3.74.72 \(\int \frac {144 x+30 x^2+27 x^3+(-160-144 x) \log (-10-9 x)}{150 x^2+135 x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{5} \left (x+\frac {16 (x+\log (x-5 (2+2 x)))}{3 x}\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {1593, 6742, 893, 2395, 36, 31, 29} \begin {gather*} \frac {x}{5}+\frac {16 \log (-9 x-10)}{15 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(144*x + 30*x^2 + 27*x^3 + (-160 - 144*x)*Log[-10 - 9*x])/(150*x^2 + 135*x^3),x]

[Out]

x/5 + (16*Log[-10 - 9*x])/(15*x)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 \frac {144 x+30 x^2+27 x^3+(-160-144 x) \log (-10-9 x)}{x^2 (150+135 x)} \, dx\\ &=\int \left (\frac {48+10 x+9 x^2}{5 x (10+9 x)}-\frac {16 \log (-10-9 x)}{15 x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {48+10 x+9 x^2}{x (10+9 x)} \, dx-\frac {16}{15} \int \frac {\log (-10-9 x)}{x^2} \, dx\\ &=\frac {16 \log (-10-9 x)}{15 x}+\frac {1}{5} \int \left (1+\frac {24}{5 x}-\frac {216}{5 (10+9 x)}\right ) \, dx+\frac {48}{5} \int \frac {1}{(-10-9 x) x} \, dx\\ &=\frac {x}{5}+\frac {16 \log (-10-9 x)}{15 x}+\frac {24 \log (x)}{25}-\frac {24}{25} \log (10+9 x)-\frac {24}{25} \int \frac {1}{x} \, dx-\frac {216}{25} \int \frac {1}{-10-9 x} \, dx\\ &=\frac {x}{5}+\frac {16 \log (-10-9 x)}{15 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.76 \begin {gather*} \frac {1}{15} \left (3 x+\frac {16 \log (-10-9 x)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(144*x + 30*x^2 + 27*x^3 + (-160 - 144*x)*Log[-10 - 9*x])/(150*x^2 + 135*x^3),x]

[Out]

(3*x + (16*Log[-10 - 9*x])/x)/15

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fricas [A]  time = 0.51, size = 19, normalized size = 0.76 \begin {gather*} \frac {3 \, x^{2} + 16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-160)*log(-9*x-10)+27*x^3+30*x^2+144*x)/(135*x^3+150*x^2),x, algorithm="fricas")

[Out]

1/15*(3*x^2 + 16*log(-9*x - 10))/x

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giac [A]  time = 0.14, size = 15, normalized size = 0.60 \begin {gather*} \frac {1}{5} \, x + \frac {16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-160)*log(-9*x-10)+27*x^3+30*x^2+144*x)/(135*x^3+150*x^2),x, algorithm="giac")

[Out]

1/5*x + 16/15*log(-9*x - 10)/x

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maple [A]  time = 0.16, size = 16, normalized size = 0.64

method	result	size
risch	\(\frac {16 \ln \left (-9 x -10\right )}{15 x}+\frac {x}{5}\)	\(16\)
norman	\(\frac {\frac {x^{2}}{5}+\frac {16 \ln \left (-9 x -10\right )}{15}}{x}\)	\(19\)
derivativedivides	\(-\frac {8 \ln \left (-9 x -10\right ) \left (-9 x -10\right )}{75 x}+\frac {x}{5}+\frac {2}{9}-\frac {24 \ln \left (-9 x -10\right )}{25}\)	\(30\)
default	\(-\frac {8 \ln \left (-9 x -10\right ) \left (-9 x -10\right )}{75 x}+\frac {x}{5}+\frac {2}{9}-\frac {24 \ln \left (-9 x -10\right )}{25}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-144*x-160)*ln(-9*x-10)+27*x^3+30*x^2+144*x)/(135*x^3+150*x^2),x,method=_RETURNVERBOSE)

[Out]

16/15/x*ln(-9*x-10)+1/5*x

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maxima [A]  time = 0.41, size = 19, normalized size = 0.76 \begin {gather*} \frac {3 \, x^{2} + 16 \, \log \left (-9 \, x - 10\right )}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-160)*log(-9*x-10)+27*x^3+30*x^2+144*x)/(135*x^3+150*x^2),x, algorithm="maxima")

[Out]

1/15*(3*x^2 + 16*log(-9*x - 10))/x

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mupad [B]  time = 0.22, size = 15, normalized size = 0.60 \begin {gather*} \frac {x}{5}+\frac {16\,\ln \left (-9\,x-10\right )}{15\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((144*x - log(- 9*x - 10)*(144*x + 160) + 30*x^2 + 27*x^3)/(150*x^2 + 135*x^3),x)

[Out]

x/5 + (16*log(- 9*x - 10))/(15*x)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.60 \begin {gather*} \frac {x}{5} + \frac {16 \log {\left (- 9 x - 10 \right )}}{15 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-144*x-160)*ln(-9*x-10)+27*x**3+30*x**2+144*x)/(135*x**3+150*x**2),x)

[Out]

x/5 + 16*log(-9*x - 10)/(15*x)

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