2\right ) \log \left (-\frac {-4+e^3+4 x}{-1+x}\right )}{1-x} \, dx\\ &=5 \int \left (\frac {x \left (-5 e^3+\left (60-17 e^3\right ) x-\left (88-7 e^3\right ) x^2-4 \left (1-2 e^3\right ) x^3+32 x^4\right )}{\left (4-e^3-4 x\right ) (1-x)}+(5+4 x) \log \left (-\frac {-4+e^3+4 x}{-1+x}\right )\right ) \, dx\\ &=5 \int \frac {x \left (-5 e^3+\left (60-17 e^3\right ) x-\left (88-7 e^3\right ) x^2-4 \left (1-2 e^3\right ) x^3+32 x^4\right )}{\left (4-e^3-4 x\right ) (1-x)} \, dx+5 \int (5+4 x) \log \left (-\frac {-4+e^3+4 x}{-1+x}\right ) \, dx\\ &=\frac {5}{8} (5+4 x)^2 \log \left (-\frac {4-e^3-4 x}{1-x}\right )+5 \int \left (-\frac {e^3}{2}-\frac {7}{-1+x}+15 x^2+8 x^3+\frac {56-18 e^3+e^6}{2 \left (-4+e^3+4 x\right )}\right ) \, dx+\frac {1}{8} \left (5 e^3\right ) \int \frac {(5+4 x)^2}{(-1+x) \left (-4+e^3+4 x\right )} \, dx\\ &=-\frac {5 e^3 x}{2}+25 x^3+10 x^4+\frac {5}{8} \left (56-18 e^3+e^6\right ) \log \left (4-e^3-4 x\right )+\frac {5}{8} (5+4 x)^2 \log \left (-\frac {4-e^3-4 x}{1-x}\right )-35 \log (1-x)+\frac {1}{8} \left (5 e^3\right ) \int \left (4+\frac {81}{e^3 (-1+x)}-\frac {4 \left (-9+e^3\right )^2}{e^3 \left (-4+e^3+4 x\right )}\right ) \, dx\\ &=25 x^3+10 x^4-\frac {5}{8} \left (9-e^3\right )^2 \log \left (4-e^3-4 x\right )+\frac {5}{8} \left (56-18 e^3+e^6\right ) \log \left (4-e^3-4 x\right )+\frac {5}{8} (5+4 x)^2 \log \left (-\frac {4-e^3-4 x}{1-x}\right )+\frac {125}{8} \log (1-x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.13, size = 162, normalized size = 6.23 \begin {gather*} 5 \left (5 x^3+2 x^4+5 \log \left (4-e^3-4 x\right )-\frac {5}{4} e^3 \log \left (4-e^3-4 x\right )-5 \log \left (-\frac {4-e^3-4 x}{1-x}\right )+\frac {5}{4} e^3 \log \left (-\frac {4-e^3-4 x}{1-x}\right )+5 x \log \left (-\frac {4-e^3-4 x}{1-x}\right )+2 x^2 \log \left (-\frac {4-e^3-4 x}{1-x}\right )-5 \log (1-x)+\frac {5}{4} e^3 \log (1-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(300*x^2 - 440*x^3 - 20*x^4 + 160*x^5 + E^3*(-25*x - 85*x^2 + 35*x^3 + 40*x^4) + (100 - 120*x - 60*x
^2 + 80*x^3 + E^3*(-25 + 5*x + 20*x^2))*Log[(4 - E^3 - 4*x)/(-1 + x)])/(4 + E^3*(-1 + x) - 8*x + 4*x^2),x]

[Out]

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

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fricas [A]  time = 0.92, size = 37, normalized size = 1.42 \begin {gather*} 10 \, x^{4} + 25 \, x^{3} + 5 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2+5*x-25)*exp(3)+80*x^3-60*x^2-120*x+100)*log((-exp(3)-4*x+4)/(x-1))+(40*x^4+35*x^3-85*x^2-2
5*x)*exp(3)+160*x^5-20*x^4-440*x^3+300*x^2)/((x-1)*exp(3)+4*x^2-8*x+4),x, algorithm="fricas")

[Out]

10*x^4 + 25*x^3 + 5*(2*x^2 + 5*x)*log(-(4*x + e^3 - 4)/(x - 1))

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giac [B]  time = 0.77, size = 546, normalized size = 21.00 \begin {gather*} \frac {5 \, {\left ({\left (e^{3} - 4\right )} e^{\left (-6\right )} + 4 \, e^{\left (-6\right )}\right )} {\left (\frac {2 \, {\left (4 \, x + e^{3} - 4\right )}^{2} e^{9} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{2}} - \frac {16 \, {\left (4 \, x + e^{3} - 4\right )} e^{9} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{x - 1} + 32 \, e^{9} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right ) + \frac {9 \, {\left (4 \, x + e^{3} - 4\right )}^{3} e^{6} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{3}} - \frac {108 \, {\left (4 \, x + e^{3} - 4\right )}^{2} e^{6} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{2}} + \frac {432 \, {\left (4 \, x + e^{3} - 4\right )} e^{6} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{x - 1} - 576 \, e^{6} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right ) + \frac {7 \, {\left (4 \, x + e^{3} - 4\right )}^{4} e^{3} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{4}} - \frac {112 \, {\left (4 \, x + e^{3} - 4\right )}^{3} e^{3} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{3}} + \frac {672 \, {\left (4 \, x + e^{3} - 4\right )}^{2} e^{3} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{{\left (x - 1\right )}^{2}} - \frac {1792 \, {\left (4 \, x + e^{3} - 4\right )} e^{3} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right )}{x - 1} + 1792 \, e^{3} \log \left (-\frac {4 \, x + e^{3} - 4}{x - 1}\right ) + \frac {13 \, {\left (4 \, x + e^{3} - 4\right )} e^{12}}{x - 1} + \frac {27 \, {\left (4 \, x + e^{3} - 4\right )}^{2} e^{9}}{{\left (x - 1\right )}^{2}} - \frac {216 \, {\left (4 \, x + e^{3} - 4\right )} e^{9}}{x - 1} + \frac {23 \, {\left (4 \, x + e^{3} - 4\right )}^{3} e^{6}}{{\left (x - 1\right )}^{3}} - \frac {276 \, {\left (4 \, x + e^{3} - 4\right )}^{2} e^{6}}{{\left (x - 1\right )}^{2}} + \frac {1104 \, {\left (4 \, x + e^{3} - 4\right )} e^{6}}{x - 1} + 2 \, e^{15} - 52 \, e^{12} + 432 \, e^{9} - 1472 \, e^{6}\right )}}{\frac {{\left (4 \, x + e^{3} - 4\right )}^{4}}{{\left (x - 1\right )}^{4}} - \frac {16 \, {\left (4 \, x + e^{3} - 4\right )}^{3}}{{\left (x - 1\right )}^{3}} + \frac {96 \, {\left (4 \, x + e^{3} - 4\right )}^{2}}{{\left (x - 1\right )}^{2}} - \frac {256 \, {\left (4 \, x + e^{3} - 4\right )}}{x - 1} + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2+5*x-25)*exp(3)+80*x^3-60*x^2-120*x+100)*log((-exp(3)-4*x+4)/(x-1))+(40*x^4+35*x^3-85*x^2-2
5*x)*exp(3)+160*x^5-20*x^4-440*x^3+300*x^2)/((x-1)*exp(3)+4*x^2-8*x+4),x, algorithm="giac")

[Out]

5*((e^3 - 4)*e^(-6) + 4*e^(-6))*(2*(4*x + e^3 - 4)^2*e^9*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1)^2 - 16*(4*x + e
^3 - 4)*e^9*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1) + 32*e^9*log(-(4*x + e^3 - 4)/(x - 1)) + 9*(4*x + e^3 - 4)^3
*e^6*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1)^3 - 108*(4*x + e^3 - 4)^2*e^6*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1)
^2 + 432*(4*x + e^3 - 4)*e^6*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1) - 576*e^6*log(-(4*x + e^3 - 4)/(x - 1)) + 7
*(4*x + e^3 - 4)^4*e^3*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1)^4 - 112*(4*x + e^3 - 4)^3*e^3*log(-(4*x + e^3 - 4
)/(x - 1))/(x - 1)^3 + 672*(4*x + e^3 - 4)^2*e^3*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1)^2 - 1792*(4*x + e^3 - 4
)*e^3*log(-(4*x + e^3 - 4)/(x - 1))/(x - 1) + 1792*e^3*log(-(4*x + e^3 - 4)/(x - 1)) + 13*(4*x + e^3 - 4)*e^12
/(x - 1) + 27*(4*x + e^3 - 4)^2*e^9/(x - 1)^2 - 216*(4*x + e^3 - 4)*e^9/(x - 1) + 23*(4*x + e^3 - 4)^3*e^6/(x
- 1)^3 - 276*(4*x + e^3 - 4)^2*e^6/(x - 1)^2 + 1104*(4*x + e^3 - 4)*e^6/(x - 1) + 2*e^15 - 52*e^12 + 432*e^9 -
1472*e^6)/((4*x + e^3 - 4)^4/(x - 1)^4 - 16*(4*x + e^3 - 4)^3/(x - 1)^3 + 96*(4*x + e^3 - 4)^2/(x - 1)^2 - 25
6*(4*x + e^3 - 4)/(x - 1) + 256)

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maple [A]  time = 0.35, size = 38, normalized size = 1.46

method	result	size
risch	\(\left (10 x^{2}+25 x \right ) \ln \left (\frac {-{\mathrm e}^{3}-4 x +4}{x -1}\right )+10 x^{4}+25 x^{3}\)	\(38\)
norman	\(25 x^{3}+10 x^{4}+25 x \ln \left (\frac {-{\mathrm e}^{3}-4 x +4}{x -1}\right )+10 x^{2} \ln \left (\frac {-{\mathrm e}^{3}-4 x +4}{x -1}\right )\)	\(52\)
derivativedivides	\({\mathrm e}^{3} \left (\frac {5 \,{\mathrm e}^{3} \ln \left (-\frac {{\mathrm e}^{3}}{x -1}\right )}{8}+\frac {5 x}{2}-\frac {5}{2}-\frac {5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )^{2} \left (x -1\right )^{2}}{8}-5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (x -1\right )^{2}-\frac {45 \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (x -1\right ) {\mathrm e}^{-3}}{4}+\frac {5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) {\mathrm e}^{6}}{8}-\frac {45 \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )}{4}+35 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )-\frac {5 \,{\mathrm e}^{-3} \ln \left (-\frac {{\mathrm e}^{3}}{x -1}\right ) {\mathrm e}^{6}}{8}-\frac {5 \,{\mathrm e}^{-6} \left (x -1\right ) {\mathrm e}^{6}}{2}+115 \left (x -1\right ) {\mathrm e}^{-3}+135 \,{\mathrm e}^{-9} {\mathrm e}^{6} \left (x -1\right )^{2}+65 \,{\mathrm e}^{-12} {\mathrm e}^{9} \left (x -1\right )^{3}+10 \,{\mathrm e}^{-15} {\mathrm e}^{12} \left (x -1\right )^{4}\right )\)	\(254\)
default	\({\mathrm e}^{3} \left (\frac {5 \,{\mathrm e}^{3} \ln \left (-\frac {{\mathrm e}^{3}}{x -1}\right )}{8}+\frac {5 x}{2}-\frac {5}{2}-\frac {5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )^{2} \left (x -1\right )^{2}}{8}-5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (x -1\right )^{2}-\frac {45 \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) \left (x -1\right ) {\mathrm e}^{-3}}{4}+\frac {5 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right ) {\mathrm e}^{6}}{8}-\frac {45 \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )}{4}+35 \,{\mathrm e}^{-3} \ln \left (-4-\frac {{\mathrm e}^{3}}{x -1}\right )-\frac {5 \,{\mathrm e}^{-3} \ln \left (-\frac {{\mathrm e}^{3}}{x -1}\right ) {\mathrm e}^{6}}{8}-\frac {5 \,{\mathrm e}^{-6} \left (x -1\right ) {\mathrm e}^{6}}{2}+115 \left (x -1\right ) {\mathrm e}^{-3}+135 \,{\mathrm e}^{-9} {\mathrm e}^{6} \left (x -1\right )^{2}+65 \,{\mathrm e}^{-12} {\mathrm e}^{9} \left (x -1\right )^{3}+10 \,{\mathrm e}^{-15} {\mathrm e}^{12} \left (x -1\right )^{4}\right )\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20*x^2+5*x-25)*exp(3)+80*x^3-60*x^2-120*x+100)*ln((-exp(3)-4*x+4)/(x-1))+(40*x^4+35*x^3-85*x^2-25*x)*ex
p(3)+160*x^5-20*x^4-440*x^3+300*x^2)/((x-1)*exp(3)+4*x^2-8*x+4),x,method=_RETURNVERBOSE)

[Out]

(10*x^2+25*x)*ln((-exp(3)-4*x+4)/(x-1))+10*x^4+25*x^3

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maxima [B]  time = 0.57, size = 641, normalized size = 24.65 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^2+5*x-25)*exp(3)+80*x^3-60*x^2-120*x+100)*log((-exp(3)-4*x+4)/(x-1))+(40*x^4+35*x^3-85*x^2-2
5*x)*exp(3)+160*x^5-20*x^4-440*x^3+300*x^2)/((x-1)*exp(3)+4*x^2-8*x+4),x, algorithm="maxima")

[Out]

10*x^4 - 10/3*x^3*(e^3 - 8) - 5/3*x^3 + 5/4*x^2*(e^6 - 12*e^3 + 48) + 5/8*x^2*(e^3 - 8) + 5/32*(e^15 - 20*e^12
+ 160*e^9 - 640*e^6 + 1280*e^3 - 1024)*e^(-3)*log(4*x + e^3 - 4) + 5/64*(e^12 - 16*e^9 + 96*e^6 - 256*e^3 + 2
56)*e^(-3)*log(4*x + e^3 - 4) - 55/8*(e^9 - 12*e^6 + 48*e^3 - 64)*e^(-3)*log(4*x + e^3 - 4) - 75/4*(e^6 - 8*e^
3 + 16)*e^(-3)*log(4*x + e^3 - 4) + 25*(e^(-3)*log(4*x + e^3 - 4) - e^(-3)*log(x - 1))*e^3*log(-4*x/(x - 1) -
e^3/(x - 1) + 4/(x - 1)) - 55*x^2 - 5/8*x*(e^9 - 16*e^6 + 96*e^3 - 256) - 5/16*x*(e^6 - 12*e^3 + 48) + 55/2*x*
(e^3 - 8) + 5/96*(64*x^3 - 24*x^2*(e^3 - 8) - 3*(e^12 - 16*e^9 + 96*e^6 - 256*e^3 + 256)*e^(-3)*log(4*x + e^3
- 4) + 12*x*(e^6 - 12*e^3 + 48) + 768*e^(-3)*log(x - 1))*e^3 + 35/64*((e^9 - 12*e^6 + 48*e^3 - 64)*e^(-3)*log(
4*x + e^3 - 4) + 8*x^2 - 4*x*(e^3 - 8) + 64*e^(-3)*log(x - 1))*e^3 + 85/16*((e^6 - 8*e^3 + 16)*e^(-3)*log(4*x
+ e^3 - 4) - 16*e^(-3)*log(x - 1) - 4*x)*e^3 - 25/4*((e^3 - 4)*e^(-3)*log(4*x + e^3 - 4) + 4*e^(-3)*log(x - 1)
)*e^3 - 5/8*(20*(e^3 - 4)*log(x - 1)^2 + 20*(e^3 - 4)*log(-4*x - e^3 + 4)^2 - 4*x*e^6 + 8*(2*x^2*e^3 + 5*x*e^3
- 7*e^3)*log(x - 1) - (16*x^2*e^3 + 40*x*e^3 + 40*(e^3 - 4)*log(x - 1) - e^9 + 18*e^6 - 56*e^3)*log(-4*x - e^
3 + 4))*e^(-3) + 50*(log(4*x + e^3 - 4)^2 - 2*log(4*x + e^3 - 4)*log(x - 1) + log(x - 1)^2)*e^(-3) - 25/2*log(
4*x + e^3 - 4)^2 + 25*log(4*x + e^3 - 4)*log(x - 1) - 25/2*log(x - 1)^2 - 100*(e^(-3)*log(4*x + e^3 - 4) - e^(
-3)*log(x - 1))*log(-4*x/(x - 1) - e^3/(x - 1) + 4/(x - 1)) + 75*x

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mupad [B]  time = 1.31, size = 27, normalized size = 1.04 \begin {gather*} 5\,x\,\left (2\,x+5\right )\,\left (\ln \left (-\frac {4\,x+{\mathrm {e}}^3-4}{x-1}\right )+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*(25*x + 85*x^2 - 35*x^3 - 40*x^4) - log(-(4*x + exp(3) - 4)/(x - 1))*(exp(3)*(5*x + 20*x^2 - 25)
- 120*x - 60*x^2 + 80*x^3 + 100) - 300*x^2 + 440*x^3 + 20*x^4 - 160*x^5)/(exp(3)*(x - 1) - 8*x + 4*x^2 + 4),x)

[Out]

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

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sympy [A]  time = 0.23, size = 31, normalized size = 1.19 \begin {gather*} 10 x^{4} + 25 x^{3} + \left (10 x^{2} + 25 x\right ) \log {\left (\frac {- 4 x - e^{3} + 4}{x - 1} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x**2+5*x-25)*exp(3)+80*x**3-60*x**2-120*x+100)*ln((-exp(3)-4*x+4)/(x-1))+(40*x**4+35*x**3-85*x
**2-25*x)*exp(3)+160*x**5-20*x**4-440*x**3+300*x**2)/((x-1)*exp(3)+4*x**2-8*x+4),x)

[Out]

10*x**4 + 25*x**3 + (10*x**2 + 25*x)*log((-4*x - exp(3) + 4)/(x - 1))

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