3.2.85 \(\int \frac {-90-60 x-10 x^2+(-405-270 x-120 x^2) \log (4)+e^x (-225+75 x+75 x^2) \log (4)}{54 x^2+36 x^3+6 x^4} \, dx\)

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

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Rubi [A]  time = 0.82, antiderivative size = 54, normalized size of antiderivative = 1.86, number of steps used = 19, number of rules used = 8, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1594, 27, 12, 6742, 44, 2177, 2178, 893} \begin {gather*} \frac {5}{3 x}-\frac {25 e^x \log (4)}{6 (x+3)}+\frac {25 \log (4)}{2 (x+3)}+\frac {25 e^x \log (4)}{6 x}+\frac {15 \log (4)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-90 - 60*x - 10*x^2 + (-405 - 270*x - 120*x^2)*Log[4] + E^x*(-225 + 75*x + 75*x^2)*Log[4])/(54*x^2 + 36*x
^3 + 6*x^4),x]

[Out]

5/(3*x) + (15*Log[4])/(2*x) + (25*E^x*Log[4])/(6*x) + (25*Log[4])/(2*(3 + x)) - (25*E^x*Log[4])/(6*(3 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 {-90-60 x-10 x^2+\left (-405-270 x-120 x^2\right ) \log (4)+e^x \left (-225+75 x+75 x^2\right ) \log (4)}{x^2 \left (54+36 x+6 x^2\right )} \, dx\\ &=\int \frac {-90-60 x-10 x^2+\left (-405-270 x-120 x^2\right ) \log (4)+e^x \left (-225+75 x+75 x^2\right ) \log (4)}{6 x^2 (3+x)^2} \, dx\\ &=\frac {1}{6} \int \frac {-90-60 x-10 x^2+\left (-405-270 x-120 x^2\right ) \log (4)+e^x \left (-225+75 x+75 x^2\right ) \log (4)}{x^2 (3+x)^2} \, dx\\ &=\frac {1}{6} \int \left (-\frac {10}{(3+x)^2}-\frac {90}{x^2 (3+x)^2}-\frac {60}{x (3+x)^2}+\frac {75 e^x \left (-3+x+x^2\right ) \log (4)}{x^2 (3+x)^2}-\frac {15 \left (27+18 x+8 x^2\right ) \log (4)}{x^2 (3+x)^2}\right ) \, dx\\ &=\frac {5}{3 (3+x)}-10 \int \frac {1}{x (3+x)^2} \, dx-15 \int \frac {1}{x^2 (3+x)^2} \, dx-\frac {1}{2} (5 \log (4)) \int \frac {27+18 x+8 x^2}{x^2 (3+x)^2} \, dx+\frac {1}{2} (25 \log (4)) \int \frac {e^x \left (-3+x+x^2\right )}{x^2 (3+x)^2} \, dx\\ &=\frac {5}{3 (3+x)}-10 \int \left (\frac {1}{9 x}-\frac {1}{3 (3+x)^2}-\frac {1}{9 (3+x)}\right ) \, dx-15 \int \left (\frac {1}{9 x^2}-\frac {2}{27 x}+\frac {1}{9 (3+x)^2}+\frac {2}{27 (3+x)}\right ) \, dx-\frac {1}{2} (5 \log (4)) \int \left (\frac {3}{x^2}+\frac {5}{(3+x)^2}\right ) \, dx+\frac {1}{2} (25 \log (4)) \int \left (-\frac {e^x}{3 x^2}+\frac {e^x}{3 x}+\frac {e^x}{3 (3+x)^2}-\frac {e^x}{3 (3+x)}\right ) \, dx\\ &=\frac {5}{3 x}+\frac {15 \log (4)}{2 x}+\frac {25 \log (4)}{2 (3+x)}-\frac {1}{6} (25 \log (4)) \int \frac {e^x}{x^2} \, dx+\frac {1}{6} (25 \log (4)) \int \frac {e^x}{x} \, dx+\frac {1}{6} (25 \log (4)) \int \frac {e^x}{(3+x)^2} \, dx-\frac {1}{6} (25 \log (4)) \int \frac {e^x}{3+x} \, dx\\ &=\frac {5}{3 x}+\frac {15 \log (4)}{2 x}+\frac {25 e^x \log (4)}{6 x}+\frac {25 \log (4)}{2 (3+x)}-\frac {25 e^x \log (4)}{6 (3+x)}+\frac {25}{6} \text {Ei}(x) \log (4)-\frac {25 \text {Ei}(3+x) \log (4)}{6 e^3}-\frac {1}{6} (25 \log (4)) \int \frac {e^x}{x} \, dx+\frac {1}{6} (25 \log (4)) \int \frac {e^x}{3+x} \, dx\\ &=\frac {5}{3 x}+\frac {15 \log (4)}{2 x}+\frac {25 e^x \log (4)}{6 x}+\frac {25 \log (4)}{2 (3+x)}-\frac {25 e^x \log (4)}{6 (3+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 33, normalized size = 1.14 \begin {gather*} \frac {5 \left (6+15 e^x \log (4)+x (2+24 \log (4))+6 \log (512)\right )}{6 x (3+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-90 - 60*x - 10*x^2 + (-405 - 270*x - 120*x^2)*Log[4] + E^x*(-225 + 75*x + 75*x^2)*Log[4])/(54*x^2
+ 36*x^3 + 6*x^4),x]

[Out]

(5*(6 + 15*E^x*Log[4] + x*(2 + 24*Log[4]) + 6*Log[512]))/(6*x*(3 + x))

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fricas [A]  time = 0.80, size = 29, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (3 \, {\left (8 \, x + 9\right )} \log \relax (2) + 15 \, e^{x} \log \relax (2) + x + 3\right )}}{3 \, {\left (x^{2} + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(75*x^2+75*x-225)*log(2)*exp(x)+2*(-120*x^2-270*x-405)*log(2)-10*x^2-60*x-90)/(6*x^4+36*x^3+54*x^
2),x, algorithm="fricas")

[Out]

5/3*(3*(8*x + 9)*log(2) + 15*e^x*log(2) + x + 3)/(x^2 + 3*x)

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giac [A]  time = 0.24, size = 29, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (24 \, x \log \relax (2) + 15 \, e^{x} \log \relax (2) + x + 27 \, \log \relax (2) + 3\right )}}{3 \, {\left (x^{2} + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(75*x^2+75*x-225)*log(2)*exp(x)+2*(-120*x^2-270*x-405)*log(2)-10*x^2-60*x-90)/(6*x^4+36*x^3+54*x^
2),x, algorithm="giac")

[Out]

5/3*(24*x*log(2) + 15*e^x*log(2) + x + 27*log(2) + 3)/(x^2 + 3*x)

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maple [A]  time = 0.06, size = 30, normalized size = 1.03

method	result	size
norman	\(\frac {\left (\frac {5}{3}+40 \ln \relax (2)\right ) x +25 \,{\mathrm e}^{x} \ln \relax (2)+5+45 \ln \relax (2)}{\left (3+x \right ) x}\)	\(30\)
risch	\(\frac {\left (\frac {5}{3}+40 \ln \relax (2)\right ) x +5+45 \ln \relax (2)}{\left (3+x \right ) x}+\frac {25 \ln \relax (2) {\mathrm e}^{x}}{\left (3+x \right ) x}\)	\(39\)
default	\(\frac {5}{3 x}+\frac {15 \ln \relax (2)}{x}+\frac {25 \ln \relax (2)}{3+x}+\frac {25 \ln \relax (2) {\mathrm e}^{x}}{3 x}-\frac {25 \ln \relax (2) {\mathrm e}^{x}}{3 \left (3+x \right )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(75*x^2+75*x-225)*ln(2)*exp(x)+2*(-120*x^2-270*x-405)*ln(2)-10*x^2-60*x-90)/(6*x^4+36*x^3+54*x^2),x,met
hod=_RETURNVERBOSE)

[Out]

((5/3+40*ln(2))*x+25*exp(x)*ln(2)+5+45*ln(2))/(3+x)/x

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maxima [B]  time = 0.93, size = 99, normalized size = 3.41 \begin {gather*} 5 \, {\left (\frac {3 \, {\left (2 \, x + 3\right )}}{x^{2} + 3 \, x} - 2 \, \log \left (x + 3\right ) + 2 \, \log \relax (x)\right )} \log \relax (2) - 10 \, {\left (\frac {3}{x + 3} - \log \left (x + 3\right ) + \log \relax (x)\right )} \log \relax (2) + \frac {25 \, e^{x} \log \relax (2)}{x^{2} + 3 \, x} + \frac {5 \, {\left (2 \, x + 3\right )}}{3 \, {\left (x^{2} + 3 \, x\right )}} + \frac {40 \, \log \relax (2)}{x + 3} - \frac {5}{3 \, {\left (x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(75*x^2+75*x-225)*log(2)*exp(x)+2*(-120*x^2-270*x-405)*log(2)-10*x^2-60*x-90)/(6*x^4+36*x^3+54*x^
2),x, algorithm="maxima")

[Out]

5*(3*(2*x + 3)/(x^2 + 3*x) - 2*log(x + 3) + 2*log(x))*log(2) - 10*(3/(x + 3) - log(x + 3) + log(x))*log(2) + 2
5*e^x*log(2)/(x^2 + 3*x) + 5/3*(2*x + 3)/(x^2 + 3*x) + 40*log(2)/(x + 3) - 5/3/(x + 3)

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mupad [B]  time = 0.35, size = 35, normalized size = 1.21 \begin {gather*} \frac {135\,\ln \relax (2)-x^2\,\left (40\,\ln \relax (2)+\frac {5}{3}\right )+75\,{\mathrm {e}}^x\,\ln \relax (2)+15}{3\,x^2+9\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*x + 2*log(2)*(270*x + 120*x^2 + 405) + 10*x^2 - 2*exp(x)*log(2)*(75*x + 75*x^2 - 225) + 90)/(54*x^2 +
36*x^3 + 6*x^4),x)

[Out]

(135*log(2) - x^2*(40*log(2) + 5/3) + 75*exp(x)*log(2) + 15)/(9*x + 3*x^2)

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sympy [A]  time = 0.46, size = 39, normalized size = 1.34 \begin {gather*} - \frac {x \left (- 120 \log {\relax (2 )} - 5\right ) - 135 \log {\relax (2 )} - 15}{3 x^{2} + 9 x} + \frac {25 e^{x} \log {\relax (2 )}}{x^{2} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(75*x**2+75*x-225)*ln(2)*exp(x)+2*(-120*x**2-270*x-405)*ln(2)-10*x**2-60*x-90)/(6*x**4+36*x**3+54
*x**2),x)

[Out]

-(x*(-120*log(2) - 5) - 135*log(2) - 15)/(3*x**2 + 9*x) + 25*exp(x)*log(2)/(x**2 + 3*x)

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