3.61.86 \(\int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} (225 x^2+420 x^3-75 x^4-30 x^5)+(225-30 x-15 x^2) \log (3-x)+(225 x^2-30 x^3-15 x^4) \log (5+x)}{-75 x^2+10 x^3+5 x^4} \, dx\)

Optimal. Leaf size=36 \[ 3 \left (-\frac {x}{5}+\frac {\log (3-x)}{x}-x \left (e^{2 x}-x^2+\log (5+x)\right )\right ) \]

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Rubi [A]  time = 1.99, antiderivative size = 60, normalized size of antiderivative = 1.67, number of steps used = 24, number of rules used = 11, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1594, 6728, 36, 31, 72, 2176, 2194, 2395, 29, 2389, 2295} \begin {gather*} 3 x^3-\frac {3 x}{5}+\frac {3 e^{2 x}}{2}-\frac {3}{2} e^{2 x} (2 x+1)-3 (x+5) \log (x+5)+15 \log (x+5)+\frac {3 \log (3-x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*x + 60*x^2 + 39*x^3 - 693*x^4 + 90*x^5 + 45*x^6 + E^(2*x)*(225*x^2 + 420*x^3 - 75*x^4 - 30*x^5) + (225
- 30*x - 15*x^2)*Log[3 - x] + (225*x^2 - 30*x^3 - 15*x^4)*Log[5 + x])/(-75*x^2 + 10*x^3 + 5*x^4),x]

[Out]

(3*E^(2*x))/2 - (3*x)/5 + 3*x^3 - (3*E^(2*x)*(1 + 2*x))/2 + (3*Log[3 - x])/x + 15*Log[5 + x] - 3*(5 + x)*Log[5
+ 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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1594

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {75 x+60 x^2+39 x^3-693 x^4+90 x^5+45 x^6+e^{2 x} \left (225 x^2+420 x^3-75 x^4-30 x^5\right )+\left (225-30 x-15 x^2\right ) \log (3-x)+\left (225 x^2-30 x^3-15 x^4\right ) \log (5+x)}{x^2 \left (-75+10 x+5 x^2\right )} \, dx\\ &=\int \left (\frac {12}{(-3+x) (5+x)}+\frac {15}{(-3+x) x (5+x)}+\frac {39 x}{5 (-3+x) (5+x)}-\frac {693 x^2}{5 (-3+x) (5+x)}+\frac {18 x^3}{(-3+x) (5+x)}+\frac {9 x^4}{(-3+x) (5+x)}-3 e^{2 x} (1+2 x)-\frac {3 \log (3-x)}{x^2}-3 \log (5+x)\right ) \, dx\\ &=-\left (3 \int e^{2 x} (1+2 x) \, dx\right )-3 \int \frac {\log (3-x)}{x^2} \, dx-3 \int \log (5+x) \, dx+\frac {39}{5} \int \frac {x}{(-3+x) (5+x)} \, dx+9 \int \frac {x^4}{(-3+x) (5+x)} \, dx+12 \int \frac {1}{(-3+x) (5+x)} \, dx+15 \int \frac {1}{(-3+x) x (5+x)} \, dx+18 \int \frac {x^3}{(-3+x) (5+x)} \, dx-\frac {693}{5} \int \frac {x^2}{(-3+x) (5+x)} \, dx\\ &=-\frac {3}{2} e^{2 x} (1+2 x)+\frac {3 \log (3-x)}{x}+\frac {3}{2} \int \frac {1}{-3+x} \, dx-\frac {3}{2} \int \frac {1}{5+x} \, dx+3 \int e^{2 x} \, dx+3 \int \frac {1}{(3-x) x} \, dx-3 \operatorname {Subst}(\int \log (x) \, dx,x,5+x)+\frac {39}{5} \int \left (\frac {3}{8 (-3+x)}+\frac {5}{8 (5+x)}\right ) \, dx+9 \int \left (19+\frac {81}{8 (-3+x)}-2 x+x^2-\frac {625}{8 (5+x)}\right ) \, dx+15 \int \left (\frac {1}{24 (-3+x)}-\frac {1}{15 x}+\frac {1}{40 (5+x)}\right ) \, dx+18 \int \left (-2+\frac {27}{8 (-3+x)}+x+\frac {125}{8 (5+x)}\right ) \, dx-\frac {693}{5} \int \left (1+\frac {9}{8 (-3+x)}-\frac {25}{8 (5+x)}\right ) \, dx\\ &=\frac {3 e^{2 x}}{2}-\frac {3 x}{5}+3 x^3-\frac {3}{2} e^{2 x} (1+2 x)+\log (3-x)+\frac {3 \log (3-x)}{x}-\log (x)+15 \log (5+x)-3 (5+x) \log (5+x)+\int \frac {1}{3-x} \, dx+\int \frac {1}{x} \, dx\\ &=\frac {3 e^{2 x}}{2}-\frac {3 x}{5}+3 x^3-\frac {3}{2} e^{2 x} (1+2 x)+\frac {3 \log (3-x)}{x}+15 \log (5+x)-3 (5+x) \log (5+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 63, normalized size = 1.75 \begin {gather*} \frac {3}{5} \left (\frac {5}{3} \log (3-x)-\frac {5}{3} \log (-3+x)\right )+\frac {3}{5} \left (-x-5 e^{2 x} x+5 x^3+\frac {5 \log (3-x)}{x}-5 x \log (5+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*x + 60*x^2 + 39*x^3 - 693*x^4 + 90*x^5 + 45*x^6 + E^(2*x)*(225*x^2 + 420*x^3 - 75*x^4 - 30*x^5)
+ (225 - 30*x - 15*x^2)*Log[3 - x] + (225*x^2 - 30*x^3 - 15*x^4)*Log[5 + x])/(-75*x^2 + 10*x^3 + 5*x^4),x]

[Out]

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

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fricas [A]  time = 0.59, size = 42, normalized size = 1.17 \begin {gather*} \frac {3 \, {\left (5 \, x^{4} - 5 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} \log \left (x + 5\right ) - x^{2} + 5 \, \log \left (-x + 3\right )\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(3-x)+(-30*x^5-75*x^4+420*x^3+225*x^2)*exp(
2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60*x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 42, normalized size = 1.17 \begin {gather*} \frac {3 \, {\left (5 \, x^{4} - 5 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} \log \left (x + 5\right ) - x^{2} + 5 \, \log \left (-x + 3\right )\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(3-x)+(-30*x^5-75*x^4+420*x^3+225*x^2)*exp(
2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60*x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 42, normalized size = 1.17

method	result	size
risch	\(-3 x \ln \left (5+x \right )+\frac {3 x^{4}-3 \,{\mathrm e}^{2 x} x^{2}-\frac {3 x^{2}}{5}+3 \ln \left (3-x \right )}{x}\)	\(42\)
default	\(-3 x \,{\mathrm e}^{2 x}+\ln \left (-x \right )+\frac {\ln \left (3-x \right ) \left (3-x \right )}{x}+3 x^{3}-\frac {3 x}{5}+15 \ln \left (5+x \right )+\ln \left (x -3\right )-\ln \relax (x )-3 \left (5+x \right ) \ln \left (5+x \right )+15\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^4-30*x^3+225*x^2)*ln(5+x)+(-15*x^2-30*x+225)*ln(3-x)+(-30*x^5-75*x^4+420*x^3+225*x^2)*exp(2*x)+45*
x^6+90*x^5-693*x^4+39*x^3+60*x^2+75*x)/(5*x^4+10*x^3-75*x^2),x,method=_RETURNVERBOSE)

[Out]

-3*x*ln(5+x)+3/5*(5*x^4-5*exp(2*x)*x^2-x^2+5*ln(3-x))/x

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maxima [A]  time = 0.40, size = 42, normalized size = 1.17 \begin {gather*} \frac {3 \, {\left (5 \, x^{4} - 5 \, x^{2} e^{\left (2 \, x\right )} - 5 \, x^{2} \log \left (x + 5\right ) - x^{2} + 5 \, \log \left (-x + 3\right )\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^4-30*x^3+225*x^2)*log(5+x)+(-15*x^2-30*x+225)*log(3-x)+(-30*x^5-75*x^4+420*x^3+225*x^2)*exp(
2*x)+45*x^6+90*x^5-693*x^4+39*x^3+60*x^2+75*x)/(5*x^4+10*x^3-75*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.43, size = 57, normalized size = 1.58 \begin {gather*} 3\,x^3-3\,x\,{\mathrm {e}}^{2\,x}-3\,x\,\ln \left (x+5\right )-\frac {3\,x}{5}+\frac {\ln \left (3-x\right )\,\left (3\,x^3+6\,x^2-45\,x\right )}{x^2\,\left (x-3\right )\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((75*x - log(3 - x)*(30*x + 15*x^2 - 225) - log(x + 5)*(30*x^3 - 225*x^2 + 15*x^4) + exp(2*x)*(225*x^2 + 42
0*x^3 - 75*x^4 - 30*x^5) + 60*x^2 + 39*x^3 - 693*x^4 + 90*x^5 + 45*x^6)/(10*x^3 - 75*x^2 + 5*x^4),x)

[Out]

3*x^3 - 3*x*exp(2*x) - 3*x*log(x + 5) - (3*x)/5 + (log(3 - x)*(6*x^2 - 45*x + 3*x^3))/(x^2*(x - 3)*(x + 5))

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sympy [A]  time = 0.58, size = 34, normalized size = 0.94 \begin {gather*} 3 x^{3} - 3 x e^{2 x} - 3 x \log {\left (x + 5 \right )} - \frac {3 x}{5} + \frac {3 \log {\left (3 - x \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**4-30*x**3+225*x**2)*ln(5+x)+(-15*x**2-30*x+225)*ln(3-x)+(-30*x**5-75*x**4+420*x**3+225*x**2
)*exp(2*x)+45*x**6+90*x**5-693*x**4+39*x**3+60*x**2+75*x)/(5*x**4+10*x**3-75*x**2),x)

[Out]

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

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