3.48.19 \(\int \frac {-50+185 x-60 x^2-100 x^3}{3 x+14 x^2-7 x^3-10 x^4+e^4 (15-5 x-10 x^2)+(3-x-2 x^2) \log (-1+x)+(-15+5 x+10 x^2) \log (3+2 x)} \, dx\)

Optimal. Leaf size=28 \[ 5 \log \left (e^4+x^2+\frac {1}{5} (x+\log (-1+x))-\log (3+2 x)\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6688, 12, 6684} \begin {gather*} 5 \log \left (5 x^2+x+\log (x-1)-5 \log (2 x+3)+5 e^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 + 185*x - 60*x^2 - 100*x^3)/(3*x + 14*x^2 - 7*x^3 - 10*x^4 + E^4*(15 - 5*x - 10*x^2) + (3 - x - 2*x^2
)*Log[-1 + x] + (-15 + 5*x + 10*x^2)*Log[3 + 2*x]),x]

[Out]

5*Log[5*E^4 + x + 5*x^2 + Log[-1 + x] - 5*Log[3 + 2*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 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 {5 \left (-10+37 x-12 x^2-20 x^3\right )}{\left (3-x-2 x^2\right ) \left (5 e^4+x+5 x^2+\log (-1+x)-5 \log (3+2 x)\right )} \, dx\\ &=5 \int \frac {-10+37 x-12 x^2-20 x^3}{\left (3-x-2 x^2\right ) \left (5 e^4+x+5 x^2+\log (-1+x)-5 \log (3+2 x)\right )} \, dx\\ &=5 \log \left (5 e^4+x+5 x^2+\log (-1+x)-5 \log (3+2 x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.69, size = 27, normalized size = 0.96 \begin {gather*} 5 \log \left (5 e^4+x+5 x^2+\log (-1+x)-5 \log (3+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 + 185*x - 60*x^2 - 100*x^3)/(3*x + 14*x^2 - 7*x^3 - 10*x^4 + E^4*(15 - 5*x - 10*x^2) + (3 - x -
2*x^2)*Log[-1 + x] + (-15 + 5*x + 10*x^2)*Log[3 + 2*x]),x]

[Out]

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

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fricas [A]  time = 0.49, size = 30, normalized size = 1.07 \begin {gather*} 5 \, \log \left (-5 \, x^{2} - x - 5 \, e^{4} + 5 \, \log \left (2 \, x + 3\right ) - \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*x^3-60*x^2+185*x-50)/((10*x^2+5*x-15)*log(2*x+3)+(-2*x^2-x+3)*log(x-1)+(-10*x^2-5*x+15)*exp(4)
-10*x^4-7*x^3+14*x^2+3*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 26, normalized size = 0.93 \begin {gather*} 5 \, \log \left (5 \, x^{2} + x + 5 \, e^{4} - 5 \, \log \left (2 \, x + 3\right ) + \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*x^3-60*x^2+185*x-50)/((10*x^2+5*x-15)*log(2*x+3)+(-2*x^2-x+3)*log(x-1)+(-10*x^2-5*x+15)*exp(4)
-10*x^4-7*x^3+14*x^2+3*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 29, normalized size = 1.04

method	result	size
risch	\(5 \ln \left (\ln \left (2 x +3\right )-x^{2}-{\mathrm e}^{4}-\frac {x}{5}-\frac {\ln \left (x -1\right )}{5}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-100*x^3-60*x^2+185*x-50)/((10*x^2+5*x-15)*ln(2*x+3)+(-2*x^2-x+3)*ln(x-1)+(-10*x^2-5*x+15)*exp(4)-10*x^4-
7*x^3+14*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 28, normalized size = 1.00 \begin {gather*} 5 \, \log \left (-x^{2} - \frac {1}{5} \, x - e^{4} + \log \left (2 \, x + 3\right ) - \frac {1}{5} \, \log \left (x - 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*x^3-60*x^2+185*x-50)/((10*x^2+5*x-15)*log(2*x+3)+(-2*x^2-x+3)*log(x-1)+(-10*x^2-5*x+15)*exp(4)
-10*x^4-7*x^3+14*x^2+3*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.92, size = 26, normalized size = 0.93 \begin {gather*} 5\,\ln \left (\frac {x}{5}+\frac {\ln \left (x-1\right )}{5}+{\mathrm {e}}^4-\ln \left (2\,x+3\right )+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x^2 - 185*x + 100*x^3 + 50)/(exp(4)*(5*x + 10*x^2 - 15) - 3*x - log(2*x + 3)*(5*x + 10*x^2 - 15) + log
(x - 1)*(x + 2*x^2 - 3) - 14*x^2 + 7*x^3 + 10*x^4),x)

[Out]

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

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sympy [A]  time = 0.52, size = 26, normalized size = 0.93 \begin {gather*} 5 \log {\left (- x^{2} - \frac {x}{5} - \frac {\log {\left (x - 1 \right )}}{5} + \log {\left (2 x + 3 \right )} - e^{4} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*x**3-60*x**2+185*x-50)/((10*x**2+5*x-15)*ln(2*x+3)+(-2*x**2-x+3)*ln(x-1)+(-10*x**2-5*x+15)*exp
(4)-10*x**4-7*x**3+14*x**2+3*x),x)

[Out]

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

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