3.74.84 \(\int \frac {-144-240 x-144 x^2-40 x^3-12 x^4-6 x^5-x^6+e^{10+4 x} (-16+16 x+16 x^2)+e^{5+2 x} (-96-32 x+64 x^2+32 x^3)}{8 x^3+12 x^4+6 x^5+x^6} \, dx\)

Optimal. Leaf size=34 \[ -x+\left (2+\frac {2 \left (-4-\frac {3+e^{5+2 x}}{x}-x\right )}{2+x}\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 1.82, antiderivative size = 139, normalized size of antiderivative = 4.09, number of steps used = 42, number of rules used = 7, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {6741, 6742, 44, 37, 43, 2177, 2178} \begin {gather*} -\frac {3 x^2}{(x+2)^2}+\frac {6 e^{2 x+5}}{x^2}+\frac {e^{4 x+10}}{x^2}+\frac {9}{x^2}-x+\frac {2 e^{2 x+5}}{x+2}+\frac {e^{4 x+10}}{x+2}-\frac {15}{x+2}-\frac {2 e^{2 x+5}}{(x+2)^2}+\frac {e^{4 x+10}}{(x+2)^2}+\frac {13}{(x+2)^2}-\frac {2 e^{2 x+5}}{x}-\frac {e^{4 x+10}}{x}+\frac {3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-144 - 240*x - 144*x^2 - 40*x^3 - 12*x^4 - 6*x^5 - x^6 + E^(10 + 4*x)*(-16 + 16*x + 16*x^2) + E^(5 + 2*x)
*(-96 - 32*x + 64*x^2 + 32*x^3))/(8*x^3 + 12*x^4 + 6*x^5 + x^6),x]

[Out]

9/x^2 + (6*E^(5 + 2*x))/x^2 + E^(10 + 4*x)/x^2 + 3/x - (2*E^(5 + 2*x))/x - E^(10 + 4*x)/x - x + 13/(2 + x)^2 -
(2*E^(5 + 2*x))/(2 + x)^2 + E^(10 + 4*x)/(2 + x)^2 - (3*x^2)/(2 + x)^2 - 15/(2 + x) + (2*E^(5 + 2*x))/(2 + x)
+ E^(10 + 4*x)/(2 + x)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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 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 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 {-144-240 x-144 x^2-40 x^3-12 x^4-6 x^5-x^6+e^{10+4 x} \left (-16+16 x+16 x^2\right )+e^{5+2 x} \left (-96-32 x+64 x^2+32 x^3\right )}{x^3 (2+x)^3} \, dx\\ &=\int \left (-\frac {40}{(2+x)^3}-\frac {144}{x^3 (2+x)^3}-\frac {240}{x^2 (2+x)^3}-\frac {144}{x (2+x)^3}-\frac {12 x}{(2+x)^3}-\frac {6 x^2}{(2+x)^3}-\frac {x^3}{(2+x)^3}+\frac {16 e^{10+4 x} \left (-1+x+x^2\right )}{x^3 (2+x)^3}+\frac {32 e^{5+2 x} \left (-3-x+2 x^2+x^3\right )}{x^3 (2+x)^3}\right ) \, dx\\ &=\frac {20}{(2+x)^2}-6 \int \frac {x^2}{(2+x)^3} \, dx-12 \int \frac {x}{(2+x)^3} \, dx+16 \int \frac {e^{10+4 x} \left (-1+x+x^2\right )}{x^3 (2+x)^3} \, dx+32 \int \frac {e^{5+2 x} \left (-3-x+2 x^2+x^3\right )}{x^3 (2+x)^3} \, dx-144 \int \frac {1}{x^3 (2+x)^3} \, dx-144 \int \frac {1}{x (2+x)^3} \, dx-240 \int \frac {1}{x^2 (2+x)^3} \, dx-\int \frac {x^3}{(2+x)^3} \, dx\\ &=\frac {20}{(2+x)^2}-\frac {3 x^2}{(2+x)^2}-6 \int \left (\frac {4}{(2+x)^3}-\frac {4}{(2+x)^2}+\frac {1}{2+x}\right ) \, dx+16 \int \left (-\frac {e^{10+4 x}}{8 x^3}+\frac {5 e^{10+4 x}}{16 x^2}-\frac {e^{10+4 x}}{4 x}-\frac {e^{10+4 x}}{8 (2+x)^3}+\frac {3 e^{10+4 x}}{16 (2+x)^2}+\frac {e^{10+4 x}}{4 (2+x)}\right ) \, dx+32 \int \left (-\frac {3 e^{5+2 x}}{8 x^3}+\frac {7 e^{5+2 x}}{16 x^2}-\frac {e^{5+2 x}}{8 x}+\frac {e^{5+2 x}}{8 (2+x)^3}-\frac {3 e^{5+2 x}}{16 (2+x)^2}+\frac {e^{5+2 x}}{8 (2+x)}\right ) \, dx-144 \int \left (\frac {1}{8 x^3}-\frac {3}{16 x^2}+\frac {3}{16 x}-\frac {1}{8 (2+x)^3}-\frac {3}{16 (2+x)^2}-\frac {3}{16 (2+x)}\right ) \, dx-144 \int \left (\frac {1}{8 x}-\frac {1}{2 (2+x)^3}-\frac {1}{4 (2+x)^2}-\frac {1}{8 (2+x)}\right ) \, dx-240 \int \left (\frac {1}{8 x^2}-\frac {3}{16 x}+\frac {1}{4 (2+x)^3}+\frac {1}{4 (2+x)^2}+\frac {3}{16 (2+x)}\right ) \, dx-\int \left (1-\frac {8}{(2+x)^3}+\frac {12}{(2+x)^2}-\frac {6}{2+x}\right ) \, dx\\ &=\frac {9}{x^2}+\frac {3}{x}-x+\frac {13}{(2+x)^2}-\frac {3 x^2}{(2+x)^2}-\frac {15}{2+x}-2 \int \frac {e^{10+4 x}}{x^3} \, dx-2 \int \frac {e^{10+4 x}}{(2+x)^3} \, dx+3 \int \frac {e^{10+4 x}}{(2+x)^2} \, dx-4 \int \frac {e^{5+2 x}}{x} \, dx-4 \int \frac {e^{10+4 x}}{x} \, dx+4 \int \frac {e^{5+2 x}}{(2+x)^3} \, dx+4 \int \frac {e^{5+2 x}}{2+x} \, dx+4 \int \frac {e^{10+4 x}}{2+x} \, dx+5 \int \frac {e^{10+4 x}}{x^2} \, dx-6 \int \frac {e^{5+2 x}}{(2+x)^2} \, dx-12 \int \frac {e^{5+2 x}}{x^3} \, dx+14 \int \frac {e^{5+2 x}}{x^2} \, dx\\ &=\frac {9}{x^2}+\frac {6 e^{5+2 x}}{x^2}+\frac {e^{10+4 x}}{x^2}+\frac {3}{x}-\frac {14 e^{5+2 x}}{x}-\frac {5 e^{10+4 x}}{x}-x+\frac {13}{(2+x)^2}-\frac {2 e^{5+2 x}}{(2+x)^2}+\frac {e^{10+4 x}}{(2+x)^2}-\frac {3 x^2}{(2+x)^2}-\frac {15}{2+x}+\frac {6 e^{5+2 x}}{2+x}-\frac {3 e^{10+4 x}}{2+x}-4 e^5 \text {Ei}(2 x)-4 e^{10} \text {Ei}(4 x)+4 e \text {Ei}(2 (2+x))+4 e^2 \text {Ei}(4 (2+x))-4 \int \frac {e^{10+4 x}}{x^2} \, dx+4 \int \frac {e^{5+2 x}}{(2+x)^2} \, dx-4 \int \frac {e^{10+4 x}}{(2+x)^2} \, dx-12 \int \frac {e^{5+2 x}}{x^2} \, dx-12 \int \frac {e^{5+2 x}}{2+x} \, dx+12 \int \frac {e^{10+4 x}}{2+x} \, dx+20 \int \frac {e^{10+4 x}}{x} \, dx+28 \int \frac {e^{5+2 x}}{x} \, dx\\ &=\frac {9}{x^2}+\frac {6 e^{5+2 x}}{x^2}+\frac {e^{10+4 x}}{x^2}+\frac {3}{x}-\frac {2 e^{5+2 x}}{x}-\frac {e^{10+4 x}}{x}-x+\frac {13}{(2+x)^2}-\frac {2 e^{5+2 x}}{(2+x)^2}+\frac {e^{10+4 x}}{(2+x)^2}-\frac {3 x^2}{(2+x)^2}-\frac {15}{2+x}+\frac {2 e^{5+2 x}}{2+x}+\frac {e^{10+4 x}}{2+x}+24 e^5 \text {Ei}(2 x)+16 e^{10} \text {Ei}(4 x)-8 e \text {Ei}(2 (2+x))+16 e^2 \text {Ei}(4 (2+x))+8 \int \frac {e^{5+2 x}}{2+x} \, dx-16 \int \frac {e^{10+4 x}}{x} \, dx-16 \int \frac {e^{10+4 x}}{2+x} \, dx-24 \int \frac {e^{5+2 x}}{x} \, dx\\ &=\frac {9}{x^2}+\frac {6 e^{5+2 x}}{x^2}+\frac {e^{10+4 x}}{x^2}+\frac {3}{x}-\frac {2 e^{5+2 x}}{x}-\frac {e^{10+4 x}}{x}-x+\frac {13}{(2+x)^2}-\frac {2 e^{5+2 x}}{(2+x)^2}+\frac {e^{10+4 x}}{(2+x)^2}-\frac {3 x^2}{(2+x)^2}-\frac {15}{2+x}+\frac {2 e^{5+2 x}}{2+x}+\frac {e^{10+4 x}}{2+x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 56, normalized size = 1.65 \begin {gather*} -\frac {-36-4 e^{10+4 x}-48 x-4 x^2+16 x^3+7 x^4+x^5-8 e^{5+2 x} (3+2 x)}{x^2 (2+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-144 - 240*x - 144*x^2 - 40*x^3 - 12*x^4 - 6*x^5 - x^6 + E^(10 + 4*x)*(-16 + 16*x + 16*x^2) + E^(5
+ 2*x)*(-96 - 32*x + 64*x^2 + 32*x^3))/(8*x^3 + 12*x^4 + 6*x^5 + x^6),x]

[Out]

-((-36 - 4*E^(10 + 4*x) - 48*x - 4*x^2 + 16*x^3 + 7*x^4 + x^5 - 8*E^(5 + 2*x)*(3 + 2*x))/(x^2*(2 + x)^2))

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 62, normalized size = 1.82 \begin {gather*} -\frac {x^{5} + 4 \, x^{4} + 4 \, x^{3} - 16 \, x^{2} - 8 \, {\left (2 \, x + 3\right )} e^{\left (2 \, x + 5\right )} - 48 \, x - 4 \, e^{\left (4 \, x + 10\right )} - 36}{x^{4} + 4 \, x^{3} + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2+16*x-16)*exp(5+2*x)^2+(32*x^3+64*x^2-32*x-96)*exp(5+2*x)-x^6-6*x^5-12*x^4-40*x^3-144*x^2-24
0*x-144)/(x^6+6*x^5+12*x^4+8*x^3),x, algorithm="fricas")

[Out]

-(x^5 + 4*x^4 + 4*x^3 - 16*x^2 - 8*(2*x + 3)*e^(2*x + 5) - 48*x - 4*e^(4*x + 10) - 36)/(x^4 + 4*x^3 + 4*x^2)

________________________________________________________________________________________

giac [B]  time = 0.13, size = 66, normalized size = 1.94 \begin {gather*} -\frac {x^{5} + 4 \, x^{4} + 4 \, x^{3} - 16 \, x^{2} - 16 \, x e^{\left (2 \, x + 5\right )} - 48 \, x - 4 \, e^{\left (4 \, x + 10\right )} - 24 \, e^{\left (2 \, x + 5\right )} - 36}{x^{4} + 4 \, x^{3} + 4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2+16*x-16)*exp(5+2*x)^2+(32*x^3+64*x^2-32*x-96)*exp(5+2*x)-x^6-6*x^5-12*x^4-40*x^3-144*x^2-24
0*x-144)/(x^6+6*x^5+12*x^4+8*x^3),x, algorithm="giac")

[Out]

-(x^5 + 4*x^4 + 4*x^3 - 16*x^2 - 16*x*e^(2*x + 5) - 48*x - 4*e^(4*x + 10) - 24*e^(2*x + 5) - 36)/(x^4 + 4*x^3
+ 4*x^2)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 66, normalized size = 1.94

method	result	size
risch	\(-x +\frac {16 x^{2}+48 x +36}{x^{2} \left (x^{2}+4 x +4\right )}+\frac {4 \,{\mathrm e}^{4 x +10}}{x^{2} \left (2+x \right )^{2}}+\frac {8 \left (2 x +3\right ) {\mathrm e}^{5+2 x}}{x^{2} \left (2+x \right )^{2}}\)	\(66\)
derivativedivides	\(\frac {4}{\left (2 x +4\right )^{2}}-\frac {6}{2 x +4}+\frac {9}{x^{2}}+\frac {3}{x}-\frac {5}{2}-x +\frac {4 \,{\mathrm e}^{5+2 x}}{2 x +4}-\frac {2 \,{\mathrm e}^{5+2 x}}{x}+\frac {6 \,{\mathrm e}^{5+2 x}}{x^{2}}-\frac {8 \,{\mathrm e}^{5+2 x}}{\left (2 x +4\right )^{2}}+\frac {2 \,{\mathrm e}^{4 x +10}}{2 x +4}-\frac {{\mathrm e}^{4 x +10}}{x}+\frac {{\mathrm e}^{4 x +10}}{x^{2}}+\frac {4 \,{\mathrm e}^{4 x +10}}{\left (2 x +4\right )^{2}}\)	\(145\)
default	\(\frac {4}{\left (2 x +4\right )^{2}}-\frac {6}{2 x +4}+\frac {9}{x^{2}}+\frac {3}{x}-\frac {5}{2}-x +\frac {4 \,{\mathrm e}^{5+2 x}}{2 x +4}-\frac {2 \,{\mathrm e}^{5+2 x}}{x}+\frac {6 \,{\mathrm e}^{5+2 x}}{x^{2}}-\frac {8 \,{\mathrm e}^{5+2 x}}{\left (2 x +4\right )^{2}}+\frac {2 \,{\mathrm e}^{4 x +10}}{2 x +4}-\frac {{\mathrm e}^{4 x +10}}{x}+\frac {{\mathrm e}^{4 x +10}}{x^{2}}+\frac {4 \,{\mathrm e}^{4 x +10}}{\left (2 x +4\right )^{2}}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^2+16*x-16)*exp(5+2*x)^2+(32*x^3+64*x^2-32*x-96)*exp(5+2*x)-x^6-6*x^5-12*x^4-40*x^3-144*x^2-240*x-14
4)/(x^6+6*x^5+12*x^4+8*x^3),x,method=_RETURNVERBOSE)

[Out]

-x+(16*x^2+48*x+36)/x^2/(x^2+4*x+4)+4/x^2/(2+x)^2*exp(4*x+10)+8*(2*x+3)/x^2/(2+x)^2*exp(5+2*x)

________________________________________________________________________________________

maxima [B]  time = 0.59, size = 180, normalized size = 5.29 \begin {gather*} -x - \frac {18 \, {\left (3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2\right )}}{x^{4} + 4 \, x^{3} + 4 \, x^{2}} + \frac {30 \, {\left (3 \, x^{2} + 9 \, x + 4\right )}}{x^{3} + 4 \, x^{2} + 4 \, x} + \frac {4 \, {\left (2 \, {\left (2 \, x e^{5} + 3 \, e^{5}\right )} e^{\left (2 \, x\right )} + e^{\left (4 \, x + 10\right )}\right )}}{x^{4} + 4 \, x^{3} + 4 \, x^{2}} + \frac {4 \, {\left (3 \, x + 5\right )}}{x^{2} + 4 \, x + 4} - \frac {12 \, {\left (2 \, x + 3\right )}}{x^{2} + 4 \, x + 4} - \frac {36 \, {\left (x + 3\right )}}{x^{2} + 4 \, x + 4} + \frac {12 \, {\left (x + 1\right )}}{x^{2} + 4 \, x + 4} + \frac {20}{x^{2} + 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2+16*x-16)*exp(5+2*x)^2+(32*x^3+64*x^2-32*x-96)*exp(5+2*x)-x^6-6*x^5-12*x^4-40*x^3-144*x^2-24
0*x-144)/(x^6+6*x^5+12*x^4+8*x^3),x, algorithm="maxima")

[Out]

-x - 18*(3*x^3 + 9*x^2 + 4*x - 2)/(x^4 + 4*x^3 + 4*x^2) + 30*(3*x^2 + 9*x + 4)/(x^3 + 4*x^2 + 4*x) + 4*(2*(2*x
*e^5 + 3*e^5)*e^(2*x) + e^(4*x + 10))/(x^4 + 4*x^3 + 4*x^2) + 4*(3*x + 5)/(x^2 + 4*x + 4) - 12*(2*x + 3)/(x^2
+ 4*x + 4) - 36*(x + 3)/(x^2 + 4*x + 4) + 12*(x + 1)/(x^2 + 4*x + 4) + 20/(x^2 + 4*x + 4)

________________________________________________________________________________________

mupad [B]  time = 4.57, size = 48, normalized size = 1.41 \begin {gather*} \frac {24\,{\mathrm {e}}^{2\,x+5}+4\,{\mathrm {e}}^{4\,x+10}+x\,\left (16\,{\mathrm {e}}^{2\,x+5}+48\right )+16\,x^2+36}{x^2\,{\left (x+2\right )}^2}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(240*x - exp(4*x + 10)*(16*x + 16*x^2 - 16) + exp(2*x + 5)*(32*x - 64*x^2 - 32*x^3 + 96) + 144*x^2 + 40*x
^3 + 12*x^4 + 6*x^5 + x^6 + 144)/(8*x^3 + 12*x^4 + 6*x^5 + x^6),x)

[Out]

(24*exp(2*x + 5) + 4*exp(4*x + 10) + x*(16*exp(2*x + 5) + 48) + 16*x^2 + 36)/(x^2*(x + 2)^2) - x

________________________________________________________________________________________

sympy [B]  time = 0.21, size = 99, normalized size = 2.91 \begin {gather*} - x + \frac {\left (4 x^{4} + 16 x^{3} + 16 x^{2}\right ) e^{4 x + 10} + \left (16 x^{5} + 88 x^{4} + 160 x^{3} + 96 x^{2}\right ) e^{2 x + 5}}{x^{8} + 8 x^{7} + 24 x^{6} + 32 x^{5} + 16 x^{4}} - \frac {- 16 x^{2} - 48 x - 36}{x^{4} + 4 x^{3} + 4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**2+16*x-16)*exp(5+2*x)**2+(32*x**3+64*x**2-32*x-96)*exp(5+2*x)-x**6-6*x**5-12*x**4-40*x**3-14
4*x**2-240*x-144)/(x**6+6*x**5+12*x**4+8*x**3),x)

[Out]

-x + ((4*x**4 + 16*x**3 + 16*x**2)*exp(4*x + 10) + (16*x**5 + 88*x**4 + 160*x**3 + 96*x**2)*exp(2*x + 5))/(x**
8 + 8*x**7 + 24*x**6 + 32*x**5 + 16*x**4) - (-16*x**2 - 48*x - 36)/(x**4 + 4*x**3 + 4*x**2)

________________________________________________________________________________________