3.82.37 \(\int \frac {-5+2 x+(3+2 x) (i \pi +\log (2))}{i \pi +\log (2)} \, dx\)

Optimal. Leaf size=23 \[ 9+3 x+x \left (x+\frac {-5+x}{i \pi +\log (2)}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.70, number of steps used = 2, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {12} \begin {gather*} \frac {x^2}{\log (2)+i \pi }+\frac {1}{4} (2 x+3)^2-\frac {5 x}{\log (2)+i \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 2*x + (3 + 2*x)*(I*Pi + Log[2]))/(I*Pi + Log[2]),x]

[Out]

(3 + 2*x)^2/4 - (5*x)/(I*Pi + Log[2]) + x^2/(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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int (-5+2 x+(3+2 x) (i \pi +\log (2))) \, dx}{i \pi +\log (2)}\\ &=\frac {1}{4} (3+2 x)^2-\frac {5 x}{i \pi +\log (2)}+\frac {x^2}{i \pi +\log (2)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 43, normalized size = 1.87 \begin {gather*} \frac {-5 x+3 i \pi x+x^2+i \pi x^2+3 x \log (2)+x^2 \log (2)}{i \pi +\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 2*x + (3 + 2*x)*(I*Pi + Log[2]))/(I*Pi + Log[2]),x]

[Out]

(-5*x + (3*I)*Pi*x + x^2 + I*Pi*x^2 + 3*x*Log[2] + x^2*Log[2])/(I*Pi + Log[2])

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 36, normalized size = 1.57 \begin {gather*} \frac {{\left (i \, \pi + 1\right )} x^{2} + {\left (3 i \, \pi - 5\right )} x + {\left (x^{2} + 3 \, x\right )} \log \relax (2)}{i \, \pi + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*(log(2)+I*pi)+2*x-5)/(log(2)+I*pi),x, algorithm="fricas")

[Out]

((I*pi + 1)*x^2 + (3*I*pi - 5)*x + (x^2 + 3*x)*log(2))/(I*pi + log(2))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 30, normalized size = 1.30 \begin {gather*} \frac {{\left (i \, \pi + \log \relax (2)\right )} {\left (x^{2} + 3 \, x\right )} + x^{2} - 5 \, x}{i \, \pi + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*(log(2)+I*pi)+2*x-5)/(log(2)+I*pi),x, algorithm="giac")

[Out]

((I*pi + log(2))*(x^2 + 3*x) + x^2 - 5*x)/(I*pi + log(2))

________________________________________________________________________________________

maple [A]  time = 0.09, size = 33, normalized size = 1.43

method	result	size
default	\(\frac {\left (\ln \relax (2)+i \pi \right ) \left (x^{2}+3 x \right )+x^{2}-5 x}{\ln \relax (2)+i \pi }\)	\(33\)
norman	\(\frac {\left (\pi ^{2}-i \pi +\ln \relax (2)^{2}+\ln \relax (2)\right ) x^{2}}{\ln \relax (2)^{2}+\pi ^{2}}+\frac {\left (3 \pi ^{2}+5 i \pi +3 \ln \relax (2)^{2}-5 \ln \relax (2)\right ) x}{\ln \relax (2)^{2}+\pi ^{2}}\)	\(62\)
gosper	\(\frac {x \left (-i x \ln \relax (2)+\pi x -3 i \ln \relax (2)-i x +3 \pi +5 i\right ) \left (2 i \pi x +3 i \pi +2 x \ln \relax (2)+3 \ln \relax (2)+2 x -5\right )}{\left (-2 i x \ln \relax (2)+2 \pi x -3 i \ln \relax (2)-2 i x +3 \pi +5 i\right ) \left (\ln \relax (2)+i \pi \right )}\)	\(86\)
risch	\(\frac {i \pi \,x^{2}}{\ln \relax (2)+i \pi }+\frac {3 i \pi x}{\ln \relax (2)+i \pi }+\frac {x^{2} \ln \relax (2)}{\ln \relax (2)+i \pi }+\frac {3 x \ln \relax (2)}{\ln \relax (2)+i \pi }+\frac {x^{2}}{\ln \relax (2)+i \pi }-\frac {5 x}{\ln \relax (2)+i \pi }\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+3)*(ln(2)+I*Pi)+2*x-5)/(ln(2)+I*Pi),x,method=_RETURNVERBOSE)

[Out]

1/(ln(2)+I*Pi)*((ln(2)+I*Pi)*(x^2+3*x)+x^2-5*x)

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 30, normalized size = 1.30 \begin {gather*} \frac {{\left (i \, \pi + \log \relax (2)\right )} {\left (x^{2} + 3 \, x\right )} + x^{2} - 5 \, x}{i \, \pi + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*(log(2)+I*pi)+2*x-5)/(log(2)+I*pi),x, algorithm="maxima")

[Out]

((I*pi + log(2))*(x^2 + 3*x) + x^2 - 5*x)/(I*pi + log(2))

________________________________________________________________________________________

mupad [B]  time = 5.89, size = 48, normalized size = 2.09 \begin {gather*} \frac {x\,\left (3\,\Pi -\ln \relax (2)\,3{}\mathrm {i}+5{}\mathrm {i}\right )}{\Pi -\ln \relax (2)\,1{}\mathrm {i}}-\frac {x^2\,\left (-\Pi +\ln \relax (2)\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{\Pi -\ln \relax (2)\,1{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + (2*x + 3)*(Pi*1i + log(2)) - 5)/(Pi*1i + log(2)),x)

[Out]

(x*(3*Pi - log(2)*3i + 5i))/(Pi - log(2)*1i) - (x^2*(log(2)*1i - Pi + 1i))/(Pi - log(2)*1i)

________________________________________________________________________________________

sympy [B]  time = 0.08, size = 37, normalized size = 1.61 \begin {gather*} \frac {x^{2} \left (\log {\relax (2 )} + 1 + i \pi \right )}{\log {\relax (2 )} + i \pi } + \frac {x \left (-5 + 3 \log {\relax (2 )} + 3 i \pi \right )}{\log {\relax (2 )} + i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*(ln(2)+I*pi)+2*x-5)/(ln(2)+I*pi),x)

[Out]

x**2*(log(2) + 1 + I*pi)/(log(2) + I*pi) + x*(-5 + 3*log(2) + 3*I*pi)/(log(2) + I*pi)

________________________________________________________________________________________