3.54.16 \(\int \frac {78125+78125 x^3+e^3 (-15625-15625 x^3)+(-234375 x^3+46875 e^3 x^3) \log (\frac {x}{2})}{x+2 x^4+x^7} \, dx\) [5316]

Optimal. Leaf size=23 \[ \frac {15625 \left (5-e^3\right ) x \log \left (\frac {x}{2}\right )}{x+x^4} \]

[Out]

15625*ln(1/2*x)/(x^4+x)*x*(-exp(3)+5)

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Rubi [A]
time = 0.38, antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 12, number of rules used = 10, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1608, 28, 6820, 12, 6874, 272, 46, 267, 2373, 266} \begin {gather*} 15625 \left (5-e^3\right ) \log (x)-\frac {15625 \left (5-e^3\right ) x^3 \log \left (\frac {x}{2}\right )}{x^3+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(78125 + 78125*x^3 + E^3*(-15625 - 15625*x^3) + (-234375*x^3 + 46875*E^3*x^3)*Log[x/2])/(x + 2*x^4 + x^7),
x]

[Out]

(-15625*(5 - E^3)*x^3*Log[x/2])/(1 + x^3) + 15625*(5 - E^3)*Log[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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1608

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

Rule 2373

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

Rule 6820

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

Rule 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {78125+78125 x^3+e^3 \left (-15625-15625 x^3\right )+\left (-234375 x^3+46875 e^3 x^3\right ) \log \left (\frac {x}{2}\right )}{x \left (1+2 x^3+x^6\right )} \, dx\\ &=\int \frac {78125+78125 x^3+e^3 \left (-15625-15625 x^3\right )+\left (-234375 x^3+46875 e^3 x^3\right ) \log \left (\frac {x}{2}\right )}{x \left (1+x^3\right )^2} \, dx\\ &=\int \frac {15625 \left (5-e^3\right ) \left (1+x^3-3 x^3 \log \left (\frac {x}{2}\right )\right )}{x \left (1+x^3\right )^2} \, dx\\ &=\left (15625 \left (5-e^3\right )\right ) \int \frac {1+x^3-3 x^3 \log \left (\frac {x}{2}\right )}{x \left (1+x^3\right )^2} \, dx\\ &=\left (15625 \left (5-e^3\right )\right ) \int \left (\frac {1}{x \left (1+x^3\right )^2}+\frac {x^2}{\left (1+x^3\right )^2}-\frac {3 x^2 \log \left (\frac {x}{2}\right )}{\left (1+x^3\right )^2}\right ) \, dx\\ &=\left (15625 \left (5-e^3\right )\right ) \int \frac {1}{x \left (1+x^3\right )^2} \, dx+\left (15625 \left (5-e^3\right )\right ) \int \frac {x^2}{\left (1+x^3\right )^2} \, dx-\left (46875 \left (5-e^3\right )\right ) \int \frac {x^2 \log \left (\frac {x}{2}\right )}{\left (1+x^3\right )^2} \, dx\\ &=-\frac {15625 \left (5-e^3\right )}{3 \left (1+x^3\right )}-\frac {15625 \left (5-e^3\right ) x^3 \log \left (\frac {x}{2}\right )}{1+x^3}+\frac {1}{3} \left (15625 \left (5-e^3\right )\right ) \text {Subst}\left (\int \frac {1}{x (1+x)^2} \, dx,x,x^3\right )+\left (15625 \left (5-e^3\right )\right ) \int \frac {x^2}{1+x^3} \, dx\\ &=-\frac {15625 \left (5-e^3\right )}{3 \left (1+x^3\right )}-\frac {15625 \left (5-e^3\right ) x^3 \log \left (\frac {x}{2}\right )}{1+x^3}+\frac {15625}{3} \left (5-e^3\right ) \log \left (1+x^3\right )+\frac {1}{3} \left (15625 \left (5-e^3\right )\right ) \text {Subst}\left (\int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^2}\right ) \, dx,x,x^3\right )\\ &=-\frac {15625 \left (5-e^3\right ) x^3 \log \left (\frac {x}{2}\right )}{1+x^3}+15625 \left (5-e^3\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.05, size = 20, normalized size = 0.87 \begin {gather*} -\frac {15625 \left (-5+e^3\right ) \log \left (\frac {x}{2}\right )}{1+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(78125 + 78125*x^3 + E^3*(-15625 - 15625*x^3) + (-234375*x^3 + 46875*E^3*x^3)*Log[x/2])/(x + 2*x^4 +
x^7),x]

[Out]

(-15625*(-5 + E^3)*Log[x/2])/(1 + x^3)

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Maple [C] Result contains complex when optimal does not.
time = 0.43, size = 798, normalized size = 34.70 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((46875*x^3*exp(3)-234375*x^3)*ln(1/2*x)+(-15625*x^3-15625)*exp(3)+78125*x^3+78125)/(x^7+2*x^4+x),x,method
=_RETURNVERBOSE)

[Out]

-15625/3*exp(3)*ln(1/2*x)/(x^2-x+1)*x+15625/9*I*exp(3)*3^(1/2)*ln(1/2*x)*ln((I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))+1
5625/9*I*exp(3)*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))*x+15625/3*exp(3)*ln(1/2*x)*x/(
x+1)+31250/3*exp(3)*ln(1/2*x)/(x^2-x+1)*x^2-15625/9*I*exp(3)*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/
(1+I*3^(1/2)))+78125/9*I*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))*x^2-78125/9*I*ln(1/2*
x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))*x+78125/9*I*3^(1/2)*ln(1/2*x)*ln((I*3^(1/2)+2*x-1)/(I
*3^(1/2)-1))-15625/9*I*exp(3)*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))*x+78125/9*I*ln(1
/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))*x-156250/3*ln(1/2*x)/(x^2-x+1)*x^2+78125/3*ln(1/2*
x)/(x^2-x+1)*x+15625/9*I*exp(3)*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))*x^2-15625/9*I*
exp(3)*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))*x^2+15625/9*I*exp(3)*ln(1/2*x)/(x^2-x+1
)*3^(1/2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))+78125/9*I*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)-2*x+1)/(1+I*
3^(1/2)))+78125*ln(1/2*x)-15625*exp(3)*ln(1/2*x)-78125/3*ln(1/2*x)*x/(x+1)-78125/9*I*ln(1/2*x)/(x^2-x+1)*3^(1/
2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))-78125/9*I*ln(1/2*x)/(x^2-x+1)*3^(1/2)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1
))*x^2-15625/9*I*exp(3)*3^(1/2)*ln(1/2*x)*ln((I*3^(1/2)+2*x-1)/(I*3^(1/2)-1))-78125/9*I*3^(1/2)*ln(1/2*x)*ln((
I*3^(1/2)-2*x+1)/(1+I*3^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (18) = 36\).
time = 0.48, size = 123, normalized size = 5.35 \begin {gather*} -\frac {15625}{3} \, {\left (\frac {3 \, \log \left (\frac {1}{2} \, x\right )}{x^{3} + 1} + \log \left (x^{3} + 1\right ) - \log \left (x^{3}\right )\right )} e^{3} - \frac {15625}{3} \, {\left (\frac {1}{x^{3} + 1} - \log \left (x^{2} - x + 1\right ) - \log \left (x + 1\right ) + 3 \, \log \left (x\right )\right )} e^{3} + \frac {15625 \, e^{3}}{3 \, {\left (x^{3} + 1\right )}} + \frac {78125 \, \log \left (\frac {1}{2} \, x\right )}{x^{3} + 1} + \frac {78125}{3} \, \log \left (x^{3} + 1\right ) - \frac {78125}{3} \, \log \left (x^{3}\right ) - \frac {78125}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {78125}{3} \, \log \left (x + 1\right ) + 78125 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((46875*x^3*exp(3)-234375*x^3)*log(1/2*x)+(-15625*x^3-15625)*exp(3)+78125*x^3+78125)/(x^7+2*x^4+x),x
, algorithm="maxima")

[Out]

-15625/3*(3*log(1/2*x)/(x^3 + 1) + log(x^3 + 1) - log(x^3))*e^3 - 15625/3*(1/(x^3 + 1) - log(x^2 - x + 1) - lo
g(x + 1) + 3*log(x))*e^3 + 15625/3*e^3/(x^3 + 1) + 78125*log(1/2*x)/(x^3 + 1) + 78125/3*log(x^3 + 1) - 78125/3
*log(x^3) - 78125/3*log(x^2 - x + 1) - 78125/3*log(x + 1) + 78125*log(x)

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Fricas [A]
time = 0.37, size = 17, normalized size = 0.74 \begin {gather*} -\frac {15625 \, {\left (e^{3} - 5\right )} \log \left (\frac {1}{2} \, x\right )}{x^{3} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((46875*x^3*exp(3)-234375*x^3)*log(1/2*x)+(-15625*x^3-15625)*exp(3)+78125*x^3+78125)/(x^7+2*x^4+x),x
, algorithm="fricas")

[Out]

-15625*(e^3 - 5)*log(1/2*x)/(x^3 + 1)

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Sympy [A]
time = 0.07, size = 15, normalized size = 0.65 \begin {gather*} \frac {\left (78125 - 15625 e^{3}\right ) \log {\left (\frac {x}{2} \right )}}{x^{3} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((46875*x**3*exp(3)-234375*x**3)*ln(1/2*x)+(-15625*x**3-15625)*exp(3)+78125*x**3+78125)/(x**7+2*x**4
+x),x)

[Out]

(78125 - 15625*exp(3))*log(x/2)/(x**3 + 1)

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Giac [A]
time = 0.39, size = 23, normalized size = 1.00 \begin {gather*} -\frac {15625 \, {\left (e^{3} \log \left (\frac {1}{2} \, x\right ) - 5 \, \log \left (\frac {1}{2} \, x\right )\right )}}{x^{3} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((46875*x^3*exp(3)-234375*x^3)*log(1/2*x)+(-15625*x^3-15625)*exp(3)+78125*x^3+78125)/(x^7+2*x^4+x),x
, algorithm="giac")

[Out]

-15625*(e^3*log(1/2*x) - 5*log(1/2*x))/(x^3 + 1)

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Mupad [B]
time = 3.68, size = 24, normalized size = 1.04 \begin {gather*} -\frac {x^2\,\ln \left (\frac {x}{2}\right )\,\left (15625\,{\mathrm {e}}^3-78125\right )}{x^5+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/2)*(46875*x^3*exp(3) - 234375*x^3) - exp(3)*(15625*x^3 + 15625) + 78125*x^3 + 78125)/(x + 2*x^4 + x
^7),x)

[Out]

-(x^2*log(x/2)*(15625*exp(3) - 78125))/(x^2 + x^5)

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