3.61.85 \(\int \frac {4+i \pi -9 x+2 x^2+\log (3)}{-4+x} \, dx\)

Optimal. Leaf size=22 \[ -x+x^2+(i \pi +\log (3)) \log (4-x) \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {698} \begin {gather*} x^2-x+(\log (3)+i \pi ) \log (4-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + I*Pi - 9*x + 2*x^2 + Log[3])/(-4 + x),x]

[Out]

-x + x^2 + (I*Pi + Log[3])*Log[4 - x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.95 \begin {gather*} -12-x+x^2+(i \pi +\log (3)) \log (-4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + I*Pi - 9*x + 2*x^2 + Log[3])/(-4 + x),x]

[Out]

-12 - x + x^2 + (I*Pi + Log[3])*Log[-4 + x]

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fricas [A]  time = 0.49, size = 18, normalized size = 0.82 \begin {gather*} x^{2} + {\left (i \, \pi + \log \relax (3)\right )} \log \left (x - 4\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+I*pi+2*x^2-9*x+4)/(x-4),x, algorithm="fricas")

[Out]

x^2 + (I*pi + log(3))*log(x - 4) - x

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giac [A]  time = 0.21, size = 19, normalized size = 0.86 \begin {gather*} x^{2} + {\left (i \, \pi + \log \relax (3)\right )} \log \left ({\left | x - 4 \right |}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+I*pi+2*x^2-9*x+4)/(x-4),x, algorithm="giac")

[Out]

x^2 + (I*pi + log(3))*log(abs(x - 4)) - x

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maple [A]  time = 0.39, size = 20, normalized size = 0.91

method	result	size
default	\(x^{2}-x +\left (\ln \relax (3)+i \pi \right ) \ln \left (x -4\right )\)	\(20\)
norman	\(x^{2}-x +\left (\ln \relax (3)+i \pi \right ) \ln \left (x -4\right )\)	\(20\)
risch	\(x^{2}-x +\ln \relax (3) \ln \left (x -4\right )+i \pi \ln \left (x -4\right )\)	\(23\)
meijerg	\(\ln \relax (3) \ln \left (-\frac {x}{4}+1\right )+i \pi \ln \left (-\frac {x}{4}+1\right )+\frac {4 x \left (\frac {3 x}{4}+6\right )}{3}-9 x\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(3)+I*Pi+2*x^2-9*x+4)/(x-4),x,method=_RETURNVERBOSE)

[Out]

x^2-x+(ln(3)+I*Pi)*ln(x-4)

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maxima [A]  time = 0.35, size = 18, normalized size = 0.82 \begin {gather*} x^{2} + {\left (i \, \pi + \log \relax (3)\right )} \log \left (x - 4\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(3)+I*pi+2*x^2-9*x+4)/(x-4),x, algorithm="maxima")

[Out]

x^2 + (I*pi + log(3))*log(x - 4) - x

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mupad [B]  time = 0.09, size = 19, normalized size = 0.86 \begin {gather*} x^2+\ln \left (x-4\right )\,\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*1i - 9*x + log(3) + 2*x^2 + 4)/(x - 4),x)

[Out]

log(x - 4)*(Pi*1i + log(3)) - x + x^2

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sympy [A]  time = 0.14, size = 15, normalized size = 0.68 \begin {gather*} x^{2} - x + \left (\log {\relax (3 )} + i \pi \right ) \log {\left (x - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(3)+I*pi+2*x**2-9*x+4)/(x-4),x)

[Out]

x**2 - x + (log(3) + I*pi)*log(x - 4)

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