3.37.26 \(\int \frac {625+75 e^{2 x}-5 e^{3 x}+e^x (-575+80 x-8 x^2)}{-3125+1250 x-125 x^2+e^{2 x} (-375+150 x-15 x^2)+e^{3 x} (25-10 x+x^2)+e^x (1875-750 x+75 x^2)} \, dx\)

Optimal. Leaf size=19 \[ \frac {4}{\left (5-e^x\right )^2}+\frac {x}{-5+x} \]

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Rubi [A]  time = 0.50, antiderivative size = 21, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 4, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {6688, 6742, 2282, 44} \begin {gather*} \frac {4}{\left (5-e^x\right )^2}-\frac {5}{5-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(625 + 75*E^(2*x) - 5*E^(3*x) + E^x*(-575 + 80*x - 8*x^2))/(-3125 + 1250*x - 125*x^2 + E^(2*x)*(-375 + 150
*x - 15*x^2) + E^(3*x)*(25 - 10*x + x^2) + E^x*(1875 - 750*x + 75*x^2)),x]

[Out]

4/(5 - E^x)^2 - 5/(5 - 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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {-625-75 e^{2 x}+5 e^{3 x}-e^x \left (-575+80 x-8 x^2\right )}{\left (5-e^x\right )^3 (5-x)^2} \, dx\\ &=\int \left (-\frac {40}{\left (-5+e^x\right )^3}-\frac {8}{\left (-5+e^x\right )^2}-\frac {5}{(-5+x)^2}\right ) \, dx\\ &=-\frac {5}{5-x}-8 \int \frac {1}{\left (-5+e^x\right )^2} \, dx-40 \int \frac {1}{\left (-5+e^x\right )^3} \, dx\\ &=-\frac {5}{5-x}-8 \operatorname {Subst}\left (\int \frac {1}{(-5+x)^2 x} \, dx,x,e^x\right )-40 \operatorname {Subst}\left (\int \frac {1}{(-5+x)^3 x} \, dx,x,e^x\right )\\ &=-\frac {5}{5-x}-8 \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx,x,e^x\right )-40 \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x)^3}-\frac {1}{25 (-5+x)^2}+\frac {1}{125 (-5+x)}-\frac {1}{125 x}\right ) \, dx,x,e^x\right )\\ &=\frac {4}{\left (5-e^x\right )^2}-\frac {5}{5-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 17, normalized size = 0.89 \begin {gather*} \frac {4}{\left (-5+e^x\right )^2}+\frac {5}{-5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(625 + 75*E^(2*x) - 5*E^(3*x) + E^x*(-575 + 80*x - 8*x^2))/(-3125 + 1250*x - 125*x^2 + E^(2*x)*(-375
+ 150*x - 15*x^2) + E^(3*x)*(25 - 10*x + x^2) + E^x*(1875 - 750*x + 75*x^2)),x]

[Out]

4/(-5 + E^x)^2 + 5/(-5 + x)

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fricas [B]  time = 0.85, size = 38, normalized size = 2.00 \begin {gather*} \frac {4 \, x + 5 \, e^{\left (2 \, x\right )} - 50 \, e^{x} + 105}{{\left (x - 5\right )} e^{\left (2 \, x\right )} - 10 \, {\left (x - 5\right )} e^{x} + 25 \, x - 125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)^3+75*exp(x)^2+(-8*x^2+80*x-575)*exp(x)+625)/((x^2-10*x+25)*exp(x)^3+(-15*x^2+150*x-375)*e
xp(x)^2+(75*x^2-750*x+1875)*exp(x)-125*x^2+1250*x-3125),x, algorithm="fricas")

[Out]

(4*x + 5*e^(2*x) - 50*e^x + 105)/((x - 5)*e^(2*x) - 10*(x - 5)*e^x + 25*x - 125)

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giac [B]  time = 0.19, size = 44, normalized size = 2.32 \begin {gather*} \frac {4 \, x + 5 \, e^{\left (2 \, x\right )} - 50 \, e^{x} + 105}{x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 25 \, x - 5 \, e^{\left (2 \, x\right )} + 50 \, e^{x} - 125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)^3+75*exp(x)^2+(-8*x^2+80*x-575)*exp(x)+625)/((x^2-10*x+25)*exp(x)^3+(-15*x^2+150*x-375)*e
xp(x)^2+(75*x^2-750*x+1875)*exp(x)-125*x^2+1250*x-3125),x, algorithm="giac")

[Out]

(4*x + 5*e^(2*x) - 50*e^x + 105)/(x*e^(2*x) - 10*x*e^x + 25*x - 5*e^(2*x) + 50*e^x - 125)

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maple [A]  time = 0.08, size = 17, normalized size = 0.89

method	result	size
risch	\(\frac {5}{x -5}+\frac {4}{\left ({\mathrm e}^{x}-5\right )^{2}}\)	\(17\)
norman	\(\frac {-50 \,{\mathrm e}^{x}+4 x +5 \,{\mathrm e}^{2 x}+105}{\left ({\mathrm e}^{x}-5\right )^{2} \left (x -5\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*exp(x)^3+75*exp(x)^2+(-8*x^2+80*x-575)*exp(x)+625)/((x^2-10*x+25)*exp(x)^3+(-15*x^2+150*x-375)*exp(x)^
2+(75*x^2-750*x+1875)*exp(x)-125*x^2+1250*x-3125),x,method=_RETURNVERBOSE)

[Out]

5/(x-5)+4/(exp(x)-5)^2

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maxima [B]  time = 0.48, size = 38, normalized size = 2.00 \begin {gather*} \frac {4 \, x + 5 \, e^{\left (2 \, x\right )} - 50 \, e^{x} + 105}{{\left (x - 5\right )} e^{\left (2 \, x\right )} - 10 \, {\left (x - 5\right )} e^{x} + 25 \, x - 125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)^3+75*exp(x)^2+(-8*x^2+80*x-575)*exp(x)+625)/((x^2-10*x+25)*exp(x)^3+(-15*x^2+150*x-375)*e
xp(x)^2+(75*x^2-750*x+1875)*exp(x)-125*x^2+1250*x-3125),x, algorithm="maxima")

[Out]

(4*x + 5*e^(2*x) - 50*e^x + 105)/((x - 5)*e^(2*x) - 10*(x - 5)*e^x + 25*x - 125)

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mupad [B]  time = 2.18, size = 27, normalized size = 1.42 \begin {gather*} \frac {4\,x+5\,{\mathrm {e}}^{2\,x}-50\,{\mathrm {e}}^x+105}{{\left ({\mathrm {e}}^x-5\right )}^2\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((75*exp(2*x) - 5*exp(3*x) - exp(x)*(8*x^2 - 80*x + 575) + 625)/(1250*x - exp(2*x)*(15*x^2 - 150*x + 375) +
exp(3*x)*(x^2 - 10*x + 25) + exp(x)*(75*x^2 - 750*x + 1875) - 125*x^2 - 3125),x)

[Out]

(4*x + 5*exp(2*x) - 50*exp(x) + 105)/((exp(x) - 5)^2*(x - 5))

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sympy [A]  time = 0.17, size = 17, normalized size = 0.89 \begin {gather*} \frac {4}{e^{2 x} - 10 e^{x} + 25} + \frac {5}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*exp(x)**3+75*exp(x)**2+(-8*x**2+80*x-575)*exp(x)+625)/((x**2-10*x+25)*exp(x)**3+(-15*x**2+150*x-
375)*exp(x)**2+(75*x**2-750*x+1875)*exp(x)-125*x**2+1250*x-3125),x)

[Out]

4/(exp(2*x) - 10*exp(x) + 25) + 5/(x - 5)

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