3.28.80 \(\int \frac {-15 x+(-135+15 x+15 \log (25)) \log (-9+x+\log (25))+(-18+2 x+2 \log (25)+e^x (-27+3 x+3 \log (25))) \log ^2(-9+x+\log (25))}{(-27+3 x+3 \log (25)) \log ^2(-9+x+\log (25))} \, dx\)

Optimal. Leaf size=24 \[ e^x+\frac {1}{3} \left (5+2 x+\frac {15 x}{\log (-9+x+\log (25))}\right ) \]

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Rubi [A]  time = 0.41, antiderivative size = 44, normalized size of antiderivative = 1.83, number of steps used = 12, number of rules used = 9, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.127, Rules used = {6688, 2194, 2411, 2353, 2297, 2298, 2302, 30, 2389} \begin {gather*} \frac {2 x}{3}+e^x-\frac {5 (-x+9-\log (25))}{\log (x-9+\log (25))}+\frac {5 (9-\log (25))}{\log (x-9+\log (25))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x + (-135 + 15*x + 15*Log[25])*Log[-9 + x + Log[25]] + (-18 + 2*x + 2*Log[25] + E^x*(-27 + 3*x + 3*Lo
g[25]))*Log[-9 + x + Log[25]]^2)/((-27 + 3*x + 3*Log[25])*Log[-9 + x + Log[25]]^2),x]

[Out]

E^x + (2*x)/3 + (5*(9 - Log[25]))/Log[-9 + x + Log[25]] - (5*(9 - x - Log[25]))/Log[-9 + x + Log[25]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

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

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2}{3}+e^x-\frac {5 x}{(-9+x+\log (25)) \log ^2(-9+x+\log (25))}+\frac {5}{\log (-9+x+\log (25))}\right ) \, dx\\ &=\frac {2 x}{3}-5 \int \frac {x}{(-9+x+\log (25)) \log ^2(-9+x+\log (25))} \, dx+5 \int \frac {1}{\log (-9+x+\log (25))} \, dx+\int e^x \, dx\\ &=e^x+\frac {2 x}{3}-5 \operatorname {Subst}\left (\int \frac {9+x-\log (25)}{x \log ^2(x)} \, dx,x,-9+x+\log (25)\right )+5 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-9+x+\log (25)\right )\\ &=e^x+\frac {2 x}{3}+5 \text {li}(-9+x+\log (25))-5 \operatorname {Subst}\left (\int \left (\frac {1}{\log ^2(x)}+\frac {9-\log (25)}{x \log ^2(x)}\right ) \, dx,x,-9+x+\log (25)\right )\\ &=e^x+\frac {2 x}{3}+5 \text {li}(-9+x+\log (25))-5 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-9+x+\log (25)\right )-(5 (9-\log (25))) \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-9+x+\log (25)\right )\\ &=e^x+\frac {2 x}{3}-\frac {5 (9-x-\log (25))}{\log (-9+x+\log (25))}+5 \text {li}(-9+x+\log (25))-5 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-9+x+\log (25)\right )-(5 (9-\log (25))) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-9+x+\log (25))\right )\\ &=e^x+\frac {2 x}{3}+\frac {5 (9-\log (25))}{\log (-9+x+\log (25))}-\frac {5 (9-x-\log (25))}{\log (-9+x+\log (25))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 0.83 \begin {gather*} e^x+\frac {2 x}{3}+\frac {5 x}{\log (-9+x+\log (25))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x + (-135 + 15*x + 15*Log[25])*Log[-9 + x + Log[25]] + (-18 + 2*x + 2*Log[25] + E^x*(-27 + 3*x
+ 3*Log[25]))*Log[-9 + x + Log[25]]^2)/((-27 + 3*x + 3*Log[25])*Log[-9 + x + Log[25]]^2),x]

[Out]

E^x + (2*x)/3 + (5*x)/Log[-9 + x + Log[25]]

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fricas [A]  time = 0.51, size = 33, normalized size = 1.38 \begin {gather*} \frac {{\left (2 \, x + 3 \, e^{x}\right )} \log \left (x + 2 \, \log \relax (5) - 9\right ) + 15 \, x}{3 \, \log \left (x + 2 \, \log \relax (5) - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*log(5)+3*x-27)*exp(x)+4*log(5)+2*x-18)*log(2*log(5)+x-9)^2+(30*log(5)+15*x-135)*log(2*log(5)+x-
9)-15*x)/(6*log(5)+3*x-27)/log(2*log(5)+x-9)^2,x, algorithm="fricas")

[Out]

1/3*((2*x + 3*e^x)*log(x + 2*log(5) - 9) + 15*x)/log(x + 2*log(5) - 9)

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giac [A]  time = 0.28, size = 39, normalized size = 1.62 \begin {gather*} \frac {2 \, x \log \left (x + 2 \, \log \relax (5) - 9\right ) + 3 \, e^{x} \log \left (x + 2 \, \log \relax (5) - 9\right ) + 15 \, x}{3 \, \log \left (x + 2 \, \log \relax (5) - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*log(5)+3*x-27)*exp(x)+4*log(5)+2*x-18)*log(2*log(5)+x-9)^2+(30*log(5)+15*x-135)*log(2*log(5)+x-
9)-15*x)/(6*log(5)+3*x-27)/log(2*log(5)+x-9)^2,x, algorithm="giac")

[Out]

1/3*(2*x*log(x + 2*log(5) - 9) + 3*e^x*log(x + 2*log(5) - 9) + 15*x)/log(x + 2*log(5) - 9)

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maple [A]  time = 0.46, size = 20, normalized size = 0.83

method	result	size
risch	\(\frac {2 x}{3}+{\mathrm e}^{x}+\frac {5 x}{\ln \left (2 \ln \relax (5)+x -9\right )}\)	\(20\)
norman	\(\frac {{\mathrm e}^{x} \ln \left (2 \ln \relax (5)+x -9\right )+5 x +\frac {2 \ln \left (2 \ln \relax (5)+x -9\right ) x}{3}}{\ln \left (2 \ln \relax (5)+x -9\right )}\)	\(38\)
default	\(\frac {2 x}{3}+\frac {10 \ln \relax (5)+5 x -45}{\ln \left (2 \ln \relax (5)+x -9\right )}-\frac {10 \ln \relax (5)}{\ln \left (2 \ln \relax (5)+x -9\right )}+\frac {45}{\ln \left (2 \ln \relax (5)+x -9\right )}+{\mathrm e}^{x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*ln(5)+3*x-27)*exp(x)+4*ln(5)+2*x-18)*ln(2*ln(5)+x-9)^2+(30*ln(5)+15*x-135)*ln(2*ln(5)+x-9)-15*x)/(6*l
n(5)+3*x-27)/ln(2*ln(5)+x-9)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.92, size = 33, normalized size = 1.38 \begin {gather*} \frac {{\left (2 \, x + 3 \, e^{x}\right )} \log \left (x + 2 \, \log \relax (5) - 9\right ) + 15 \, x}{3 \, \log \left (x + 2 \, \log \relax (5) - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*log(5)+3*x-27)*exp(x)+4*log(5)+2*x-18)*log(2*log(5)+x-9)^2+(30*log(5)+15*x-135)*log(2*log(5)+x-
9)-15*x)/(6*log(5)+3*x-27)/log(2*log(5)+x-9)^2,x, algorithm="maxima")

[Out]

1/3*((2*x + 3*e^x)*log(x + 2*log(5) - 9) + 15*x)/log(x + 2*log(5) - 9)

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mupad [B]  time = 2.08, size = 34, normalized size = 1.42 \begin {gather*} \frac {17\,x}{3}+{\mathrm {e}}^x+\frac {5\,x-\ln \left (x+\ln \left (25\right )-9\right )\,\left (5\,x+\ln \left (9765625\right )-45\right )}{\ln \left (x+\ln \left (25\right )-9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 2*log(5) - 9)^2*(2*x + 4*log(5) + exp(x)*(3*x + 6*log(5) - 27) - 18) - 15*x + log(x + 2*log(5) -
9)*(15*x + 30*log(5) - 135))/(log(x + 2*log(5) - 9)^2*(3*x + 6*log(5) - 27)),x)

[Out]

(17*x)/3 + exp(x) + (5*x - log(x + log(25) - 9)*(5*x + log(9765625) - 45))/log(x + log(25) - 9)

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sympy [A]  time = 0.39, size = 20, normalized size = 0.83 \begin {gather*} \frac {2 x}{3} + \frac {5 x}{\log {\left (x - 9 + 2 \log {\relax (5 )} \right )}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*ln(5)+3*x-27)*exp(x)+4*ln(5)+2*x-18)*ln(2*ln(5)+x-9)**2+(30*ln(5)+15*x-135)*ln(2*ln(5)+x-9)-15*
x)/(6*ln(5)+3*x-27)/ln(2*ln(5)+x-9)**2,x)

[Out]

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

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