3.72.38 \(\int \frac {-37-61 x-99 x^2+27 x^3+e^3 (-45-81 x+18 x^2)+(-90-162 x+36 x^2) \log (5-x)}{-10+2 x} \, dx\)

Optimal. Leaf size=25 \[ x+\frac {9}{2} \left (1+x+x^2\right ) \left (e^3+x+2 \log (5-x)\right ) \]

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Rubi [B]  time = 0.20, antiderivative size = 69, normalized size of antiderivative = 2.76, number of steps used = 7, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6742, 1850, 2395, 43} \begin {gather*} \frac {9 x^3}{2}+\frac {9}{2} \left (2+e^3\right ) x^2-\frac {9 x^2}{2}+\frac {1}{2} \left (119+9 e^3\right ) x-54 x+\frac {9}{4} (2 x+1)^2 \log (5-x)+\frac {27}{4} \log (5-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-37 - 61*x - 99*x^2 + 27*x^3 + E^3*(-45 - 81*x + 18*x^2) + (-90 - 162*x + 36*x^2)*Log[5 - x])/(-10 + 2*x)
,x]

[Out]

-54*x + ((119 + 9*E^3)*x)/2 - (9*x^2)/2 + (9*(2 + E^3)*x^2)/2 + (9*x^3)/2 + (27*Log[5 - x])/4 + (9*(1 + 2*x)^2
*Log[5 - x])/4

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2395

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

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 \left (\frac {37+45 e^3+\left (61+81 e^3\right ) x+9 \left (11-2 e^3\right ) x^2-27 x^3}{2 (5-x)}+9 (1+2 x) \log (5-x)\right ) \, dx\\ &=\frac {1}{2} \int \frac {37+45 e^3+\left (61+81 e^3\right ) x+9 \left (11-2 e^3\right ) x^2-27 x^3}{5-x} \, dx+9 \int (1+2 x) \log (5-x) \, dx\\ &=\frac {9}{4} (1+2 x)^2 \log (5-x)+\frac {1}{2} \int \left (119 \left (1+\frac {9 e^3}{119}\right )+\frac {558}{-5+x}+18 \left (2+e^3\right ) x+27 x^2\right ) \, dx+\frac {9}{4} \int \frac {(1+2 x)^2}{5-x} \, dx\\ &=\frac {1}{2} \left (119+9 e^3\right ) x+\frac {9}{2} \left (2+e^3\right ) x^2+\frac {9 x^3}{2}+279 \log (5-x)+\frac {9}{4} (1+2 x)^2 \log (5-x)+\frac {9}{4} \int \left (-24-\frac {121}{-5+x}-4 x\right ) \, dx\\ &=-54 x+\frac {1}{2} \left (119+9 e^3\right ) x-\frac {9 x^2}{2}+\frac {9}{2} \left (2+e^3\right ) x^2+\frac {9 x^3}{2}+\frac {27}{4} \log (5-x)+\frac {9}{4} (1+2 x)^2 \log (5-x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 60, normalized size = 2.40 \begin {gather*} \frac {1}{2} \left (11 x+9 e^3 x+9 x^2+9 e^3 x^2+9 x^3+18 \log (5-x)+18 x \log (5-x)+18 x^2 \log (5-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-37 - 61*x - 99*x^2 + 27*x^3 + E^3*(-45 - 81*x + 18*x^2) + (-90 - 162*x + 36*x^2)*Log[5 - x])/(-10
+ 2*x),x]

[Out]

(11*x + 9*E^3*x + 9*x^2 + 9*E^3*x^2 + 9*x^3 + 18*Log[5 - x] + 18*x*Log[5 - x] + 18*x^2*Log[5 - x])/2

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fricas [A]  time = 0.63, size = 37, normalized size = 1.48 \begin {gather*} \frac {9}{2} \, x^{3} + \frac {9}{2} \, x^{2} + \frac {9}{2} \, {\left (x^{2} + x\right )} e^{3} + 9 \, {\left (x^{2} + x + 1\right )} \log \left (-x + 5\right ) + \frac {11}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2-162*x-90)*log(5-x)+(18*x^2-81*x-45)*exp(3)+27*x^3-99*x^2-61*x-37)/(2*x-10),x, algorithm="fr
icas")

[Out]

9/2*x^3 + 9/2*x^2 + 9/2*(x^2 + x)*e^3 + 9*(x^2 + x + 1)*log(-x + 5) + 11/2*x

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giac [B]  time = 0.18, size = 52, normalized size = 2.08 \begin {gather*} \frac {9}{2} \, x^{3} + \frac {9}{2} \, x^{2} e^{3} + 9 \, x^{2} \log \left (-x + 5\right ) + \frac {9}{2} \, x^{2} + \frac {9}{2} \, x e^{3} + 9 \, x \log \left (-x + 5\right ) + \frac {11}{2} \, x + 9 \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2-162*x-90)*log(5-x)+(18*x^2-81*x-45)*exp(3)+27*x^3-99*x^2-61*x-37)/(2*x-10),x, algorithm="gi
ac")

[Out]

9/2*x^3 + 9/2*x^2*e^3 + 9*x^2*log(-x + 5) + 9/2*x^2 + 9/2*x*e^3 + 9*x*log(-x + 5) + 11/2*x + 9*log(x - 5)

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maple [B]  time = 0.12, size = 49, normalized size = 1.96

method	result	size
risch	\(\left (9 x^{2}+9 x \right ) \ln \left (5-x \right )+\frac {9 x^{2} {\mathrm e}^{3}}{2}+\frac {9 x^{3}}{2}+\frac {9 x \,{\mathrm e}^{3}}{2}+\frac {9 x^{2}}{2}+\frac {11 x}{2}+9 \ln \left (x -5\right )\)	\(49\)
norman	\(9 \ln \left (5-x \right )+\left (\frac {9}{2}+\frac {9 \,{\mathrm e}^{3}}{2}\right ) x^{2}+\left (\frac {11}{2}+\frac {9 \,{\mathrm e}^{3}}{2}\right ) x +\frac {9 x^{3}}{2}+9 \ln \left (5-x \right ) x +9 \ln \left (5-x \right ) x^{2}\)	\(53\)
derivativedivides	\(\frac {9 \,{\mathrm e}^{3} \left (5-x \right )^{2}}{2}+9 \ln \left (5-x \right ) \left (5-x \right )^{2}+72 \left (5-x \right )^{2}-\frac {9 \left (5-x \right )^{3}}{2}-\frac {99 \,{\mathrm e}^{3} \left (5-x \right )}{2}-99 \left (5-x \right ) \ln \left (5-x \right )-1940+388 x +279 \ln \left (5-x \right )\)	\(80\)
default	\(\frac {9 \,{\mathrm e}^{3} \left (5-x \right )^{2}}{2}+9 \ln \left (5-x \right ) \left (5-x \right )^{2}+72 \left (5-x \right )^{2}-\frac {9 \left (5-x \right )^{3}}{2}-\frac {99 \,{\mathrm e}^{3} \left (5-x \right )}{2}-99 \left (5-x \right ) \ln \left (5-x \right )-1940+388 x +279 \ln \left (5-x \right )\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x^2-162*x-90)*ln(5-x)+(18*x^2-81*x-45)*exp(3)+27*x^3-99*x^2-61*x-37)/(2*x-10),x,method=_RETURNVERBOSE
)

[Out]

(9*x^2+9*x)*ln(5-x)+9/2*x^2*exp(3)+9/2*x^3+9/2*x*exp(3)+9/2*x^2+11/2*x+9*ln(x-5)

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maxima [B]  time = 0.36, size = 112, normalized size = 4.48 \begin {gather*} \frac {9}{2} \, x^{3} + \frac {9}{2} \, x^{2} + \frac {9}{2} \, {\left (x^{2} + 10 \, x + 50 \, \log \left (x - 5\right )\right )} e^{3} - \frac {81}{2} \, {\left (x + 5 \, \log \left (x - 5\right )\right )} e^{3} - \frac {45}{2} \, e^{3} \log \left (x - 5\right ) - \frac {45}{2} \, \log \left (x - 5\right )^{2} + 9 \, {\left (x^{2} + 10 \, x + 50 \, \log \left (x - 5\right )\right )} \log \left (-x + 5\right ) - 81 \, {\left (x + 5 \, \log \left (x - 5\right )\right )} \log \left (-x + 5\right ) - \frac {45}{2} \, \log \left (-x + 5\right )^{2} + \frac {11}{2} \, x + 9 \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2-162*x-90)*log(5-x)+(18*x^2-81*x-45)*exp(3)+27*x^3-99*x^2-61*x-37)/(2*x-10),x, algorithm="ma
xima")

[Out]

9/2*x^3 + 9/2*x^2 + 9/2*(x^2 + 10*x + 50*log(x - 5))*e^3 - 81/2*(x + 5*log(x - 5))*e^3 - 45/2*e^3*log(x - 5) -
45/2*log(x - 5)^2 + 9*(x^2 + 10*x + 50*log(x - 5))*log(-x + 5) - 81*(x + 5*log(x - 5))*log(-x + 5) - 45/2*log
(-x + 5)^2 + 11/2*x + 9*log(x - 5)

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mupad [B]  time = 4.28, size = 46, normalized size = 1.84 \begin {gather*} 9\,\ln \left (x-5\right )+x^2\,\left (\frac {9\,{\mathrm {e}}^3}{2}+9\,\ln \left (5-x\right )+\frac {9}{2}\right )+\frac {9\,x^3}{2}+x\,\left (\frac {9\,{\mathrm {e}}^3}{2}+9\,\ln \left (5-x\right )+\frac {11}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(61*x + exp(3)*(81*x - 18*x^2 + 45) + log(5 - x)*(162*x - 36*x^2 + 90) + 99*x^2 - 27*x^3 + 37)/(2*x - 10)
,x)

[Out]

9*log(x - 5) + x^2*((9*exp(3))/2 + 9*log(5 - x) + 9/2) + (9*x^3)/2 + x*((9*exp(3))/2 + 9*log(5 - x) + 11/2)

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sympy [A]  time = 0.22, size = 51, normalized size = 2.04 \begin {gather*} \frac {9 x^{3}}{2} + x^{2} \left (\frac {9}{2} + \frac {9 e^{3}}{2}\right ) + x \left (\frac {11}{2} + \frac {9 e^{3}}{2}\right ) + \left (9 x^{2} + 9 x\right ) \log {\left (5 - x \right )} + 9 \log {\left (x - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x**2-162*x-90)*ln(5-x)+(18*x**2-81*x-45)*exp(3)+27*x**3-99*x**2-61*x-37)/(2*x-10),x)

[Out]

9*x**3/2 + x**2*(9/2 + 9*exp(3)/2) + x*(11/2 + 9*exp(3)/2) + (9*x**2 + 9*x)*log(5 - x) + 9*log(x - 5)

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