3.88.97 \(\int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+(-60+4 x+8 x^2) \log ((3+x) \log ^2(5))}{15+5 x} \, dx\)

Optimal. Leaf size=29 \[ \left (-2+e^2+x-\frac {x^2}{5}\right ) \left (x-4 \left (2+\log \left ((3+x) \log ^2(5)\right )\right )\right ) \]

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Rubi [B]  time = 0.18, antiderivative size = 89, normalized size of antiderivative = 3.07, number of steps used = 7, number of rules used = 4, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6742, 1850, 2395, 43} \begin {gather*} -\frac {x^3}{5}+3 x^2-\frac {1}{5} \left (82-5 e^2\right ) x+\frac {22 x}{5}-\frac {1}{10} (5-2 x)^2+\frac {1}{5} (5-2 x)^2 \log \left (x \log ^2(5)+3 \log ^2(5)\right )+\frac {4}{5} \left (34-5 e^2\right ) \log (x+3)-\frac {121}{5} \log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-110 + 8*x + 21*x^2 - 3*x^3 + E^2*(-5 + 5*x) + (-60 + 4*x + 8*x^2)*Log[(3 + x)*Log[5]^2])/(15 + 5*x),x]

[Out]

-1/10*(5 - 2*x)^2 + (22*x)/5 - ((82 - 5*E^2)*x)/5 + 3*x^2 - x^3/5 - (121*Log[3 + x])/5 + (4*(34 - 5*E^2)*Log[3
+ x])/5 + ((5 - 2*x)^2*Log[3*Log[5]^2 + x*Log[5]^2])/5

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 {-5 \left (22+e^2\right )+\left (8+5 e^2\right ) x+21 x^2-3 x^3}{5 (3+x)}+\frac {4}{5} (-5+2 x) \log \left (3 \log ^2(5)+x \log ^2(5)\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {-5 \left (22+e^2\right )+\left (8+5 e^2\right ) x+21 x^2-3 x^3}{3+x} \, dx+\frac {4}{5} \int (-5+2 x) \log \left (3 \log ^2(5)+x \log ^2(5)\right ) \, dx\\ &=\frac {1}{5} (5-2 x)^2 \log \left (3 \log ^2(5)+x \log ^2(5)\right )+\frac {1}{5} \int \left (-82 \left (1-\frac {5 e^2}{82}\right )+30 x-3 x^2-\frac {4 \left (-34+5 e^2\right )}{3+x}\right ) \, dx-\frac {1}{5} \log ^2(5) \int \frac {(-5+2 x)^2}{3 \log ^2(5)+x \log ^2(5)} \, dx\\ &=-\frac {1}{5} \left (82-5 e^2\right ) x+3 x^2-\frac {x^3}{5}+\frac {4}{5} \left (34-5 e^2\right ) \log (3+x)+\frac {1}{5} (5-2 x)^2 \log \left (3 \log ^2(5)+x \log ^2(5)\right )-\frac {1}{5} \log ^2(5) \int \left (-\frac {22}{\log ^2(5)}+\frac {2 (-5+2 x)}{\log ^2(5)}+\frac {121}{3 \log ^2(5)+x \log ^2(5)}\right ) \, dx\\ &=-\frac {1}{10} (5-2 x)^2+\frac {22 x}{5}-\frac {1}{5} \left (82-5 e^2\right ) x+3 x^2-\frac {x^3}{5}-\frac {121}{5} \log (3+x)+\frac {4}{5} \left (34-5 e^2\right ) \log (3+x)+\frac {1}{5} (5-2 x)^2 \log \left (3 \log ^2(5)+x \log ^2(5)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 86, normalized size = 2.97 \begin {gather*} \frac {1}{5} \left (-50 x+5 e^2 x+13 x^2-x^3+100 \log (3+x)-20 e^2 \log (3+x)+4 x^2 \log \left ((3+x) \log ^2(5)\right )-60 \log \left (3 \log ^2(5)+x \log ^2(5)\right )-20 x \log \left (3 \log ^2(5)+x \log ^2(5)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-110 + 8*x + 21*x^2 - 3*x^3 + E^2*(-5 + 5*x) + (-60 + 4*x + 8*x^2)*Log[(3 + x)*Log[5]^2])/(15 + 5*x
),x]

[Out]

(-50*x + 5*E^2*x + 13*x^2 - x^3 + 100*Log[3 + x] - 20*E^2*Log[3 + x] + 4*x^2*Log[(3 + x)*Log[5]^2] - 60*Log[3*
Log[5]^2 + x*Log[5]^2] - 20*x*Log[3*Log[5]^2 + x*Log[5]^2])/5

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fricas [A]  time = 0.70, size = 41, normalized size = 1.41 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {13}{5} \, x^{2} + x e^{2} + \frac {4}{5} \, {\left (x^{2} - 5 \, x - 5 \, e^{2} + 10\right )} \log \left ({\left (x + 3\right )} \log \relax (5)^{2}\right ) - 10 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-110)/(5*x+15),x, algorithm="fric
as")

[Out]

-1/5*x^3 + 13/5*x^2 + x*e^2 + 4/5*(x^2 - 5*x - 5*e^2 + 10)*log((x + 3)*log(5)^2) - 10*x

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giac [B]  time = 0.16, size = 68, normalized size = 2.34 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {4}{5} \, x^{2} \log \left (x \log \relax (5)^{2} + 3 \, \log \relax (5)^{2}\right ) + \frac {13}{5} \, x^{2} + x e^{2} - 4 \, x \log \left (x \log \relax (5)^{2} + 3 \, \log \relax (5)^{2}\right ) - 4 \, e^{2} \log \left (x + 3\right ) - 10 \, x + 8 \, \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-110)/(5*x+15),x, algorithm="giac
")

[Out]

-1/5*x^3 + 4/5*x^2*log(x*log(5)^2 + 3*log(5)^2) + 13/5*x^2 + x*e^2 - 4*x*log(x*log(5)^2 + 3*log(5)^2) - 4*e^2*
log(x + 3) - 10*x + 8*log(x + 3)

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maple [B]  time = 0.43, size = 52, normalized size = 1.79

method	result	size
risch	\(\left (\frac {4}{5} x^{2}-4 x \right ) \ln \left (\left (3+x \right ) \ln \relax (5)^{2}\right )-\frac {x^{3}}{5}+{\mathrm e}^{2} x +\frac {13 x^{2}}{5}-10 x -4 \ln \left (3+x \right ) {\mathrm e}^{2}+8 \ln \left (3+x \right )\)	\(52\)
norman	\(\left (-10+{\mathrm e}^{2}\right ) x +\left (8-4 \,{\mathrm e}^{2}\right ) \ln \left (\left (3+x \right ) \ln \relax (5)^{2}\right )+\frac {13 x^{2}}{5}-\frac {x^{3}}{5}-4 \ln \left (\left (3+x \right ) \ln \relax (5)^{2}\right ) x +\frac {4 \ln \left (\left (3+x \right ) \ln \relax (5)^{2}\right ) x^{2}}{5}\)	\(60\)
derivativedivides	\(\frac {-20 \ln \relax (5)^{2} {\mathrm e}^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )+5 \,{\mathrm e}^{2} \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )+136 \ln \relax (5)^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-44 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right ) \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-155 x \ln \relax (5)^{2}-465 \ln \relax (5)^{2}+\frac {4 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-2 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2}}{\ln \relax (5)^{2}}+\frac {24 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2}}{\ln \relax (5)^{2}}-\frac {\left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{3}}{\ln \relax (5)^{4}}}{5 \ln \relax (5)^{2}}\)	\(206\)
default	\(\frac {-20 \ln \relax (5)^{2} {\mathrm e}^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )+5 \,{\mathrm e}^{2} \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )+136 \ln \relax (5)^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-44 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right ) \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-155 x \ln \relax (5)^{2}-465 \ln \relax (5)^{2}+\frac {4 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2} \ln \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )-2 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2}}{\ln \relax (5)^{2}}+\frac {24 \left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{2}}{\ln \relax (5)^{2}}-\frac {\left (x \ln \relax (5)^{2}+3 \ln \relax (5)^{2}\right )^{3}}{\ln \relax (5)^{4}}}{5 \ln \relax (5)^{2}}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+4*x-60)*ln((3+x)*ln(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-110)/(5*x+15),x,method=_RETURNVERBOSE)

[Out]

(4/5*x^2-4*x)*ln((3+x)*ln(5)^2)-1/5*x^3+exp(2)*x+13/5*x^2-10*x-4*ln(3+x)*exp(2)+8*ln(3+x)

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maxima [B]  time = 0.45, size = 109, normalized size = 3.76 \begin {gather*} -\frac {1}{5} \, x^{3} + \frac {13}{5} \, x^{2} + {\left (x - 3 \, \log \left (x + 3\right )\right )} e^{2} + \frac {4}{5} \, {\left (x^{2} - 6 \, x + 18 \, \log \left (x + 3\right )\right )} \log \left (x \log \relax (5)^{2} + 3 \, \log \relax (5)^{2}\right ) + \frac {4}{5} \, {\left (x - 3 \, \log \left (x + 3\right )\right )} \log \left (x \log \relax (5)^{2} + 3 \, \log \relax (5)^{2}\right ) - e^{2} \log \left (x + 3\right ) - 12 \, \log \left (x + 3\right )^{2} - 24 \, \log \left (x + 3\right ) \log \left (\log \relax (5)\right ) - 10 \, x + 8 \, \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-110)/(5*x+15),x, algorithm="maxi
ma")

[Out]

-1/5*x^3 + 13/5*x^2 + (x - 3*log(x + 3))*e^2 + 4/5*(x^2 - 6*x + 18*log(x + 3))*log(x*log(5)^2 + 3*log(5)^2) +
4/5*(x - 3*log(x + 3))*log(x*log(5)^2 + 3*log(5)^2) - e^2*log(x + 3) - 12*log(x + 3)^2 - 24*log(x + 3)*log(log
(5)) - 10*x + 8*log(x + 3)

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mupad [B]  time = 5.81, size = 58, normalized size = 2.00 \begin {gather*} 8\,\ln \left (x+3\right )-10\,x-4\,\ln \left (x+3\right )\,{\mathrm {e}}^2+x\,{\mathrm {e}}^2-4\,x\,\ln \left ({\ln \relax (5)}^2\,\left (x+3\right )\right )+\frac {4\,x^2\,\ln \left ({\ln \relax (5)}^2\,\left (x+3\right )\right )}{5}+\frac {13\,x^2}{5}-\frac {x^3}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(log(5)^2*(x + 3))*(4*x + 8*x^2 - 60) + 21*x^2 - 3*x^3 + exp(2)*(5*x - 5) - 110)/(5*x + 15),x)

[Out]

8*log(x + 3) - 10*x - 4*log(x + 3)*exp(2) + x*exp(2) - 4*x*log(log(5)^2*(x + 3)) + (4*x^2*log(log(5)^2*(x + 3)
))/5 + (13*x^2)/5 - x^3/5

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sympy [A]  time = 0.23, size = 49, normalized size = 1.69 \begin {gather*} - \frac {x^{3}}{5} + \frac {13 x^{2}}{5} - x \left (10 - e^{2}\right ) + \left (\frac {4 x^{2}}{5} - 4 x\right ) \log {\left (\left (x + 3\right ) \log {\relax (5 )}^{2} \right )} - 4 \left (-2 + e^{2}\right ) \log {\left (x + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+4*x-60)*ln((3+x)*ln(5)**2)+(5*x-5)*exp(2)-3*x**3+21*x**2+8*x-110)/(5*x+15),x)

[Out]

-x**3/5 + 13*x**2/5 - x*(10 - exp(2)) + (4*x**2/5 - 4*x)*log((x + 3)*log(5)**2) - 4*(-2 + exp(2))*log(x + 3)

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