3.87.1 \(\int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx\)

Optimal. Leaf size=26 \[ e^5-\frac {4 e^3 x}{3 (7-i \pi +x-\log (2))} \]

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {4 e^3 (\pi +i (7-\log (2)))}{3 (i x+\pi +i (7-\log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-28*E^3 + 4*E^3*(I*Pi + Log[2]))/(147 + 42*x + 3*x^2 + (-42 - 6*x)*(I*Pi + Log[2]) + 3*(I*Pi + Log[2])^2)
,x]

[Out]

(4*E^3*(Pi + I*(7 - Log[2])))/(3*(Pi + I*x + I*(7 - 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (4 e^3 (7-i \pi -\log (2))\right ) \int \frac {1}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx\right )\\ &=-\left (\left (4 e^3 (7-i \pi -\log (2))\right ) \int \frac {1}{3 x^2-3 (\pi +i (7-\log (2)))^2+6 x (7-i \pi -\log (2))} \, dx\right )\\ &=-\left (\left (4 e^3 (7-i \pi -\log (2))\right ) \int -\frac {1}{3 (7 i+\pi +i x-i \log (2))^2} \, dx\right )\\ &=\frac {1}{3} \left (4 e^3 (7-i \pi -\log (2))\right ) \int \frac {1}{(7 i+\pi +i x-i \log (2))^2} \, dx\\ &=\frac {4 e^3 (\pi +i (7-\log (2)))}{3 (\pi +i x+i (7-\log (2)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.35 \begin {gather*} \frac {4 i e^3 (7-i \pi -\log (2))}{3 (\pi +i (7+x-\log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-28*E^3 + 4*E^3*(I*Pi + Log[2]))/(147 + 42*x + 3*x^2 + (-42 - 6*x)*(I*Pi + Log[2]) + 3*(I*Pi + Log[
2])^2),x]

[Out]

(((4*I)/3)*E^3*(7 - I*Pi - Log[2]))/(Pi + I*(7 + x - Log[2]))

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fricas [A]  time = 0.64, size = 31, normalized size = 1.19 \begin {gather*} -\frac {4 \, {\left ({\left (-i \, \pi + 7\right )} e^{3} - e^{3} \log \relax (2)\right )}}{3 i \, \pi - 3 \, x + 3 \, \log \relax (2) - 21} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)*(log(2)+I*pi)+3*x^2+42*x+147),x, alg
orithm="fricas")

[Out]

-4*((-I*pi + 7)*e^3 - e^3*log(2))/(3*I*pi - 3*x + 3*log(2) - 21)

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giac [A]  time = 0.14, size = 28, normalized size = 1.08 \begin {gather*} -\frac {4 \, {\left ({\left (i \, \pi + \log \relax (2)\right )} e^{3} - 7 \, e^{3}\right )}}{3 \, {\left (-i \, \pi + x - \log \relax (2) + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)*(log(2)+I*pi)+3*x^2+42*x+147),x, alg
orithm="giac")

[Out]

-4/3*((I*pi + log(2))*e^3 - 7*e^3)/(-I*pi + x - log(2) + 7)

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maple [A]  time = 0.70, size = 32, normalized size = 1.23

method	result	size
default	\(-\frac {4 \,{\mathrm e}^{3} \left (\ln \relax (2)+i \pi \right )-28 \,{\mathrm e}^{3}}{3 \left (-i \pi -\ln \relax (2)+x +7\right )}\)	\(32\)
risch	\(\frac {4 \pi \,{\mathrm e}^{3}}{3 \left (-i \ln \relax (2)+i x +\pi +7 i\right )}-\frac {4 i {\mathrm e}^{3} \ln \relax (2)}{3 \left (-i \ln \relax (2)+i x +\pi +7 i\right )}+\frac {28 i {\mathrm e}^{3}}{3 \left (-i \ln \relax (2)+i x +\pi +7 i\right )}\)	\(64\)
norman	\(\frac {\left (-\frac {4 i {\mathrm e}^{3} \pi }{3}-\frac {4 \,{\mathrm e}^{3} \ln \relax (2)}{3}+\frac {28 \,{\mathrm e}^{3}}{3}\right ) x +\frac {4 \,{\mathrm e}^{3} \pi ^{2}}{3}+\frac {4 \ln \relax (2)^{2} {\mathrm e}^{3}}{3}-\frac {56 \,{\mathrm e}^{3} \ln \relax (2)}{3}+\frac {196 \,{\mathrm e}^{3}}{3}}{\pi ^{2}+\ln \relax (2)^{2}-2 x \ln \relax (2)+x^{2}-14 \ln \relax (2)+14 x +49}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(3)*(ln(2)+I*Pi)-28*exp(3))/(3*(ln(2)+I*Pi)^2+(-6*x-42)*(ln(2)+I*Pi)+3*x^2+42*x+147),x,method=_RETUR
NVERBOSE)

[Out]

-1/3*(4*exp(3)*(ln(2)+I*Pi)-28*exp(3))/(-I*Pi-ln(2)+x+7)

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maxima [A]  time = 0.35, size = 28, normalized size = 1.08 \begin {gather*} -\frac {4 \, {\left ({\left (i \, \pi + \log \relax (2)\right )} e^{3} - 7 \, e^{3}\right )}}{3 \, {\left (-i \, \pi + x - \log \relax (2) + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)*(log(2)+I*pi)+3*x^2+42*x+147),x, alg
orithm="maxima")

[Out]

-4/3*((I*pi + log(2))*e^3 - 7*e^3)/(-I*pi + x - log(2) + 7)

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mupad [B]  time = 5.55, size = 31, normalized size = 1.19 \begin {gather*} \frac {\frac {4\,{\mathrm {e}}^3\,\ln \relax (2)}{3}-\frac {28\,{\mathrm {e}}^3}{3}+\frac {\Pi \,{\mathrm {e}}^3\,4{}\mathrm {i}}{3}}{\ln \relax (2)-x-7+\Pi \,1{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(28*exp(3) - 4*exp(3)*(Pi*1i + log(2)))/(42*x - (6*x + 42)*(Pi*1i + log(2)) + 3*x^2 + 3*(Pi*1i + log(2))^
2 + 147),x)

[Out]

((Pi*exp(3)*4i)/3 - (28*exp(3))/3 + (4*exp(3)*log(2))/3)/(Pi*1i - x + log(2) - 7)

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sympy [A]  time = 0.29, size = 36, normalized size = 1.38 \begin {gather*} \frac {- 4 e^{3} \log {\relax (2 )} + 28 e^{3} - 4 i \pi e^{3}}{3 x - 3 \log {\relax (2 )} + 21 - 3 i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(3)*(ln(2)+I*pi)-28*exp(3))/(3*(ln(2)+I*pi)**2+(-6*x-42)*(ln(2)+I*pi)+3*x**2+42*x+147),x)

[Out]

(-4*exp(3)*log(2) + 28*exp(3) - 4*I*pi*exp(3))/(3*x - 3*log(2) + 21 - 3*I*pi)

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