3.99.53 \(\int \frac {48-98 x+53 x^2+15 x^3+(-440+314 x+286 x^2+45 x^3) \log (4+x)}{48+12 x} \, dx\)

Optimal. Leaf size=21 \[ x-\left (\frac {11}{6}-\frac {5 x}{4}\right ) x (5+x) \log (4+x) \]

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Rubi [B]  time = 0.16, antiderivative size = 43, normalized size of antiderivative = 2.05, number of steps used = 14, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6742, 1850, 2417, 2389, 2295, 2395, 43} \begin {gather*} \frac {5}{4} x^3 \log (x+4)+\frac {53}{12} x^2 \log (x+4)+x-\frac {55}{6} (x+4) \log (x+4)+\frac {110}{3} \log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(48 - 98*x + 53*x^2 + 15*x^3 + (-440 + 314*x + 286*x^2 + 45*x^3)*Log[4 + x])/(48 + 12*x),x]

[Out]

x + (110*Log[4 + x])/3 + (53*x^2*Log[4 + x])/12 + (5*x^3*Log[4 + x])/4 - (55*(4 + x)*Log[4 + x])/6

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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 2417

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Poly
x*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && PolynomialQ[Polyx, x]

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 {48-98 x+53 x^2+15 x^3}{12 (4+x)}+\frac {1}{12} \left (-110+106 x+45 x^2\right ) \log (4+x)\right ) \, dx\\ &=\frac {1}{12} \int \frac {48-98 x+53 x^2+15 x^3}{4+x} \, dx+\frac {1}{12} \int \left (-110+106 x+45 x^2\right ) \log (4+x) \, dx\\ &=\frac {1}{12} \int \left (-70-7 x+15 x^2+\frac {328}{4+x}\right ) \, dx+\frac {1}{12} \int \left (-110 \log (4+x)+106 x \log (4+x)+45 x^2 \log (4+x)\right ) \, dx\\ &=-\frac {35 x}{6}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {15}{4} \int x^2 \log (4+x) \, dx+\frac {53}{6} \int x \log (4+x) \, dx-\frac {55}{6} \int \log (4+x) \, dx\\ &=-\frac {35 x}{6}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {5}{4} \int \frac {x^3}{4+x} \, dx-\frac {53}{12} \int \frac {x^2}{4+x} \, dx-\frac {55}{6} \operatorname {Subst}(\int \log (x) \, dx,x,4+x)\\ &=\frac {10 x}{3}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {55}{6} (4+x) \log (4+x)-\frac {5}{4} \int \left (16-4 x+x^2-\frac {64}{4+x}\right ) \, dx-\frac {53}{12} \int \left (-4+x+\frac {16}{4+x}\right ) \, dx\\ &=x+\frac {110}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {55}{6} (4+x) \log (4+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 33, normalized size = 1.57 \begin {gather*} \frac {1}{12} \left (12 x-110 x \log (4+x)+53 x^2 \log (4+x)+15 x^3 \log (4+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48 - 98*x + 53*x^2 + 15*x^3 + (-440 + 314*x + 286*x^2 + 45*x^3)*Log[4 + x])/(48 + 12*x),x]

[Out]

(12*x - 110*x*Log[4 + x] + 53*x^2*Log[4 + x] + 15*x^3*Log[4 + x])/12

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fricas [A]  time = 0.54, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{12} \, {\left (15 \, x^{3} + 53 \, x^{2} - 110 \, x\right )} \log \left (x + 4\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*x^3+286*x^2+314*x-440)*log(4+x)+15*x^3+53*x^2-98*x+48)/(12*x+48),x, algorithm="fricas")

[Out]

1/12*(15*x^3 + 53*x^2 - 110*x)*log(x + 4) + x

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giac [A]  time = 0.23, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{12} \, {\left (15 \, x^{3} + 53 \, x^{2} - 110 \, x\right )} \log \left (x + 4\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*x^3+286*x^2+314*x-440)*log(4+x)+15*x^3+53*x^2-98*x+48)/(12*x+48),x, algorithm="giac")

[Out]

1/12*(15*x^3 + 53*x^2 - 110*x)*log(x + 4) + x

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maple [A]  time = 0.10, size = 22, normalized size = 1.05

method	result	size
risch	\(\left (\frac {5}{4} x^{3}+\frac {53}{12} x^{2}-\frac {55}{6} x \right ) \ln \left (4+x \right )+x\)	\(22\)
norman	\(x +\frac {53 x^{2} \ln \left (4+x \right )}{12}+\frac {5 x^{3} \ln \left (4+x \right )}{4}-\frac {55 \ln \left (4+x \right ) x}{6}\)	\(28\)
derivativedivides	\(\frac {5 \ln \left (4+x \right ) \left (4+x \right )^{3}}{4}-\frac {127 \ln \left (4+x \right ) \left (4+x \right )^{2}}{12}+\frac {31 \left (4+x \right ) \ln \left (4+x \right )}{2}+4+x +\frac {82 \ln \left (4+x \right )}{3}\)	\(41\)
default	\(\frac {5 \ln \left (4+x \right ) \left (4+x \right )^{3}}{4}-\frac {127 \ln \left (4+x \right ) \left (4+x \right )^{2}}{12}+\frac {31 \left (4+x \right ) \ln \left (4+x \right )}{2}+4+x +\frac {82 \ln \left (4+x \right )}{3}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((45*x^3+286*x^2+314*x-440)*ln(4+x)+15*x^3+53*x^2-98*x+48)/(12*x+48),x,method=_RETURNVERBOSE)

[Out]

(5/4*x^3+53/12*x^2-55/6*x)*ln(4+x)+x

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maxima [B]  time = 0.35, size = 67, normalized size = 3.19 \begin {gather*} \frac {5}{4} \, {\left (x^{3} - 6 \, x^{2} + 48 \, x - 192 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) + \frac {143}{12} \, {\left (x^{2} - 8 \, x + 32 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) + \frac {157}{6} \, {\left (x - 4 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - \frac {110}{3} \, \log \left (x + 4\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*x^3+286*x^2+314*x-440)*log(4+x)+15*x^3+53*x^2-98*x+48)/(12*x+48),x, algorithm="maxima")

[Out]

5/4*(x^3 - 6*x^2 + 48*x - 192*log(x + 4))*log(x + 4) + 143/12*(x^2 - 8*x + 32*log(x + 4))*log(x + 4) + 157/6*(
x - 4*log(x + 4))*log(x + 4) - 110/3*log(x + 4)^2 + x

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mupad [B]  time = 5.61, size = 30, normalized size = 1.43 \begin {gather*} \frac {53\,x^2\,\ln \left (x+4\right )}{12}+\frac {5\,x^3\,\ln \left (x+4\right )}{4}-x\,\left (\frac {55\,\ln \left (x+4\right )}{6}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 4)*(314*x + 286*x^2 + 45*x^3 - 440) - 98*x + 53*x^2 + 15*x^3 + 48)/(12*x + 48),x)

[Out]

(53*x^2*log(x + 4))/12 + (5*x^3*log(x + 4))/4 - x*((55*log(x + 4))/6 - 1)

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sympy [A]  time = 0.15, size = 24, normalized size = 1.14 \begin {gather*} x + \left (\frac {5 x^{3}}{4} + \frac {53 x^{2}}{12} - \frac {55 x}{6}\right ) \log {\left (x + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*x**3+286*x**2+314*x-440)*ln(4+x)+15*x**3+53*x**2-98*x+48)/(12*x+48),x)

[Out]

x + (5*x**3/4 + 53*x**2/12 - 55*x/6)*log(x + 4)

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