3.26.60 \(\int \frac {-428-15 e^2-48 x+(-130-30 x) \log (\frac {8}{13+3 x})}{65+15 x} \, dx\)

Optimal. Leaf size=29 \[ \left (6+e^2-2 x\right ) \left (\frac {13}{5}+\log \left (\frac {2}{3+\frac {1}{4} (1+3 x)}\right )\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 42, normalized size of antiderivative = 1.45, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6741, 6742, 43, 2389, 2295} \begin {gather*} -\frac {26 x}{5}-\frac {2}{3} (3 x+13) \log \left (\frac {8}{3 x+13}\right )-\frac {1}{3} \left (44+3 e^2\right ) \log (3 x+13) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-428 - 15*E^2 - 48*x + (-130 - 30*x)*Log[8/(13 + 3*x)])/(65 + 15*x),x]

[Out]

(-26*x)/5 - (2*(13 + 3*x)*Log[8/(13 + 3*x)])/3 - ((44 + 3*E^2)*Log[13 + 3*x])/3

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 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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

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 {-428 \left (1+\frac {15 e^2}{428}\right )-48 x+(-130-30 x) \log \left (\frac {8}{13+3 x}\right )}{65+15 x} \, dx\\ &=\int \left (\frac {-428-15 e^2-48 x}{5 (13+3 x)}-2 \log \left (\frac {8}{13+3 x}\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {-428-15 e^2-48 x}{13+3 x} \, dx-2 \int \log \left (\frac {8}{13+3 x}\right ) \, dx\\ &=\frac {1}{5} \int \left (-16-\frac {5 \left (44+3 e^2\right )}{13+3 x}\right ) \, dx-\frac {2}{3} \operatorname {Subst}\left (\int \log \left (\frac {8}{x}\right ) \, dx,x,13+3 x\right )\\ &=-\frac {26 x}{5}-\frac {2}{3} (13+3 x) \log \left (\frac {8}{13+3 x}\right )-\frac {1}{3} \left (44+3 e^2\right ) \log (13+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 40, normalized size = 1.38 \begin {gather*} -\frac {26 x}{5}-\left (\frac {26}{3}+2 x\right ) \log \left (\frac {8}{13+3 x}\right )-\left (\frac {44}{3}+e^2\right ) \log (13+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-428 - 15*E^2 - 48*x + (-130 - 30*x)*Log[8/(13 + 3*x)])/(65 + 15*x),x]

[Out]

(-26*x)/5 - (26/3 + 2*x)*Log[8/(13 + 3*x)] - (44/3 + E^2)*Log[13 + 3*x]

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fricas [A]  time = 0.54, size = 25, normalized size = 0.86 \begin {gather*} -{\left (2 \, x - e^{2} - 6\right )} \log \left (\frac {8}{3 \, x + 13}\right ) - \frac {26}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x-130)*log(8/(3*x+13))-15*exp(2)-48*x-428)/(15*x+65),x, algorithm="fricas")

[Out]

-(2*x - e^2 - 6)*log(8/(3*x + 13)) - 26/5*x

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giac [B]  time = 0.25, size = 61, normalized size = 2.10 \begin {gather*} \frac {1}{15} \, {\left (3 \, x + 13\right )} {\left (\frac {15 \, e^{2} \log \left (\frac {8}{3 \, x + 13}\right )}{3 \, x + 13} + \frac {220 \, \log \left (\frac {8}{3 \, x + 13}\right )}{3 \, x + 13} - 10 \, \log \left (\frac {8}{3 \, x + 13}\right ) - 26\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x-130)*log(8/(3*x+13))-15*exp(2)-48*x-428)/(15*x+65),x, algorithm="giac")

[Out]

1/15*(3*x + 13)*(15*e^2*log(8/(3*x + 13))/(3*x + 13) + 220*log(8/(3*x + 13))/(3*x + 13) - 10*log(8/(3*x + 13))
- 26)

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maple [A]  time = 0.39, size = 33, normalized size = 1.14

method	result	size
norman	\(\left ({\mathrm e}^{2}+6\right ) \ln \left (\frac {8}{3 x +13}\right )-\frac {26 x}{5}-2 \ln \left (\frac {8}{3 x +13}\right ) x\)	\(33\)
risch	\(-2 \ln \left (\frac {8}{3 x +13}\right ) x -\frac {26 x}{5}-6 \ln \left (3 x +13\right )-\ln \left (3 x +13\right ) {\mathrm e}^{2}\)	\(36\)
derivativedivides	\({\mathrm e}^{2} \ln \left (\frac {8}{3 x +13}\right )-\frac {2 \left (3 x +13\right ) \ln \left (\frac {8}{3 x +13}\right )}{3}-\frac {26 x}{5}-\frac {338}{15}+\frac {44 \ln \left (\frac {8}{3 x +13}\right )}{3}\)	\(48\)
default	\({\mathrm e}^{2} \ln \left (\frac {8}{3 x +13}\right )-\frac {2 \left (3 x +13\right ) \ln \left (\frac {8}{3 x +13}\right )}{3}-\frac {26 x}{5}-\frac {338}{15}+\frac {44 \ln \left (\frac {8}{3 x +13}\right )}{3}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-30*x-130)*ln(8/(3*x+13))-15*exp(2)-48*x-428)/(15*x+65),x,method=_RETURNVERBOSE)

[Out]

(exp(2)+6)*ln(8/(3*x+13))-26/5*x-2*ln(8/(3*x+13))*x

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maxima [B]  time = 0.42, size = 70, normalized size = 2.41 \begin {gather*} -e^{2} \log \left (3 \, x + 13\right ) + \frac {13}{3} \, \log \left (3 \, x + 13\right )^{2} - \frac {2}{3} \, {\left (3 \, x - 13 \, \log \left (3 \, x + 13\right )\right )} \log \left (\frac {8}{3 \, x + 13}\right ) + \frac {13}{3} \, \log \left (\frac {8}{3 \, x + 13}\right )^{2} - \frac {26}{5} \, x - 6 \, \log \left (3 \, x + 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x-130)*log(8/(3*x+13))-15*exp(2)-48*x-428)/(15*x+65),x, algorithm="maxima")

[Out]

-e^2*log(3*x + 13) + 13/3*log(3*x + 13)^2 - 2/3*(3*x - 13*log(3*x + 13))*log(8/(3*x + 13)) + 13/3*log(8/(3*x +
13))^2 - 26/5*x - 6*log(3*x + 13)

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mupad [B]  time = 1.55, size = 38, normalized size = 1.31 \begin {gather*} 6\,\ln \left (\frac {1}{3\,x+13}\right )-\frac {26\,x}{5}-2\,x\,\ln \left (\frac {8}{3\,x+13}\right )+\ln \left (\frac {1}{3\,x+13}\right )\,{\mathrm {e}}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x + 15*exp(2) + log(8/(3*x + 13))*(30*x + 130) + 428)/(15*x + 65),x)

[Out]

6*log(1/(3*x + 13)) - (26*x)/5 - 2*x*log(8/(3*x + 13)) + log(1/(3*x + 13))*exp(2)

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sympy [A]  time = 0.19, size = 29, normalized size = 1.00 \begin {gather*} - 2 x \log {\left (\frac {8}{3 x + 13} \right )} - \frac {26 x}{5} - \left (6 + e^{2}\right ) \log {\left (3 x + 13 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-30*x-130)*ln(8/(3*x+13))-15*exp(2)-48*x-428)/(15*x+65),x)

[Out]

-2*x*log(8/(3*x + 13)) - 26*x/5 - (6 + exp(2))*log(3*x + 13)

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