3.2.53 \(\int \frac {-40 x^3+8 x^4+(-60 x^2-120 x^3+12 x^4+e (-20 x^6+4 x^7)) \log (5+10 x-x^2)+e (-30 x^5-60 x^6+6 x^7) \log ^2(5+10 x-x^2)}{e (-5-10 x+x^2)} \, dx\)

Optimal. Leaf size=22 \[ \left (\frac {2}{e}+x^3 \log (5+(10-x) x)\right )^2 \]

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Rubi [A]  time = 0.22, antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 6688, 6686} \begin {gather*} \frac {\left (e x^3 \log \left (-x^2+10 x+5\right )+2\right )^2}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40*x^3 + 8*x^4 + (-60*x^2 - 120*x^3 + 12*x^4 + E*(-20*x^6 + 4*x^7))*Log[5 + 10*x - x^2] + E*(-30*x^5 - 6
0*x^6 + 6*x^7)*Log[5 + 10*x - x^2]^2)/(E*(-5 - 10*x + x^2)),x]

[Out]

(2 + E*x^3*Log[5 + 10*x - x^2])^2/E^2

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} &=\frac {\int \frac {-40 x^3+8 x^4+\left (-60 x^2-120 x^3+12 x^4+e \left (-20 x^6+4 x^7\right )\right ) \log \left (5+10 x-x^2\right )+e \left (-30 x^5-60 x^6+6 x^7\right ) \log ^2\left (5+10 x-x^2\right )}{-5-10 x+x^2} \, dx}{e}\\ &=\frac {\int \frac {2 x^2 \left (2+e x^3 \log \left (5+10 x-x^2\right )\right ) \left (-2 (-5+x) x-3 \left (-5-10 x+x^2\right ) \log \left (5+10 x-x^2\right )\right )}{5+10 x-x^2} \, dx}{e}\\ &=\frac {2 \int \frac {x^2 \left (2+e x^3 \log \left (5+10 x-x^2\right )\right ) \left (-2 (-5+x) x-3 \left (-5-10 x+x^2\right ) \log \left (5+10 x-x^2\right )\right )}{5+10 x-x^2} \, dx}{e}\\ &=\frac {\left (2+e x^3 \log \left (5+10 x-x^2\right )\right )^2}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 1.09 \begin {gather*} \frac {\left (2+e x^3 \log \left (5+10 x-x^2\right )\right )^2}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*x^3 + 8*x^4 + (-60*x^2 - 120*x^3 + 12*x^4 + E*(-20*x^6 + 4*x^7))*Log[5 + 10*x - x^2] + E*(-30*x
^5 - 60*x^6 + 6*x^7)*Log[5 + 10*x - x^2]^2)/(E*(-5 - 10*x + x^2)),x]

[Out]

(2 + E*x^3*Log[5 + 10*x - x^2])^2/E^2

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fricas [A]  time = 0.88, size = 39, normalized size = 1.77 \begin {gather*} {\left (x^{6} e \log \left (-x^{2} + 10 \, x + 5\right )^{2} + 4 \, x^{3} \log \left (-x^{2} + 10 \, x + 5\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^7-60*x^6-30*x^5)*exp(1)*log(-x^2+10*x+5)^2+((4*x^7-20*x^6)*exp(1)+12*x^4-120*x^3-60*x^2)*log(-
x^2+10*x+5)+8*x^4-40*x^3)/(x^2-10*x-5)/exp(1),x, algorithm="fricas")

[Out]

(x^6*e*log(-x^2 + 10*x + 5)^2 + 4*x^3*log(-x^2 + 10*x + 5))*e^(-1)

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giac [A]  time = 0.61, size = 39, normalized size = 1.77 \begin {gather*} {\left (x^{6} e \log \left (-x^{2} + 10 \, x + 5\right )^{2} + 4 \, x^{3} \log \left (-x^{2} + 10 \, x + 5\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^7-60*x^6-30*x^5)*exp(1)*log(-x^2+10*x+5)^2+((4*x^7-20*x^6)*exp(1)+12*x^4-120*x^3-60*x^2)*log(-
x^2+10*x+5)+8*x^4-40*x^3)/(x^2-10*x-5)/exp(1),x, algorithm="giac")

[Out]

(x^6*e*log(-x^2 + 10*x + 5)^2 + 4*x^3*log(-x^2 + 10*x + 5))*e^(-1)

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maple [A]  time = 0.22, size = 37, normalized size = 1.68




method result size



risch \(x^{6} \ln \left (-x^{2}+10 x +5\right )^{2}+4 \,{\mathrm e}^{-1} x^{3} \ln \left (-x^{2}+10 x +5\right )\) \(37\)
norman \(x^{6} \ln \left (-x^{2}+10 x +5\right )^{2}+4 \,{\mathrm e}^{-1} x^{3} \ln \left (-x^{2}+10 x +5\right )\) \(39\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^7-60*x^6-30*x^5)*exp(1)*ln(-x^2+10*x+5)^2+((4*x^7-20*x^6)*exp(1)+12*x^4-120*x^3-60*x^2)*ln(-x^2+10*x
+5)+8*x^4-40*x^3)/(x^2-10*x-5)/exp(1),x,method=_RETURNVERBOSE)

[Out]

x^6*ln(-x^2+10*x+5)^2+4*exp(-1)*x^3*ln(-x^2+10*x+5)

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maxima [B]  time = 0.56, size = 52, normalized size = 2.36 \begin {gather*} {\left (x^{6} e \log \left (-x^{2} + 10 \, x + 5\right )^{2} + 4 \, {\left (x^{3} - 575\right )} \log \left (-x^{2} + 10 \, x + 5\right ) + 2300 \, \log \left (x^{2} - 10 \, x - 5\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^7-60*x^6-30*x^5)*exp(1)*log(-x^2+10*x+5)^2+((4*x^7-20*x^6)*exp(1)+12*x^4-120*x^3-60*x^2)*log(-
x^2+10*x+5)+8*x^4-40*x^3)/(x^2-10*x-5)/exp(1),x, algorithm="maxima")

[Out]

(x^6*e*log(-x^2 + 10*x + 5)^2 + 4*(x^3 - 575)*log(-x^2 + 10*x + 5) + 2300*log(x^2 - 10*x - 5))*e^(-1)

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mupad [B]  time = 0.62, size = 36, normalized size = 1.64 \begin {gather*} x^6\,{\ln \left (-x^2+10\,x+5\right )}^2+4\,{\mathrm {e}}^{-1}\,x^3\,\ln \left (-x^2+10\,x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-1)*(log(10*x - x^2 + 5)*(exp(1)*(20*x^6 - 4*x^7) + 60*x^2 + 120*x^3 - 12*x^4) + 40*x^3 - 8*x^4 + exp
(1)*log(10*x - x^2 + 5)^2*(30*x^5 + 60*x^6 - 6*x^7)))/(10*x - x^2 + 5),x)

[Out]

x^6*log(10*x - x^2 + 5)^2 + 4*x^3*exp(-1)*log(10*x - x^2 + 5)

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sympy [A]  time = 0.22, size = 32, normalized size = 1.45 \begin {gather*} x^{6} \log {\left (- x^{2} + 10 x + 5 \right )}^{2} + \frac {4 x^{3} \log {\left (- x^{2} + 10 x + 5 \right )}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**7-60*x**6-30*x**5)*exp(1)*ln(-x**2+10*x+5)**2+((4*x**7-20*x**6)*exp(1)+12*x**4-120*x**3-60*x*
*2)*ln(-x**2+10*x+5)+8*x**4-40*x**3)/(x**2-10*x-5)/exp(1),x)

[Out]

x**6*log(-x**2 + 10*x + 5)**2 + 4*x**3*exp(-1)*log(-x**2 + 10*x + 5)

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