3.100.72 \(\int \frac {15-15 x+30 x^2+e^{2 x} (-3-3 x-6 x^2+6 x^3)}{1+x-2 x^2-2 x^3+x^4+x^5} \, dx\)

Optimal. Leaf size=28 \[ -3+\frac {3 \left (-5+e^{2 x}\right )}{x+x^2-\frac {x+x^2}{x^2}} \]

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Rubi [B]  time = 0.84, antiderivative size = 74, normalized size of antiderivative = 2.64, number of steps used = 25, number of rules used = 9, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6741, 6742, 44, 207, 77, 88, 2177, 2178, 2269} \begin {gather*} -\frac {3 e^{2 x}}{4 (x+1)}+\frac {15}{4 (x+1)}+\frac {3 e^{2 x}}{2 (x+1)^2}-\frac {15}{2 (x+1)^2}-\frac {3 e^{2 x}}{4 (1-x)}+\frac {15}{4 (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 - 15*x + 30*x^2 + E^(2*x)*(-3 - 3*x - 6*x^2 + 6*x^3))/(1 + x - 2*x^2 - 2*x^3 + x^4 + x^5),x]

[Out]

15/(4*(1 - x)) - (3*E^(2*x))/(4*(1 - x)) - 15/(2*(1 + x)^2) + (3*E^(2*x))/(2*(1 + x)^2) + 15/(4*(1 + x)) - (3*
E^(2*x))/(4*(1 + x))

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 2269

Int[(F_)^((g_.)*((d_.) + (e_.)*(x_))^(n_.))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[F^(g*(d +
e*x)^n), 1/(a + c*x^2), x], x] /; FreeQ[{F, a, c, d, e, g, n}, x]

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 {15-15 x+30 x^2+e^{2 x} \left (-3-3 x-6 x^2+6 x^3\right )}{(1-x)^2 (1+x)^3} \, dx\\ &=\int \left (\frac {15}{(-1+x)^2 (1+x)^3}-\frac {15 x}{(-1+x)^2 (1+x)^3}+\frac {30 x^2}{(-1+x)^2 (1+x)^3}+\frac {3 e^{2 x} \left (-1-x-2 x^2+2 x^3\right )}{(-1+x)^2 (1+x)^3}\right ) \, dx\\ &=3 \int \frac {e^{2 x} \left (-1-x-2 x^2+2 x^3\right )}{(-1+x)^2 (1+x)^3} \, dx+15 \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx-15 \int \frac {x}{(-1+x)^2 (1+x)^3} \, dx+30 \int \frac {x^2}{(-1+x)^2 (1+x)^3} \, dx\\ &=3 \int \left (-\frac {e^{2 x}}{4 (-1+x)^2}-\frac {e^{2 x}}{(1+x)^3}+\frac {5 e^{2 x}}{4 (1+x)^2}+\frac {e^{2 x}}{-1+x^2}\right ) \, dx+15 \int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}+\frac {1}{4 (1+x)^2}-\frac {3}{8 \left (-1+x^2\right )}\right ) \, dx-15 \int \left (\frac {1}{8 (-1+x)^2}-\frac {1}{4 (1+x)^3}-\frac {1}{8 \left (-1+x^2\right )}\right ) \, dx+30 \int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}-\frac {1}{4 (1+x)^2}+\frac {1}{8 \left (-1+x^2\right )}\right ) \, dx\\ &=\frac {15}{4 (1-x)}-\frac {15}{2 (1+x)^2}+\frac {15}{4 (1+x)}-\frac {3}{4} \int \frac {e^{2 x}}{(-1+x)^2} \, dx+\frac {15}{8} \int \frac {1}{-1+x^2} \, dx-3 \int \frac {e^{2 x}}{(1+x)^3} \, dx+3 \int \frac {e^{2 x}}{-1+x^2} \, dx+\frac {15}{4} \int \frac {e^{2 x}}{(1+x)^2} \, dx+\frac {15}{4} \int \frac {1}{-1+x^2} \, dx-\frac {45}{8} \int \frac {1}{-1+x^2} \, dx\\ &=\frac {15}{4 (1-x)}-\frac {3 e^{2 x}}{4 (1-x)}-\frac {15}{2 (1+x)^2}+\frac {3 e^{2 x}}{2 (1+x)^2}+\frac {15}{4 (1+x)}-\frac {15 e^{2 x}}{4 (1+x)}-\frac {3}{2} \int \frac {e^{2 x}}{-1+x} \, dx-3 \int \frac {e^{2 x}}{(1+x)^2} \, dx+3 \int \left (-\frac {e^{2 x}}{2 (1-x)}-\frac {e^{2 x}}{2 (1+x)}\right ) \, dx+\frac {15}{2} \int \frac {e^{2 x}}{1+x} \, dx\\ &=\frac {15}{4 (1-x)}-\frac {3 e^{2 x}}{4 (1-x)}-\frac {15}{2 (1+x)^2}+\frac {3 e^{2 x}}{2 (1+x)^2}+\frac {15}{4 (1+x)}-\frac {3 e^{2 x}}{4 (1+x)}-\frac {3}{2} e^2 \text {Ei}(-2 (1-x))+\frac {15 \text {Ei}(2 (1+x))}{2 e^2}-\frac {3}{2} \int \frac {e^{2 x}}{1-x} \, dx-\frac {3}{2} \int \frac {e^{2 x}}{1+x} \, dx-6 \int \frac {e^{2 x}}{1+x} \, dx\\ &=\frac {15}{4 (1-x)}-\frac {3 e^{2 x}}{4 (1-x)}-\frac {15}{2 (1+x)^2}+\frac {3 e^{2 x}}{2 (1+x)^2}+\frac {15}{4 (1+x)}-\frac {3 e^{2 x}}{4 (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 20, normalized size = 0.71 \begin {gather*} \frac {3 \left (-5+e^{2 x}\right ) x}{(-1+x) (1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 - 15*x + 30*x^2 + E^(2*x)*(-3 - 3*x - 6*x^2 + 6*x^3))/(1 + x - 2*x^2 - 2*x^3 + x^4 + x^5),x]

[Out]

(3*(-5 + E^(2*x))*x)/((-1 + x)*(1 + x)^2)

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fricas [A]  time = 1.27, size = 25, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (x e^{\left (2 \, x\right )} - 5 \, x\right )}}{x^{3} + x^{2} - x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-6*x^2-3*x-3)*exp(x)^2+30*x^2-15*x+15)/(x^5+x^4-2*x^3-2*x^2+x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 25, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (x e^{\left (2 \, x\right )} - 5 \, x\right )}}{x^{3} + x^{2} - x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-6*x^2-3*x-3)*exp(x)^2+30*x^2-15*x+15)/(x^5+x^4-2*x^3-2*x^2+x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 23, normalized size = 0.82

method	result	size
norman	\(\frac {-15 x +3 x \,{\mathrm e}^{2 x}}{\left (x -1\right ) \left (x +1\right )^{2}}\)	\(23\)
risch	\(-\frac {15 x}{x^{3}+x^{2}-x -1}+\frac {3 x \,{\mathrm e}^{2 x}}{\left (x -1\right ) \left (x +1\right )^{2}}\)	\(35\)
default	\(-\frac {15}{2 \left (x +1\right )^{2}}+\frac {15}{4 \left (x +1\right )}-\frac {15}{4 \left (x -1\right )}-\frac {3 \,{\mathrm e}^{2 x}}{4 \left (x +1\right )}+\frac {3 \,{\mathrm e}^{2 x}}{4 \left (x -1\right )}+\frac {3 \,{\mathrm e}^{2 x}}{2 \left (x +1\right )^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^3-6*x^2-3*x-3)*exp(x)^2+30*x^2-15*x+15)/(x^5+x^4-2*x^3-2*x^2+x+1),x,method=_RETURNVERBOSE)

[Out]

(-15*x+3*x*exp(x)^2)/(x-1)/(x+1)^2

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maxima [B]  time = 0.41, size = 90, normalized size = 3.21 \begin {gather*} \frac {3 \, x e^{\left (2 \, x\right )}}{x^{3} + x^{2} - x - 1} - \frac {15 \, {\left (3 \, x^{2} + 3 \, x - 2\right )}}{8 \, {\left (x^{3} + x^{2} - x - 1\right )}} + \frac {15 \, {\left (x^{2} + x + 2\right )}}{8 \, {\left (x^{3} + x^{2} - x - 1\right )}} + \frac {15 \, {\left (x^{2} - 3 \, x - 2\right )}}{4 \, {\left (x^{3} + x^{2} - x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^3-6*x^2-3*x-3)*exp(x)^2+30*x^2-15*x+15)/(x^5+x^4-2*x^3-2*x^2+x+1),x, algorithm="maxima")

[Out]

3*x*e^(2*x)/(x^3 + x^2 - x - 1) - 15/8*(3*x^2 + 3*x - 2)/(x^3 + x^2 - x - 1) + 15/8*(x^2 + x + 2)/(x^3 + x^2 -
x - 1) + 15/4*(x^2 - 3*x - 2)/(x^3 + x^2 - x - 1)

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mupad [B]  time = 8.46, size = 26, normalized size = 0.93 \begin {gather*} -\frac {x\,\left (3\,{\mathrm {e}}^{2\,x}-15\right )}{-x^3-x^2+x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + exp(2*x)*(3*x + 6*x^2 - 6*x^3 + 3) - 30*x^2 - 15)/(x - 2*x^2 - 2*x^3 + x^4 + x^5 + 1),x)

[Out]

-(x*(3*exp(2*x) - 15))/(x - x^2 - x^3 + 1)

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sympy [A]  time = 0.14, size = 31, normalized size = 1.11 \begin {gather*} \frac {3 x e^{2 x}}{x^{3} + x^{2} - x - 1} - \frac {15 x}{x^{3} + x^{2} - x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**3-6*x**2-3*x-3)*exp(x)**2+30*x**2-15*x+15)/(x**5+x**4-2*x**3-2*x**2+x+1),x)

[Out]

3*x*exp(2*x)/(x**3 + x**2 - x - 1) - 15*x/(x**3 + x**2 - x - 1)

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