\(\int \frac {9+6 x (i \pi +\log (25))+(-1+x^2) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx\) [1724]

Optimal result

Integrand size = 60, antiderivative size = 23 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {1}{4} \left (-3+x+\frac {1}{x+\frac {3}{i \pi +\log (25)}}\right ) \]

[Out]

1/4*x+1/4/(x+3/(2*ln(5)+I*Pi))-3/4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 6, 12, 1864} \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {x}{4}-\frac {\pi -i \log (25)}{4 (3 i-x (\pi -i \log (25)))} \]

[In]

Int[(9 + 6*x*(I*Pi + Log[25]) + (-1 + x^2)*(I*Pi + Log[25])^2)/(36 + 24*x*(I*Pi + Log[25]) + 4*x^2*(I*Pi + Log
[25])^2),x]

[Out]

x/4 - (Pi - I*Log[25])/(4*(3*I - x*(Pi - I*Log[25])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int -\frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{4 (-3 i+\pi x-i x \log (25))^2} \, dx \\ & = \int \frac {-9-6 x (i \pi +\log (25))-\left (-1+x^2\right ) (i \pi +\log (25))^2}{4 (-3 i+x (\pi -i \log (25)))^2} \, dx \\ & = \frac {1}{4} \int \frac {-9-6 x (i \pi +\log (25))-\left (-1+x^2\right ) (i \pi +\log (25))^2}{(-3 i+x (\pi -i \log (25)))^2} \, dx \\ & = \frac {1}{4} \int \left (1-\frac {(\pi -i \log (25))^2}{(3 i-x (\pi -i \log (25)))^2}\right ) \, dx \\ & = \frac {x}{4}-\frac {\pi -i \log (25)}{4 (3 i-x (\pi -i \log (25)))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {1}{4} \left (x-\frac {-\pi +i \log (25)}{-3 i+\pi x-i x \log (25)}\right ) \]

[In]

Integrate[(9 + 6*x*(I*Pi + Log[25]) + (-1 + x^2)*(I*Pi + Log[25])^2)/(36 + 24*x*(I*Pi + Log[25]) + 4*x^2*(I*Pi
+ Log[25])^2),x]

[Out]

(x - (-Pi + I*Log[25])/(-3*I + Pi*x - I*x*Log[25]))/4

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78

method	result	size
risch	\(\frac {x}{4}+\frac {\pi }{-8 i x \ln \left (5\right )+4 \pi x -12 i}-\frac {i \ln \left (5\right )}{2 \left (-2 i x \ln \left (5\right )+\pi x -3 i\right )}\)	\(41\)
default	\(\frac {x}{4}-\frac {-\pi ^{2}+4 i \pi \ln \left (5\right )+4 \ln \left (5\right )^{2}}{4 \left (\pi -2 i \ln \left (5\right )\right ) \left (-2 i x \ln \left (5\right )+\pi x -3 i\right )}\)	\(48\)
parallelrisch	\(-\frac {-3 \pi \,x^{2}+i x \,\pi ^{2}-4 i x \ln \left (5\right )^{2}+6 i \ln \left (5\right ) x^{2}+4 \pi \ln \left (5\right ) x +9 i x}{12 \left (-2 i x \ln \left (5\right )+\pi x -3 i\right )}\)	\(57\)
norman	\(\frac {\left (\frac {\pi ^{2}}{4}+\ln \left (5\right )^{2}\right ) x^{3}+\left (\frac {9}{4}+\frac {\pi ^{2}}{4}-\ln \left (5\right )^{2}-i \pi \ln \left (5\right )\right ) x +\left (-\frac {i \pi ^{3}}{12}-\frac {i \pi \ln \left (5\right )^{2}}{3}-\frac {\pi ^{2} \ln \left (5\right )}{6}-\frac {2 \ln \left (5\right )^{3}}{3}+3 \ln \left (5\right )\right ) x^{2}}{\pi ^{2} x^{2}+4 x^{2} \ln \left (5\right )^{2}+12 x \ln \left (5\right )+9}\)	\(99\)
gosper	\(-\frac {\left (-2 i x \ln \left (5\right )+\pi x -3 i\right ) x \left (4 \pi \ln \left (5\right )+i \pi ^{2}-4 i \ln \left (5\right )^{2}+6 i x \ln \left (5\right )-3 \pi x +9 i\right ) \left (-4 i \pi \ln \left (5\right ) x^{2}+\pi ^{2} x^{2}-4 x^{2} \ln \left (5\right )^{2}+4 i \pi \ln \left (5\right )-6 i \pi x -\pi ^{2}+4 \ln \left (5\right )^{2}-12 x \ln \left (5\right )-9\right )}{12 \left (-2 i x \ln \left (5\right )+\pi x +2 i \ln \left (5\right )-\pi -3 i\right ) \left (-2 i x \ln \left (5\right )+\pi x -2 i \ln \left (5\right )+\pi -3 i\right ) \left (-4 i \pi \ln \left (5\right ) x^{2}+\pi ^{2} x^{2}-4 x^{2} \ln \left (5\right )^{2}-6 i \pi x -12 x \ln \left (5\right )-9\right )}\)	\(182\)

[In]

int(((x^2-1)*(2*ln(5)+I*Pi)^2+6*x*(2*ln(5)+I*Pi)+9)/(4*x^2*(2*ln(5)+I*Pi)^2+24*x*(2*ln(5)+I*Pi)+36),x,method=_
RETURNVERBOSE)

[Out]

1/4*x+1/4/(-2*I*x*ln(5)+Pi*x-3*I)*Pi-1/2*I/(-2*I*x*ln(5)+Pi*x-3*I)*ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {i \, \pi + i \, \pi x^{2} + 2 \, {\left (x^{2} + 1\right )} \log \left (5\right ) + 3 \, x}{4 i \, \pi x + 8 \, x \log \left (5\right ) + 12} \]

[In]

integrate(((x^2-1)*(2*log(5)+I*pi)^2+6*x*(2*log(5)+I*pi)+9)/(4*x^2*(2*log(5)+I*pi)^2+24*x*(2*log(5)+I*pi)+36),
x, algorithm="fricas")

[Out]

(I*pi + I*pi*x^2 + 2*(x^2 + 1)*log(5) + 3*x)/(4*I*pi*x + 8*x*log(5) + 12)

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {x}{4} + \frac {2 \log {\left (5 \right )} + i \pi }{x \left (8 \log {\left (5 \right )} + 4 i \pi \right ) + 12} \]

[In]

integrate(((x**2-1)*(2*ln(5)+I*pi)**2+6*x*(2*ln(5)+I*pi)+9)/(4*x**2*(2*ln(5)+I*pi)**2+24*x*(2*ln(5)+I*pi)+36),
x)

[Out]

x/4 + (2*log(5) + I*pi)/(x*(8*log(5) + 4*I*pi) + 12)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {1}{4} \, x + \frac {i \, \pi + 2 \, \log \left (5\right )}{4 \, {\left ({\left (i \, \pi + 2 \, \log \left (5\right )\right )} x + 3\right )}} \]

[In]

integrate(((x^2-1)*(2*log(5)+I*pi)^2+6*x*(2*log(5)+I*pi)+9)/(4*x^2*(2*log(5)+I*pi)^2+24*x*(2*log(5)+I*pi)+36),
x, algorithm="maxima")

[Out]

1/4*x + 1/4*(I*pi + 2*log(5))/((I*pi + 2*log(5))*x + 3)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {\pi ^{2} x - 4 i \, \pi x \log \left (5\right ) - 4 \, x \log \left (5\right )^{2}}{4 \, {\left (\pi ^{2} - 4 i \, \pi \log \left (5\right ) - 4 \, \log \left (5\right )^{2}\right )}} + \frac {-i \, \pi - 2 \, \log \left (5\right )}{4 \, {\left (-i \, \pi x - 2 \, x \log \left (5\right ) - 3\right )}} \]

[In]

integrate(((x^2-1)*(2*log(5)+I*pi)^2+6*x*(2*log(5)+I*pi)+9)/(4*x^2*(2*log(5)+I*pi)^2+24*x*(2*log(5)+I*pi)+36),
x, algorithm="giac")

[Out]

1/4*(pi^2*x - 4*I*pi*x*log(5) - 4*x*log(5)^2)/(pi^2 - 4*I*pi*log(5) - 4*log(5)^2) + 1/4*(-I*pi - 2*log(5))/(-I
*pi*x - 2*x*log(5) - 3)

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {9+6 x (i \pi +\log (25))+\left (-1+x^2\right ) (i \pi +\log (25))^2}{36+24 x (i \pi +\log (25))+4 x^2 (i \pi +\log (25))^2} \, dx=\frac {-\Pi +x\,3{}\mathrm {i}+\ln \left (5\right )\,2{}\mathrm {i}-\Pi \,x^2+x^2\,\ln \left (5\right )\,2{}\mathrm {i}}{4\,\left (-\Pi \,x+x\,\ln \left (5\right )\,2{}\mathrm {i}+3{}\mathrm {i}\right )} \]

[In]

int((6*x*(Pi*1i + 2*log(5)) + (x^2 - 1)*(Pi*1i + 2*log(5))^2 + 9)/(4*x^2*(Pi*1i + 2*log(5))^2 + 24*x*(Pi*1i +
2*log(5)) + 36),x)

[Out]

(x*3i - Pi + log(5)*2i - Pi*x^2 + x^2*log(5)*2i)/(4*(x*log(5)*2i - Pi*x + 3i))