3.45.46 \(\int \frac {-1-11 x-5 e^5 x+5 x^2+5 x \log (\frac {3 x}{2})}{-10 x-5 e^5 x+5 x^2+5 x \log (\frac {3 x}{2})} \, dx\)

Optimal. Leaf size=22 \[ 1+x-\frac {1}{5} \log \left (-2-e^5+x+\log \left (\frac {3 x}{2}\right )\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 6741, 12, 6742, 6684} \begin {gather*} x-\frac {1}{5} \log \left (-x-\log \left (\frac {3 x}{2}\right )+e^5+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 11*x - 5*E^5*x + 5*x^2 + 5*x*Log[(3*x)/2])/(-10*x - 5*E^5*x + 5*x^2 + 5*x*Log[(3*x)/2]),x]

[Out]

x - Log[2 + E^5 - x - Log[(3*x)/2]]/5

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 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

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 {-1+\left (-11-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )}{-10 x-5 e^5 x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \frac {-1+\left (-11-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )}{\left (-10-5 e^5\right ) x+5 x^2+5 x \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \frac {1-\left (-11-5 e^5\right ) x-5 x^2-5 x \log \left (\frac {3 x}{2}\right )}{5 x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {1-\left (-11-5 e^5\right ) x-5 x^2-5 x \log \left (\frac {3 x}{2}\right )}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=\frac {1}{5} \int \left (5+\frac {1+x}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )}\right ) \, dx\\ &=x+\frac {1}{5} \int \frac {1+x}{x \left (2 \left (1+\frac {e^5}{2}\right )-x-\log \left (\frac {3 x}{2}\right )\right )} \, dx\\ &=x-\frac {1}{5} \log \left (2+e^5-x-\log \left (\frac {3 x}{2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{5} \left (5 x-\log \left (2+e^5-x-\log \left (\frac {3 x}{2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 11*x - 5*E^5*x + 5*x^2 + 5*x*Log[(3*x)/2])/(-10*x - 5*E^5*x + 5*x^2 + 5*x*Log[(3*x)/2]),x]

[Out]

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

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fricas [A]  time = 0.55, size = 16, normalized size = 0.73 \begin {gather*} x - \frac {1}{5} \, \log \left (x - e^{5} + \log \left (\frac {3}{2} \, x\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(3/2*x)-5*x*exp(5)+5*x^2-11*x-1)/(5*x*log(3/2*x)-5*x*exp(5)+5*x^2-10*x),x, algorithm="fricas
")

[Out]

x - 1/5*log(x - e^5 + log(3/2*x) - 2)

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giac [A]  time = 0.16, size = 18, normalized size = 0.82 \begin {gather*} x - \frac {1}{5} \, \log \left (-x + e^{5} - \log \left (\frac {3}{2} \, x\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(3/2*x)-5*x*exp(5)+5*x^2-11*x-1)/(5*x*log(3/2*x)-5*x*exp(5)+5*x^2-10*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 17, normalized size = 0.77

method	result	size
risch	\(x -\frac {\ln \left (x -{\mathrm e}^{5}+\ln \left (\frac {3 x}{2}\right )-2\right )}{5}\)	\(17\)
norman	\(x -\frac {\ln \left ({\mathrm e}^{5}-\ln \left (\frac {3 x}{2}\right )-x +2\right )}{5}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*ln(3/2*x)-5*x*exp(5)+5*x^2-11*x-1)/(5*x*ln(3/2*x)-5*x*exp(5)+5*x^2-10*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 20, normalized size = 0.91 \begin {gather*} x - \frac {1}{5} \, \log \left (x - e^{5} + \log \relax (3) - \log \relax (2) + \log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(3/2*x)-5*x*exp(5)+5*x^2-11*x-1)/(5*x*log(3/2*x)-5*x*exp(5)+5*x^2-10*x),x, algorithm="maxima
")

[Out]

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

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mupad [B]  time = 3.33, size = 16, normalized size = 0.73 \begin {gather*} x-\frac {\ln \left (x+\ln \left (\frac {3\,x}{2}\right )-{\mathrm {e}}^5-2\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((11*x - 5*x*log((3*x)/2) + 5*x*exp(5) - 5*x^2 + 1)/(10*x - 5*x*log((3*x)/2) + 5*x*exp(5) - 5*x^2),x)

[Out]

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

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sympy [A]  time = 0.14, size = 17, normalized size = 0.77 \begin {gather*} x - \frac {\log {\left (x + \log {\left (\frac {3 x}{2} \right )} - e^{5} - 2 \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*ln(3/2*x)-5*x*exp(5)+5*x**2-11*x-1)/(5*x*ln(3/2*x)-5*x*exp(5)+5*x**2-10*x),x)

[Out]

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

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