3.37.66 \(\int -\frac {20 e^{5 x}}{3-10 e^{5 x}+7 e^{10 x}} \, dx\)

Optimal. Leaf size=24 \[ \log \left (3+\frac {4 x^2}{x^2-e^{-5 x} x^2}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {12, 2282, 616, 31} \begin {gather*} \log \left (3-7 e^{5 x}\right )-\log \left (1-e^{5 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*E^(5*x))/(3 - 10*E^(5*x) + 7*E^(10*x)),x]

[Out]

Log[3 - 7*E^(5*x)] - Log[1 - E^(5*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 31

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (20 \int \frac {e^{5 x}}{3-10 e^{5 x}+7 e^{10 x}} \, dx\right )\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {1}{3-10 x+7 x^2} \, dx,x,e^{5 x}\right )\right )\\ &=-\left (7 \operatorname {Subst}\left (\int \frac {1}{-7+7 x} \, dx,x,e^{5 x}\right )\right )+7 \operatorname {Subst}\left (\int \frac {1}{-3+7 x} \, dx,x,e^{5 x}\right )\\ &=\log \left (3-7 e^{5 x}\right )-\log \left (1-e^{5 x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.67 \begin {gather*} 2 \tanh ^{-1}\left (\frac {1}{4} \left (-10+14 e^{5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*E^(5*x))/(3 - 10*E^(5*x) + 7*E^(10*x)),x]

[Out]

2*ArcTanh[(-10 + 14*E^(5*x))/4]

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fricas [A]  time = 0.95, size = 19, normalized size = 0.79 \begin {gather*} \log \left (7 \, e^{\left (5 \, x\right )} - 3\right ) - \log \left (e^{\left (5 \, x\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-20*exp(5*x)/(7*exp(5*x)^2-10*exp(5*x)+3),x, algorithm="fricas")

[Out]

log(7*e^(5*x) - 3) - log(e^(5*x) - 1)

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giac [A]  time = 0.25, size = 21, normalized size = 0.88 \begin {gather*} \log \left ({\left | 7 \, e^{\left (5 \, x\right )} - 3 \right |}\right ) - \log \left ({\left | e^{\left (5 \, x\right )} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-20*exp(5*x)/(7*exp(5*x)^2-10*exp(5*x)+3),x, algorithm="giac")

[Out]

log(abs(7*e^(5*x) - 3)) - log(abs(e^(5*x) - 1))

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maple [A]  time = 0.04, size = 18, normalized size = 0.75

method	result	size
risch	\(-\ln \left ({\mathrm e}^{5 x}-1\right )+\ln \left ({\mathrm e}^{5 x}-\frac {3}{7}\right )\)	\(18\)
derivativedivides	\(\ln \left (7 \,{\mathrm e}^{5 x}-3\right )-\ln \left ({\mathrm e}^{5 x}-1\right )\)	\(20\)
default	\(\ln \left (7 \,{\mathrm e}^{5 x}-3\right )-\ln \left ({\mathrm e}^{5 x}-1\right )\)	\(20\)
norman	\(\ln \left (7 \,{\mathrm e}^{5 x}-3\right )-\ln \left ({\mathrm e}^{5 x}-1\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-20*exp(5*x)/(7*exp(5*x)^2-10*exp(5*x)+3),x,method=_RETURNVERBOSE)

[Out]

-ln(exp(5*x)-1)+ln(exp(5*x)-3/7)

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maxima [A]  time = 0.39, size = 19, normalized size = 0.79 \begin {gather*} \log \left (7 \, e^{\left (5 \, x\right )} - 3\right ) - \log \left (e^{\left (5 \, x\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-20*exp(5*x)/(7*exp(5*x)^2-10*exp(5*x)+3),x, algorithm="maxima")

[Out]

log(7*e^(5*x) - 3) - log(e^(5*x) - 1)

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mupad [B]  time = 2.20, size = 19, normalized size = 0.79 \begin {gather*} \ln \left (7\,{\mathrm {e}}^{5\,x}-3\right )-\ln \left ({\mathrm {e}}^{5\,x}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*exp(5*x))/(7*exp(10*x) - 10*exp(5*x) + 3),x)

[Out]

log(7*exp(5*x) - 3) - log(exp(5*x) - 1)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.71 \begin {gather*} - \log {\left (e^{5 x} - 1 \right )} + \log {\left (e^{5 x} - \frac {3}{7} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-20*exp(5*x)/(7*exp(5*x)**2-10*exp(5*x)+3),x)

[Out]

-log(exp(5*x) - 1) + log(exp(5*x) - 3/7)

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