3.4.28 \(\int -\frac {55 e^{\frac {5}{2 \log (\frac {1}{4} (20-11 x))}} (i \pi +\log (2))}{(-40+22 x) \log ^2(\frac {1}{4} (20-11 x))} \, dx\)

Optimal. Leaf size=29 \[ e^{\frac {5}{2 \log \left (5 \left (1-\frac {3 x}{4}\right )+x\right )}} (i \pi +\log (2)) \]

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Rubi [A]  time = 0.21, antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6706} \begin {gather*} (\log (2)+i \pi ) e^{\frac {5}{2 \log \left (\frac {1}{4} (20-11 x)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-55*E^(5/(2*Log[(20 - 11*x)/4]))*(I*Pi + Log[2]))/((-40 + 22*x)*Log[(20 - 11*x)/4]^2),x]

[Out]

E^(5/(2*Log[(20 - 11*x)/4]))*(I*Pi + Log[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((55 (i \pi +\log (2))) \int \frac {e^{\frac {5}{2 \log \left (\frac {1}{4} (20-11 x)\right )}}}{(-40+22 x) \log ^2\left (\frac {1}{4} (20-11 x)\right )} \, dx\right )\\ &=e^{\frac {5}{2 \log \left (\frac {1}{4} (20-11 x)\right )}} (i \pi +\log (2))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 0.97 \begin {gather*} i e^{\frac {5}{2 \log \left (5-\frac {11 x}{4}\right )}} (\pi -i \log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-55*E^(5/(2*Log[(20 - 11*x)/4]))*(I*Pi + Log[2]))/((-40 + 22*x)*Log[(20 - 11*x)/4]^2),x]

[Out]

I*E^(5/(2*Log[5 - (11*x)/4]))*(Pi - I*Log[2])

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fricas [A]  time = 0.63, size = 18, normalized size = 0.62 \begin {gather*} {\left (i \, \pi + \log \relax (2)\right )} e^{\left (\frac {5}{2 \, \log \left (-\frac {11}{4} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-55*(log(2)+I*pi)*exp(5/2/log(-11/4*x+5))/(22*x-40)/log(-11/4*x+5)^2,x, algorithm="fricas")

[Out]

(I*pi + log(2))*e^(5/2/log(-11/4*x + 5))

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giac [A]  time = 0.49, size = 18, normalized size = 0.62 \begin {gather*} {\left (i \, \pi + \log \relax (2)\right )} e^{\left (\frac {5}{2 \, \log \left (-\frac {11}{4} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-55*(log(2)+I*pi)*exp(5/2/log(-11/4*x+5))/(22*x-40)/log(-11/4*x+5)^2,x, algorithm="giac")

[Out]

(I*pi + log(2))*e^(5/2/log(-11/4*x + 5))

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maple [A]  time = 0.26, size = 20, normalized size = 0.69

method	result	size
norman	\(\left (\ln \relax (2)+i \pi \right ) {\mathrm e}^{\frac {5}{2 \ln \left (-\frac {11 x}{4}+5\right )}}\)	\(20\)
risch	\(i \pi \,{\mathrm e}^{\frac {5}{2 \ln \left (-\frac {11 x}{4}+5\right )}}+{\mathrm e}^{\frac {5}{2 \ln \left (-\frac {11 x}{4}+5\right )}} \ln \relax (2)\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-55*(ln(2)+I*Pi)*exp(5/2/ln(-11/4*x+5))/(22*x-40)/ln(-11/4*x+5)^2,x,method=_RETURNVERBOSE)

[Out]

(ln(2)+I*Pi)*exp(5/2/ln(-11/4*x+5))

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maxima [A]  time = 0.54, size = 18, normalized size = 0.62 \begin {gather*} {\left (i \, \pi + \log \relax (2)\right )} e^{\left (\frac {5}{2 \, \log \left (-\frac {11}{4} \, x + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-55*(log(2)+I*pi)*exp(5/2/log(-11/4*x+5))/(22*x-40)/log(-11/4*x+5)^2,x, algorithm="maxima")

[Out]

(I*pi + log(2))*e^(5/2/log(-11/4*x + 5))

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mupad [B]  time = 0.89, size = 19, normalized size = 0.66 \begin {gather*} {\mathrm {e}}^{\frac {5}{2\,\ln \left (5-\frac {11\,x}{4}\right )}}\,\left (\ln \relax (2)+\Pi \,1{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(55*exp(5/(2*log(5 - (11*x)/4)))*(Pi*1i + log(2)))/(log(5 - (11*x)/4)^2*(22*x - 40)),x)

[Out]

exp(5/(2*log(5 - (11*x)/4)))*(Pi*1i + log(2))

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sympy [A]  time = 0.54, size = 22, normalized size = 0.76 \begin {gather*} - \left (- \log {\relax (2 )} - i \pi \right ) e^{\frac {5}{2 \log {\left (5 - \frac {11 x}{4} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-55*(ln(2)+I*pi)*exp(5/2/ln(-11/4*x+5))/(22*x-40)/ln(-11/4*x+5)**2,x)

[Out]

-(-log(2) - I*pi)*exp(5/(2*log(5 - 11*x/4)))

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