3.1.57 \(\int \frac {-3 x^2-6 x^3-3 x^4-2 x^3 \log (2)+(-3+6 x+13 x^2+6 x^3+4 x^2 \log (2)) \log (4)+(-3-8 x-3 x^2-2 x \log (2)) \log ^2(4)+e^x (3 x^2+2 x^3+x^4+(-6 x-4 x^2-2 x^3) \log (4)+(3+2 x+x^2) \log ^2(4))}{x^2-2 x \log (4)+\log ^2(4)} \, dx\)

Optimal. Leaf size=28 \[ \left (3+x^2\right ) \left (-4+e^x-x-\log (2)+\frac {x}{x-\log (4)}\right ) \]

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Rubi [B]  time = 1.02, antiderivative size = 267, normalized size of antiderivative = 9.54, number of steps used = 22, number of rules used = 9, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 27, 6742, 2196, 2194, 2176, 43, 698, 1850} \begin {gather*} -x^3+e^x x^2-\frac {3}{2} x^2 \log (16)+3 x^2 \log (4)-x^2 (3+\log (2))-3 x+3 e^x+\frac {3 \log ^4(4)}{x-\log (4)}-12 \log ^3(4) \log (x-\log (4))+\frac {2 (3+\log (2)) \log ^3(4)}{x-\log (4)}-12 x \log ^2(4)+2 \log (4) \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right ) \log (x-\log (4))-\log ^2(4) (8+7 \log (4)) \log (x-\log (4))-6 (3+\log (2)) \log ^2(4) \log (x-\log (4))+\frac {\log ^2(4) \left (3+4 \log ^2(4)+8 \log (4)\right )}{x-\log (4)}+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {\log (4) \left (3-6 \log ^3(4)-\log ^2(4) (13+\log (16))-6 \log (4)\right )}{x-\log (4)}+13 x \log (4) \left (1+\log \left (4\ 2^{2/13}\right )\right )-2 x (3+\log (2)) \log (16)-3 \log (16) \log (x-\log (4)) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(-3*x^2 - 6*x^3 - 3*x^4 - 2*x^3*Log[2] + (-3 + 6*x + 13*x^2 + 6*x^3 + 4*x^2*Log[2])*Log[4] + (-3 - 8*x - 3
*x^2 - 2*x*Log[2])*Log[4]^2 + E^x*(3*x^2 + 2*x^3 + x^4 + (-6*x - 4*x^2 - 2*x^3)*Log[4] + (3 + 2*x + x^2)*Log[4
]^2))/(x^2 - 2*x*Log[4] + Log[4]^2),x]

[Out]

3*E^x - 3*x + E^x*x^2 - x^3 - x^2*(3 + Log[2]) + 3*x^2*Log[4] - 12*x*Log[4]^2 + (3*Log[4]^2)/(x - Log[4]) + (2
*(3 + Log[2])*Log[4]^3)/(x - Log[4]) + (3*Log[4]^4)/(x - Log[4]) + (Log[4]^2*(3 + 8*Log[4] + 4*Log[4]^2))/(x -
Log[4]) - (3*x^2*Log[16])/2 - 2*x*(3 + Log[2])*Log[16] + (Log[4]*(3 - 6*Log[4] - 6*Log[4]^3 - Log[4]^2*(13 +
Log[16])))/(x - Log[4]) + 13*x*Log[4]*(1 + Log[4*2^(2/13)]) - 6*(3 + Log[2])*Log[4]^2*Log[x - Log[4]] - 12*Log
[4]^3*Log[x - Log[4]] - Log[4]^2*(8 + 7*Log[4])*Log[x - Log[4]] - 3*Log[16]*Log[x - Log[4]] + 2*Log[4]*(3 + 9*
Log[4]^2 + Log[4]*(13 + Log[16]))*Log[x - Log[4]]

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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

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 {-3 x^2-3 x^4+x^3 (-6-2 \log (2))+\left (-3+6 x+13 x^2+6 x^3+4 x^2 \log (2)\right ) \log (4)+\left (-3-8 x-3 x^2-2 x \log (2)\right ) \log ^2(4)+e^x \left (3 x^2+2 x^3+x^4+\left (-6 x-4 x^2-2 x^3\right ) \log (4)+\left (3+2 x+x^2\right ) \log ^2(4)\right )}{x^2-2 x \log (4)+\log ^2(4)} \, dx\\ &=\int \frac {-3 x^2-3 x^4+x^3 (-6-2 \log (2))+\left (-3+6 x+13 x^2+6 x^3+4 x^2 \log (2)\right ) \log (4)+\left (-3-8 x-3 x^2-2 x \log (2)\right ) \log ^2(4)+e^x \left (3 x^2+2 x^3+x^4+\left (-6 x-4 x^2-2 x^3\right ) \log (4)+\left (3+2 x+x^2\right ) \log ^2(4)\right )}{(x-\log (4))^2} \, dx\\ &=\int \left (e^x \left (3+2 x+x^2\right )-\frac {3 x^2}{(x-\log (4))^2}-\frac {3 x^4}{(x-\log (4))^2}-\frac {2 x^3 (3+\log (2))}{(x-\log (4))^2}+\frac {\log ^2(4) \left (-3-3 x^2-x (8+\log (4))\right )}{(x-\log (4))^2}+\frac {\log (4) \left (-3+6 x+6 x^3+x^2 (13+\log (16))\right )}{(x-\log (4))^2}\right ) \, dx\\ &=-\left (3 \int \frac {x^2}{(x-\log (4))^2} \, dx\right )-3 \int \frac {x^4}{(x-\log (4))^2} \, dx-(2 (3+\log (2))) \int \frac {x^3}{(x-\log (4))^2} \, dx+\log (4) \int \frac {-3+6 x+6 x^3+x^2 (13+\log (16))}{(x-\log (4))^2} \, dx+\log ^2(4) \int \frac {-3-3 x^2-x (8+\log (4))}{(x-\log (4))^2} \, dx+\int e^x \left (3+2 x+x^2\right ) \, dx\\ &=-\left (3 \int \left (x^2+3 \log ^2(4)+\frac {4 \log ^3(4)}{x-\log (4)}+\frac {\log ^4(4)}{(x-\log (4))^2}+x \log (16)\right ) \, dx\right )-3 \int \left (1+\frac {\log ^2(4)}{(x-\log (4))^2}+\frac {\log (16)}{x-\log (4)}\right ) \, dx-(2 (3+\log (2))) \int \left (x+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {\log ^3(4)}{(x-\log (4))^2}+\log (16)\right ) \, dx+\log (4) \int \left (6 x+\frac {2 \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right )}{x-\log (4)}+\frac {-3+6 \log (4)+6 \log ^3(4)+\log ^2(4) (13+\log (16))}{(x-\log (4))^2}+13 \left (1+\log \left (4\ 2^{2/13}\right )\right )\right ) \, dx+\log ^2(4) \int \left (-3+\frac {-8-7 \log (4)}{x-\log (4)}+\frac {-3-8 \log (4)-4 \log ^2(4)}{(x-\log (4))^2}\right ) \, dx+\int \left (3 e^x+2 e^x x+e^x x^2\right ) \, dx\\ &=-3 x-x^3-x^2 (3+\log (2))+3 x^2 \log (4)-12 x \log ^2(4)+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {2 (3+\log (2)) \log ^3(4)}{x-\log (4)}+\frac {3 \log ^4(4)}{x-\log (4)}+\frac {\log ^2(4) \left (3+8 \log (4)+4 \log ^2(4)\right )}{x-\log (4)}-\frac {3}{2} x^2 \log (16)-2 x (3+\log (2)) \log (16)+\frac {\log (4) \left (3-6 \log (4)-6 \log ^3(4)-\log ^2(4) (13+\log (16))\right )}{x-\log (4)}+13 x \log (4) \left (1+\log \left (4\ 2^{2/13}\right )\right )-6 (3+\log (2)) \log ^2(4) \log (x-\log (4))-12 \log ^3(4) \log (x-\log (4))-\log ^2(4) (8+7 \log (4)) \log (x-\log (4))-3 \log (16) \log (x-\log (4))+2 \log (4) \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right ) \log (x-\log (4))+2 \int e^x x \, dx+3 \int e^x \, dx+\int e^x x^2 \, dx\\ &=3 e^x-3 x+2 e^x x+e^x x^2-x^3-x^2 (3+\log (2))+3 x^2 \log (4)-12 x \log ^2(4)+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {2 (3+\log (2)) \log ^3(4)}{x-\log (4)}+\frac {3 \log ^4(4)}{x-\log (4)}+\frac {\log ^2(4) \left (3+8 \log (4)+4 \log ^2(4)\right )}{x-\log (4)}-\frac {3}{2} x^2 \log (16)-2 x (3+\log (2)) \log (16)+\frac {\log (4) \left (3-6 \log (4)-6 \log ^3(4)-\log ^2(4) (13+\log (16))\right )}{x-\log (4)}+13 x \log (4) \left (1+\log \left (4\ 2^{2/13}\right )\right )-6 (3+\log (2)) \log ^2(4) \log (x-\log (4))-12 \log ^3(4) \log (x-\log (4))-\log ^2(4) (8+7 \log (4)) \log (x-\log (4))-3 \log (16) \log (x-\log (4))+2 \log (4) \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right ) \log (x-\log (4))-2 \int e^x \, dx-2 \int e^x x \, dx\\ &=e^x-3 x+e^x x^2-x^3-x^2 (3+\log (2))+3 x^2 \log (4)-12 x \log ^2(4)+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {2 (3+\log (2)) \log ^3(4)}{x-\log (4)}+\frac {3 \log ^4(4)}{x-\log (4)}+\frac {\log ^2(4) \left (3+8 \log (4)+4 \log ^2(4)\right )}{x-\log (4)}-\frac {3}{2} x^2 \log (16)-2 x (3+\log (2)) \log (16)+\frac {\log (4) \left (3-6 \log (4)-6 \log ^3(4)-\log ^2(4) (13+\log (16))\right )}{x-\log (4)}+13 x \log (4) \left (1+\log \left (4\ 2^{2/13}\right )\right )-6 (3+\log (2)) \log ^2(4) \log (x-\log (4))-12 \log ^3(4) \log (x-\log (4))-\log ^2(4) (8+7 \log (4)) \log (x-\log (4))-3 \log (16) \log (x-\log (4))+2 \log (4) \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right ) \log (x-\log (4))+2 \int e^x \, dx\\ &=3 e^x-3 x+e^x x^2-x^3-x^2 (3+\log (2))+3 x^2 \log (4)-12 x \log ^2(4)+\frac {3 \log ^2(4)}{x-\log (4)}+\frac {2 (3+\log (2)) \log ^3(4)}{x-\log (4)}+\frac {3 \log ^4(4)}{x-\log (4)}+\frac {\log ^2(4) \left (3+8 \log (4)+4 \log ^2(4)\right )}{x-\log (4)}-\frac {3}{2} x^2 \log (16)-2 x (3+\log (2)) \log (16)+\frac {\log (4) \left (3-6 \log (4)-6 \log ^3(4)-\log ^2(4) (13+\log (16))\right )}{x-\log (4)}+13 x \log (4) \left (1+\log \left (4\ 2^{2/13}\right )\right )-6 (3+\log (2)) \log ^2(4) \log (x-\log (4))-12 \log ^3(4) \log (x-\log (4))-\log ^2(4) (8+7 \log (4)) \log (x-\log (4))-3 \log (16) \log (x-\log (4))+2 \log (4) \left (3+9 \log ^2(4)+\log (4) (13+\log (16))\right ) \log (x-\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.18, size = 70, normalized size = 2.50 \begin {gather*} -x^3+e^x \left (3+x^2\right )-\frac {1}{2} x^2 (6+\log (4))+x \left (-3-2 \log ^2(4)+\log (4) (1+\log (16))\right )+\frac {-2 \log ^4(4)+\log ^3(4) (1+\log (16))+\log (64)}{x-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 - 6*x^3 - 3*x^4 - 2*x^3*Log[2] + (-3 + 6*x + 13*x^2 + 6*x^3 + 4*x^2*Log[2])*Log[4] + (-3 - 8
*x - 3*x^2 - 2*x*Log[2])*Log[4]^2 + E^x*(3*x^2 + 2*x^3 + x^4 + (-6*x - 4*x^2 - 2*x^3)*Log[4] + (3 + 2*x + x^2)
*Log[4]^2))/(x^2 - 2*x*Log[4] + Log[4]^2),x]

[Out]

-x^3 + E^x*(3 + x^2) - (x^2*(6 + Log[4]))/2 + x*(-3 - 2*Log[4]^2 + Log[4]*(1 + Log[16])) + (-2*Log[4]^4 + Log[
4]^3*(1 + Log[16]) + Log[64])/(x - Log[4])

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fricas [B]  time = 0.79, size = 80, normalized size = 2.86 \begin {gather*} -\frac {x^{4} + 3 \, x^{3} - 2 \, {\left (x^{2} - 2 \, x\right )} \log \relax (2)^{2} - 8 \, \log \relax (2)^{3} + 3 \, x^{2} - {\left (x^{3} - 2 \, {\left (x^{2} + 3\right )} \log \relax (2) + 3 \, x\right )} e^{x} - {\left (x^{3} + 8 \, x^{2} + 6 \, x + 6\right )} \log \relax (2)}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(x^2+2*x+3)*log(2)^2+2*(-2*x^3-4*x^2-6*x)*log(2)+x^4+2*x^3+3*x^2)*exp(x)+4*(-2*x*log(2)-3*x^2-8*
x-3)*log(2)^2+2*(4*x^2*log(2)+6*x^3+13*x^2+6*x-3)*log(2)-2*x^3*log(2)-3*x^4-6*x^3-3*x^2)/(4*log(2)^2-4*x*log(2
)+x^2),x, algorithm="fricas")

[Out]

-(x^4 + 3*x^3 - 2*(x^2 - 2*x)*log(2)^2 - 8*log(2)^3 + 3*x^2 - (x^3 - 2*(x^2 + 3)*log(2) + 3*x)*e^x - (x^3 + 8*
x^2 + 6*x + 6)*log(2))/(x - 2*log(2))

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giac [B]  time = 0.33, size = 96, normalized size = 3.43 \begin {gather*} -\frac {x^{4} - x^{3} e^{x} - x^{3} \log \relax (2) + 2 \, x^{2} e^{x} \log \relax (2) - 2 \, x^{2} \log \relax (2)^{2} + 3 \, x^{3} - 8 \, x^{2} \log \relax (2) + 4 \, x \log \relax (2)^{2} - 8 \, \log \relax (2)^{3} + 3 \, x^{2} - 3 \, x e^{x} - 6 \, x \log \relax (2) + 6 \, e^{x} \log \relax (2) - 6 \, \log \relax (2)}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(x^2+2*x+3)*log(2)^2+2*(-2*x^3-4*x^2-6*x)*log(2)+x^4+2*x^3+3*x^2)*exp(x)+4*(-2*x*log(2)-3*x^2-8*
x-3)*log(2)^2+2*(4*x^2*log(2)+6*x^3+13*x^2+6*x-3)*log(2)-2*x^3*log(2)-3*x^4-6*x^3-3*x^2)/(4*log(2)^2-4*x*log(2
)+x^2),x, algorithm="giac")

[Out]

-(x^4 - x^3*e^x - x^3*log(2) + 2*x^2*e^x*log(2) - 2*x^2*log(2)^2 + 3*x^3 - 8*x^2*log(2) + 4*x*log(2)^2 - 8*log
(2)^3 + 3*x^2 - 3*x*e^x - 6*x*log(2) + 6*e^x*log(2) - 6*log(2))/(x - 2*log(2))

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maple [B]  time = 0.19, size = 61, normalized size = 2.18

method	result	size
risch	\(-x^{2} \ln \relax (2)-x^{3}+2 x \ln \relax (2)-3 x^{2}-3 x -\frac {4 \ln \relax (2)^{3}}{\ln \relax (2)-\frac {x}{2}}-\frac {3 \ln \relax (2)}{\ln \relax (2)-\frac {x}{2}}+\left (x^{2}+3\right ) {\mathrm e}^{x}\)	\(61\)
default	\(-3 x -x^{3}-3 x^{2}+3 \,{\mathrm e}^{x}+2 x \ln \relax (2)-x^{2} \ln \relax (2)+{\mathrm e}^{x} x^{2}+\frac {6 \ln \relax (2)}{x -2 \ln \relax (2)}+\frac {8 \ln \relax (2)^{3}}{x -2 \ln \relax (2)}\)	\(63\)
norman	\(\frac {x^{4}+\left (-\ln \relax (2)+3\right ) x^{3}+\left (-2 \ln \relax (2)^{2}-8 \ln \relax (2)+3\right ) x^{2}-3 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{3}+6 \,{\mathrm e}^{x} \ln \relax (2)+2 x^{2} \ln \relax (2) {\mathrm e}^{x}-12 \ln \relax (2)^{2}-6 \ln \relax (2)}{2 \ln \relax (2)-x}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*(x^2+2*x+3)*ln(2)^2+2*(-2*x^3-4*x^2-6*x)*ln(2)+x^4+2*x^3+3*x^2)*exp(x)+4*(-2*x*ln(2)-3*x^2-8*x-3)*ln(2
)^2+2*(4*x^2*ln(2)+6*x^3+13*x^2+6*x-3)*ln(2)-2*x^3*ln(2)-3*x^4-6*x^3-3*x^2)/(4*ln(2)^2-4*x*ln(2)+x^2),x,method
=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 8 \, {\left (\frac {2 \, \log \relax (2)}{x - 2 \, \log \relax (2)} - \log \left (x - 2 \, \log \relax (2)\right )\right )} \log \relax (2)^{3} - 96 \, \log \relax (2)^{3} \log \left (x - 2 \, \log \relax (2)\right ) - x^{3} - 6 \, x^{2} \log \relax (2) - 4 \, {\left (4 \, \log \relax (2) \log \left (x - 2 \, \log \relax (2)\right ) + x - \frac {4 \, \log \relax (2)^{2}}{x - 2 \, \log \relax (2)}\right )} \log \relax (2)^{2} - 36 \, x \log \relax (2)^{2} + 32 \, {\left (\frac {2 \, \log \relax (2)}{x - 2 \, \log \relax (2)} - \log \left (x - 2 \, \log \relax (2)\right )\right )} \log \relax (2)^{2} - 24 \, \int \frac {e^{x}}{x^{3} - 6 \, x^{2} \log \relax (2) + 12 \, x \log \relax (2)^{2} - 8 \, \log \relax (2)^{3}}\,{d x} \log \relax (2)^{2} + \frac {48 \, \log \relax (2)^{4}}{x - 2 \, \log \relax (2)} - 72 \, \log \relax (2)^{2} \log \left (x - 2 \, \log \relax (2)\right ) - 3 \, x^{2} + 5 \, {\left (24 \, \log \relax (2)^{2} \log \left (x - 2 \, \log \relax (2)\right ) + x^{2} + 8 \, x \log \relax (2) - \frac {16 \, \log \relax (2)^{3}}{x - 2 \, \log \relax (2)}\right )} \log \relax (2) + 26 \, {\left (4 \, \log \relax (2) \log \left (x - 2 \, \log \relax (2)\right ) + x - \frac {4 \, \log \relax (2)^{2}}{x - 2 \, \log \relax (2)}\right )} \log \relax (2) - 24 \, x \log \relax (2) - 12 \, {\left (\frac {2 \, \log \relax (2)}{x - 2 \, \log \relax (2)} - \log \left (x - 2 \, \log \relax (2)\right )\right )} \log \relax (2) - \frac {48 \, E_{2}\left (-x + 2 \, \log \relax (2)\right ) \log \relax (2)^{2}}{x - 2 \, \log \relax (2)} + \frac {48 \, \log \relax (2)^{3}}{x - 2 \, \log \relax (2)} - 12 \, \log \relax (2) \log \left (x - 2 \, \log \relax (2)\right ) - 3 \, x + \frac {{\left (x^{4} - 4 \, x^{3} \log \relax (2) + {\left (4 \, \log \relax (2)^{2} + 3\right )} x^{2} - 12 \, x \log \relax (2)\right )} e^{x}}{x^{2} - 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2}} + \frac {24 \, \log \relax (2)^{2}}{x - 2 \, \log \relax (2)} + \frac {6 \, \log \relax (2)}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(x^2+2*x+3)*log(2)^2+2*(-2*x^3-4*x^2-6*x)*log(2)+x^4+2*x^3+3*x^2)*exp(x)+4*(-2*x*log(2)-3*x^2-8*
x-3)*log(2)^2+2*(4*x^2*log(2)+6*x^3+13*x^2+6*x-3)*log(2)-2*x^3*log(2)-3*x^4-6*x^3-3*x^2)/(4*log(2)^2-4*x*log(2
)+x^2),x, algorithm="maxima")

[Out]

8*(2*log(2)/(x - 2*log(2)) - log(x - 2*log(2)))*log(2)^3 - 96*log(2)^3*log(x - 2*log(2)) - x^3 - 6*x^2*log(2)
- 4*(4*log(2)*log(x - 2*log(2)) + x - 4*log(2)^2/(x - 2*log(2)))*log(2)^2 - 36*x*log(2)^2 + 32*(2*log(2)/(x -
2*log(2)) - log(x - 2*log(2)))*log(2)^2 - 24*integrate(e^x/(x^3 - 6*x^2*log(2) + 12*x*log(2)^2 - 8*log(2)^3),
x)*log(2)^2 + 48*log(2)^4/(x - 2*log(2)) - 72*log(2)^2*log(x - 2*log(2)) - 3*x^2 + 5*(24*log(2)^2*log(x - 2*lo
g(2)) + x^2 + 8*x*log(2) - 16*log(2)^3/(x - 2*log(2)))*log(2) + 26*(4*log(2)*log(x - 2*log(2)) + x - 4*log(2)^
2/(x - 2*log(2)))*log(2) - 24*x*log(2) - 12*(2*log(2)/(x - 2*log(2)) - log(x - 2*log(2)))*log(2) - 48*exp_inte
gral_e(2, -x + 2*log(2))*log(2)^2/(x - 2*log(2)) + 48*log(2)^3/(x - 2*log(2)) - 12*log(2)*log(x - 2*log(2)) -
3*x + (x^4 - 4*x^3*log(2) + (4*log(2)^2 + 3)*x^2 - 12*x*log(2))*e^x/(x^2 - 4*x*log(2) + 4*log(2)^2) + 24*log(2
)^2/(x - 2*log(2)) + 6*log(2)/(x - 2*log(2))

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mupad [B]  time = 0.63, size = 67, normalized size = 2.39 \begin {gather*} x\,\left (26\,\ln \relax (2)+8\,{\ln \relax (2)}^2-4\,\ln \relax (2)\,\left (2\,\ln \relax (2)+6\right )-3\right )-x^2\,\left (\ln \relax (2)+3\right )+{\mathrm {e}}^x\,\left (x^2+3\right )+\frac {6\,\ln \relax (2)+8\,{\ln \relax (2)}^3}{x-2\,\ln \relax (2)}-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3*log(2) - exp(x)*(4*log(2)^2*(2*x + x^2 + 3) - 2*log(2)*(6*x + 4*x^2 + 2*x^3) + 3*x^2 + 2*x^3 + x^4
) - 2*log(2)*(6*x + 4*x^2*log(2) + 13*x^2 + 6*x^3 - 3) + 3*x^2 + 6*x^3 + 3*x^4 + 4*log(2)^2*(8*x + 2*x*log(2)
+ 3*x^2 + 3))/(4*log(2)^2 - 4*x*log(2) + x^2),x)

[Out]

x*(26*log(2) + 8*log(2)^2 - 4*log(2)*(2*log(2) + 6) - 3) - x^2*(log(2) + 3) + exp(x)*(x^2 + 3) + (6*log(2) + 8
*log(2)^3)/(x - 2*log(2)) - x^3

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sympy [B]  time = 0.30, size = 48, normalized size = 1.71 \begin {gather*} - x^{3} - x^{2} \left (\log {\relax (2 )} + 3\right ) - x \left (3 - 2 \log {\relax (2 )}\right ) + \left (x^{2} + 3\right ) e^{x} - \frac {- 6 \log {\relax (2 )} - 8 \log {\relax (2 )}^{3}}{x - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*(x**2+2*x+3)*ln(2)**2+2*(-2*x**3-4*x**2-6*x)*ln(2)+x**4+2*x**3+3*x**2)*exp(x)+4*(-2*x*ln(2)-3*x*
*2-8*x-3)*ln(2)**2+2*(4*x**2*ln(2)+6*x**3+13*x**2+6*x-3)*ln(2)-2*x**3*ln(2)-3*x**4-6*x**3-3*x**2)/(4*ln(2)**2-
4*x*ln(2)+x**2),x)

[Out]

-x**3 - x**2*(log(2) + 3) - x*(3 - 2*log(2)) + (x**2 + 3)*exp(x) - (-6*log(2) - 8*log(2)**3)/(x - 2*log(2))

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