3.66.90 \(\int \frac {50-80 x-465 x^2-820 x^3-330 x^4-360 x^5+e^x (-10-70 x-155 x^2-125 x^3-75 x^4-45 x^5)+(-10-60 x-95 x^2-30 x^3-45 x^4) \log (2+x^2)}{4+24 x+38 x^2+12 x^3+18 x^4} \, dx\)

Optimal. Leaf size=31 \[ \frac {5}{2} x \left (-e^x-4 x+\frac {5}{1+3 x}-\log \left (2+x^2\right )\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 45, normalized size of antiderivative = 1.45, number of steps used = 38, number of rules used = 11, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6742, 2176, 2194, 710, 801, 635, 203, 260, 1629, 2448, 321} \begin {gather*} -10 x^2-\frac {5}{2} x \log \left (x^2+2\right )+\frac {5 e^x}{2}-\frac {5}{2} e^x (x+1)-\frac {25}{6 (3 x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 - 80*x - 465*x^2 - 820*x^3 - 330*x^4 - 360*x^5 + E^x*(-10 - 70*x - 155*x^2 - 125*x^3 - 75*x^4 - 45*x^5
) + (-10 - 60*x - 95*x^2 - 30*x^3 - 45*x^4)*Log[2 + x^2])/(4 + 24*x + 38*x^2 + 12*x^3 + 18*x^4),x]

[Out]

(5*E^x)/2 - 10*x^2 - (5*E^x*(1 + x))/2 - 25/(6*(1 + 3*x)) - (5*x*Log[2 + x^2])/2

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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 \left (-\frac {5}{2} e^x (1+x)+\frac {25}{(1+3 x)^2 \left (2+x^2\right )}-\frac {40 x}{(1+3 x)^2 \left (2+x^2\right )}-\frac {465 x^2}{2 (1+3 x)^2 \left (2+x^2\right )}-\frac {410 x^3}{(1+3 x)^2 \left (2+x^2\right )}-\frac {165 x^4}{(1+3 x)^2 \left (2+x^2\right )}-\frac {180 x^5}{(1+3 x)^2 \left (2+x^2\right )}-\frac {5}{2} \log \left (2+x^2\right )\right ) \, dx\\ &=-\left (\frac {5}{2} \int e^x (1+x) \, dx\right )-\frac {5}{2} \int \log \left (2+x^2\right ) \, dx+25 \int \frac {1}{(1+3 x)^2 \left (2+x^2\right )} \, dx-40 \int \frac {x}{(1+3 x)^2 \left (2+x^2\right )} \, dx-165 \int \frac {x^4}{(1+3 x)^2 \left (2+x^2\right )} \, dx-180 \int \frac {x^5}{(1+3 x)^2 \left (2+x^2\right )} \, dx-\frac {465}{2} \int \frac {x^2}{(1+3 x)^2 \left (2+x^2\right )} \, dx-410 \int \frac {x^3}{(1+3 x)^2 \left (2+x^2\right )} \, dx\\ &=-\frac {5}{2} e^x (1+x)-\frac {75}{19 (1+3 x)}-\frac {5}{2} x \log \left (2+x^2\right )+\frac {25}{19} \int \frac {1-3 x}{(1+3 x) \left (2+x^2\right )} \, dx+\frac {5 \int e^x \, dx}{2}+5 \int \frac {x^2}{2+x^2} \, dx-40 \int \left (-\frac {3}{19 (1+3 x)^2}+\frac {51}{361 (1+3 x)}+\frac {12-17 x}{361 \left (2+x^2\right )}\right ) \, dx-165 \int \left (\frac {1}{9}+\frac {1}{171 (1+3 x)^2}-\frac {74}{3249 (1+3 x)}-\frac {4 (17+6 x)}{361 \left (2+x^2\right )}\right ) \, dx-180 \int \left (-\frac {2}{27}+\frac {x}{9}-\frac {1}{513 (1+3 x)^2}+\frac {31}{3249 (1+3 x)}-\frac {4 (-12+17 x)}{361 \left (2+x^2\right )}\right ) \, dx-\frac {465}{2} \int \left (\frac {1}{19 (1+3 x)^2}-\frac {36}{361 (1+3 x)}+\frac {2 (17+6 x)}{361 \left (2+x^2\right )}\right ) \, dx-410 \int \left (-\frac {1}{57 (1+3 x)^2}+\frac {55}{1083 (1+3 x)}+\frac {2 (-12+17 x)}{361 \left (2+x^2\right )}\right ) \, dx\\ &=\frac {5 e^x}{2}-10 x^2-\frac {5}{2} e^x (1+x)-\frac {25}{6 (1+3 x)}-\frac {150}{361} \log (1+3 x)-\frac {5}{2} x \log \left (2+x^2\right )-\frac {40}{361} \int \frac {12-17 x}{2+x^2} \, dx-\frac {465}{361} \int \frac {17+6 x}{2+x^2} \, dx+\frac {25}{19} \int \left (\frac {18}{19 (1+3 x)}+\frac {-17-6 x}{19 \left (2+x^2\right )}\right ) \, dx+\frac {660}{361} \int \frac {17+6 x}{2+x^2} \, dx+\frac {720}{361} \int \frac {-12+17 x}{2+x^2} \, dx-\frac {820}{361} \int \frac {-12+17 x}{2+x^2} \, dx-10 \int \frac {1}{2+x^2} \, dx\\ &=\frac {5 e^x}{2}-10 x^2-\frac {5}{2} e^x (1+x)-\frac {25}{6 (1+3 x)}-5 \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {5}{2} x \log \left (2+x^2\right )+\frac {25}{361} \int \frac {-17-6 x}{2+x^2} \, dx-\frac {480}{361} \int \frac {1}{2+x^2} \, dx+\frac {680}{361} \int \frac {x}{2+x^2} \, dx-\frac {2790}{361} \int \frac {x}{2+x^2} \, dx+\frac {3960}{361} \int \frac {x}{2+x^2} \, dx-\frac {7905}{361} \int \frac {1}{2+x^2} \, dx-\frac {8640}{361} \int \frac {1}{2+x^2} \, dx+\frac {9840}{361} \int \frac {1}{2+x^2} \, dx+\frac {11220}{361} \int \frac {1}{2+x^2} \, dx+\frac {12240}{361} \int \frac {x}{2+x^2} \, dx-\frac {13940}{361} \int \frac {x}{2+x^2} \, dx\\ &=\frac {5 e^x}{2}-10 x^2-\frac {5}{2} e^x (1+x)-\frac {25}{6 (1+3 x)}-\frac {7905 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{361 \sqrt {2}}+\frac {4165}{361} \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )+\frac {75}{361} \log \left (2+x^2\right )-\frac {5}{2} x \log \left (2+x^2\right )-\frac {150}{361} \int \frac {x}{2+x^2} \, dx-\frac {425}{361} \int \frac {1}{2+x^2} \, dx\\ &=\frac {5 e^x}{2}-10 x^2-\frac {5}{2} e^x (1+x)-\frac {25}{6 (1+3 x)}-\frac {5}{2} x \log \left (2+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.10 \begin {gather*} -\frac {5}{2} \left (e^x x+4 x^2+\frac {5}{3 (1+3 x)}+x \log \left (2+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 - 80*x - 465*x^2 - 820*x^3 - 330*x^4 - 360*x^5 + E^x*(-10 - 70*x - 155*x^2 - 125*x^3 - 75*x^4 -
45*x^5) + (-10 - 60*x - 95*x^2 - 30*x^3 - 45*x^4)*Log[2 + x^2])/(4 + 24*x + 38*x^2 + 12*x^3 + 18*x^4),x]

[Out]

(-5*(E^x*x + 4*x^2 + 5/(3*(1 + 3*x)) + x*Log[2 + x^2]))/2

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fricas [A]  time = 0.68, size = 47, normalized size = 1.52 \begin {gather*} -\frac {5 \, {\left (36 \, x^{3} + 12 \, x^{2} + 3 \, {\left (3 \, x^{2} + x\right )} e^{x} + 3 \, {\left (3 \, x^{2} + x\right )} \log \left (x^{2} + 2\right ) + 5\right )}}{6 \, {\left (3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-45*x^4-30*x^3-95*x^2-60*x-10)*log(x^2+2)+(-45*x^5-75*x^4-125*x^3-155*x^2-70*x-10)*exp(x)-360*x^5-
330*x^4-820*x^3-465*x^2-80*x+50)/(18*x^4+12*x^3+38*x^2+24*x+4),x, algorithm="fricas")

[Out]

-5/6*(36*x^3 + 12*x^2 + 3*(3*x^2 + x)*e^x + 3*(3*x^2 + x)*log(x^2 + 2) + 5)/(3*x + 1)

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giac [B]  time = 0.20, size = 53, normalized size = 1.71 \begin {gather*} -\frac {5 \, {\left (36 \, x^{3} + 9 \, x^{2} e^{x} + 9 \, x^{2} \log \left (x^{2} + 2\right ) + 12 \, x^{2} + 3 \, x e^{x} + 3 \, x \log \left (x^{2} + 2\right ) + 5\right )}}{6 \, {\left (3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-45*x^4-30*x^3-95*x^2-60*x-10)*log(x^2+2)+(-45*x^5-75*x^4-125*x^3-155*x^2-70*x-10)*exp(x)-360*x^5-
330*x^4-820*x^3-465*x^2-80*x+50)/(18*x^4+12*x^3+38*x^2+24*x+4),x, algorithm="giac")

[Out]

-5/6*(36*x^3 + 9*x^2*e^x + 9*x^2*log(x^2 + 2) + 12*x^2 + 3*x*e^x + 3*x*log(x^2 + 2) + 5)/(3*x + 1)

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maple [A]  time = 0.05, size = 30, normalized size = 0.97

method	result	size
default	\(-\frac {5 \,{\mathrm e}^{x} x}{2}-10 x^{2}-\frac {25}{6 \left (3 x +1\right )}-\frac {5 \ln \left (x^{2}+2\right ) x}{2}\)	\(30\)
risch	\(-\frac {5 \ln \left (x^{2}+2\right ) x}{2}-\frac {5 \left (36 x^{3}+9 \,{\mathrm e}^{x} x^{2}+12 x^{2}+3 \,{\mathrm e}^{x} x +5\right )}{6 \left (3 x +1\right )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-45*x^4-30*x^3-95*x^2-60*x-10)*ln(x^2+2)+(-45*x^5-75*x^4-125*x^3-155*x^2-70*x-10)*exp(x)-360*x^5-330*x^4
-820*x^3-465*x^2-80*x+50)/(18*x^4+12*x^3+38*x^2+24*x+4),x,method=_RETURNVERBOSE)

[Out]

-5/2*exp(x)*x-10*x^2-25/6/(3*x+1)-5/2*ln(x^2+2)*x

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maxima [A]  time = 0.53, size = 29, normalized size = 0.94 \begin {gather*} -10 \, x^{2} - \frac {5}{2} \, x e^{x} - \frac {5}{2} \, x \log \left (x^{2} + 2\right ) - \frac {25}{6 \, {\left (3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-45*x^4-30*x^3-95*x^2-60*x-10)*log(x^2+2)+(-45*x^5-75*x^4-125*x^3-155*x^2-70*x-10)*exp(x)-360*x^5-
330*x^4-820*x^3-465*x^2-80*x+50)/(18*x^4+12*x^3+38*x^2+24*x+4),x, algorithm="maxima")

[Out]

-10*x^2 - 5/2*x*e^x - 5/2*x*log(x^2 + 2) - 25/6/(3*x + 1)

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mupad [B]  time = 4.36, size = 29, normalized size = 0.94 \begin {gather*} -\frac {25}{18\,\left (x+\frac {1}{3}\right )}-\frac {5\,x\,\ln \left (x^2+2\right )}{2}-\frac {5\,x\,{\mathrm {e}}^x}{2}-10\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + exp(x)*(70*x + 155*x^2 + 125*x^3 + 75*x^4 + 45*x^5 + 10) + 465*x^2 + 820*x^3 + 330*x^4 + 360*x^5
+ log(x^2 + 2)*(60*x + 95*x^2 + 30*x^3 + 45*x^4 + 10) - 50)/(24*x + 38*x^2 + 12*x^3 + 18*x^4 + 4),x)

[Out]

- 25/(18*(x + 1/3)) - (5*x*log(x^2 + 2))/2 - (5*x*exp(x))/2 - 10*x^2

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sympy [A]  time = 0.44, size = 32, normalized size = 1.03 \begin {gather*} - 10 x^{2} - \frac {5 x e^{x}}{2} - \frac {5 x \log {\left (x^{2} + 2 \right )}}{2} - \frac {25}{18 x + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-45*x**4-30*x**3-95*x**2-60*x-10)*ln(x**2+2)+(-45*x**5-75*x**4-125*x**3-155*x**2-70*x-10)*exp(x)-3
60*x**5-330*x**4-820*x**3-465*x**2-80*x+50)/(18*x**4+12*x**3+38*x**2+24*x+4),x)

[Out]

-10*x**2 - 5*x*exp(x)/2 - 5*x*log(x**2 + 2)/2 - 25/(18*x + 6)

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