3.1.92 \(\int \frac {-75 x^3-75 x^4+(300 e^5 x^3-300 x^4-300 x^3 \log (x)) \log (-e^5+x+\log (x))}{-9 e^5+9 x+9 \log (x)+(-30 e^5 x^4+30 x^5+30 x^4 \log (x)) \log (-e^5+x+\log (x))+(-25 e^5 x^8+25 x^9+25 x^8 \log (x)) \log ^2(-e^5+x+\log (x))} \, dx\)

Optimal. Leaf size=25 \[ 3 e+\frac {15}{3+5 x^4 \log \left (-e^5+x+\log (x)\right )} \]

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Rubi [A]  time = 0.27, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 3, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 6686} \begin {gather*} \frac {15}{5 x^4 \log \left (x+\log (x)-e^5\right )+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-75*x^3 - 75*x^4 + (300*E^5*x^3 - 300*x^4 - 300*x^3*Log[x])*Log[-E^5 + x + Log[x]])/(-9*E^5 + 9*x + 9*Log
[x] + (-30*E^5*x^4 + 30*x^5 + 30*x^4*Log[x])*Log[-E^5 + x + Log[x]] + (-25*E^5*x^8 + 25*x^9 + 25*x^8*Log[x])*L
og[-E^5 + x + Log[x]]^2),x]

[Out]

15/(3 + 5*x^4*Log[-E^5 + x + 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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {75 x^3 \left (1+x+4 \left (-e^5+x+\log (x)\right ) \log \left (-e^5+x+\log (x)\right )\right )}{\left (e^5-x-\log (x)\right ) \left (3+5 x^4 \log \left (-e^5+x+\log (x)\right )\right )^2} \, dx\\ &=75 \int \frac {x^3 \left (1+x+4 \left (-e^5+x+\log (x)\right ) \log \left (-e^5+x+\log (x)\right )\right )}{\left (e^5-x-\log (x)\right ) \left (3+5 x^4 \log \left (-e^5+x+\log (x)\right )\right )^2} \, dx\\ &=\frac {15}{3+5 x^4 \log \left (-e^5+x+\log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.84 \begin {gather*} \frac {75}{15+25 x^4 \log \left (-e^5+x+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-75*x^3 - 75*x^4 + (300*E^5*x^3 - 300*x^4 - 300*x^3*Log[x])*Log[-E^5 + x + Log[x]])/(-9*E^5 + 9*x +
9*Log[x] + (-30*E^5*x^4 + 30*x^5 + 30*x^4*Log[x])*Log[-E^5 + x + Log[x]] + (-25*E^5*x^8 + 25*x^9 + 25*x^8*Log
[x])*Log[-E^5 + x + Log[x]]^2),x]

[Out]

75/(15 + 25*x^4*Log[-E^5 + x + Log[x]])

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fricas [A]  time = 0.82, size = 20, normalized size = 0.80 \begin {gather*} \frac {15}{5 \, x^{4} \log \left (x - e^{5} + \log \relax (x)\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3*log(x)+300*x^3*exp(5)-300*x^4)*log(log(x)-exp(5)+x)-75*x^4-75*x^3)/((25*x^8*log(x)-25*x^8
*exp(5)+25*x^9)*log(log(x)-exp(5)+x)^2+(30*x^4*log(x)-30*x^4*exp(5)+30*x^5)*log(log(x)-exp(5)+x)+9*log(x)-9*ex
p(5)+9*x),x, algorithm="fricas")

[Out]

15/(5*x^4*log(x - e^5 + log(x)) + 3)

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giac [A]  time = 0.89, size = 20, normalized size = 0.80 \begin {gather*} \frac {15}{5 \, x^{4} \log \left (x - e^{5} + \log \relax (x)\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3*log(x)+300*x^3*exp(5)-300*x^4)*log(log(x)-exp(5)+x)-75*x^4-75*x^3)/((25*x^8*log(x)-25*x^8
*exp(5)+25*x^9)*log(log(x)-exp(5)+x)^2+(30*x^4*log(x)-30*x^4*exp(5)+30*x^5)*log(log(x)-exp(5)+x)+9*log(x)-9*ex
p(5)+9*x),x, algorithm="giac")

[Out]

15/(5*x^4*log(x - e^5 + log(x)) + 3)

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maple [A]  time = 0.04, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {15}{5 \ln \left (\ln \relax (x )-{\mathrm e}^{5}+x \right ) x^{4}+3}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-300*x^3*ln(x)+300*x^3*exp(5)-300*x^4)*ln(ln(x)-exp(5)+x)-75*x^4-75*x^3)/((25*x^8*ln(x)-25*x^8*exp(5)+25
*x^9)*ln(ln(x)-exp(5)+x)^2+(30*x^4*ln(x)-30*x^4*exp(5)+30*x^5)*ln(ln(x)-exp(5)+x)+9*ln(x)-9*exp(5)+9*x),x,meth
od=_RETURNVERBOSE)

[Out]

15/(5*ln(ln(x)-exp(5)+x)*x^4+3)

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maxima [A]  time = 0.90, size = 20, normalized size = 0.80 \begin {gather*} \frac {15}{5 \, x^{4} \log \left (x - e^{5} + \log \relax (x)\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x^3*log(x)+300*x^3*exp(5)-300*x^4)*log(log(x)-exp(5)+x)-75*x^4-75*x^3)/((25*x^8*log(x)-25*x^8
*exp(5)+25*x^9)*log(log(x)-exp(5)+x)^2+(30*x^4*log(x)-30*x^4*exp(5)+30*x^5)*log(log(x)-exp(5)+x)+9*log(x)-9*ex
p(5)+9*x),x, algorithm="maxima")

[Out]

15/(5*x^4*log(x - e^5 + log(x)) + 3)

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mupad [B]  time = 0.82, size = 20, normalized size = 0.80 \begin {gather*} \frac {15}{5\,x^4\,\ln \left (x-{\mathrm {e}}^5+\ln \relax (x)\right )+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x - exp(5) + log(x))*(300*x^3*log(x) - 300*x^3*exp(5) + 300*x^4) + 75*x^3 + 75*x^4)/(9*x - 9*exp(5)
+ 9*log(x) + log(x - exp(5) + log(x))*(30*x^4*log(x) - 30*x^4*exp(5) + 30*x^5) + log(x - exp(5) + log(x))^2*(2
5*x^8*log(x) - 25*x^8*exp(5) + 25*x^9)),x)

[Out]

15/(5*x^4*log(x - exp(5) + log(x)) + 3)

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sympy [A]  time = 0.36, size = 17, normalized size = 0.68 \begin {gather*} \frac {15}{5 x^{4} \log {\left (x + \log {\relax (x )} - e^{5} \right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-300*x**3*ln(x)+300*x**3*exp(5)-300*x**4)*ln(ln(x)-exp(5)+x)-75*x**4-75*x**3)/((25*x**8*ln(x)-25*x
**8*exp(5)+25*x**9)*ln(ln(x)-exp(5)+x)**2+(30*x**4*ln(x)-30*x**4*exp(5)+30*x**5)*ln(ln(x)-exp(5)+x)+9*ln(x)-9*
exp(5)+9*x),x)

[Out]

15/(5*x**4*log(x + log(x) - exp(5)) + 3)

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