∫ 2x2+(−4x+2x2)(iπ+log(5 4))+(2−4x+x2)(iπ+log(5 4))2+(x3+(−2x2+x3)(iπ+log(5 4))+(x−x2)(iπ+log(5 4))2) log(x2+(−x+x2)(iπ+log(54)) _________________________________ iπ−x+log(54) ) ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ −x3+(2x2−x3)(iπ+log(5 4))+(−x+x2)(iπ+log(5 4))2+(−2x3+(4x2−2x3)(iπ+log(5 4))+(−2x+2x2)(iπ+log(5 4))2) log(x2+(−x+x2)(iπ+log(54)) _________________________________ iπ−x+log(54) )+(−x3+(2x2−x3)(iπ+log(5 4))+(−x+x2)(iπ+log(5 4))2) log2(x2+(−x+x2)(iπ+log(54)) ________________________________

3.41.10 $\int \frac{2 x^{2} + (- 4 x + 2 x^{2}) (i π + \log (\frac{5}{4})) + (2 - 4 x + x^{2}) (i π + \log (\frac{5}{4}))^{2} + (x^{3} + (- 2 x^{2} + x^{3}) (i π + \log (\frac{5}{4})) + (x - x^{2}) (i π + \log (\frac{5}{4}))^{2}) \log (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})})}{- x^{3} + (2 x^{2} - x^{3}) (i π + \log (\frac{5}{4})) + (- x + x^{2}) (i π + \log (\frac{5}{4}))^{2} + (- 2 x^{3} + (4 x^{2} - 2 x^{3}) (i π + \log (\frac{5}{4})) + (- 2 x + 2 x^{2}) (i π + \log (\frac{5}{4}))^{2}) \log (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})}) + (- x^{3} + (2 x^{2} - x^{3}) (i π + \log (\frac{5}{4})) + (- x + x^{2}) (i π + \log (\frac{5}{4}))^{2}) \log^{2} (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})})} d x$

Optimal. Leaf size=38 $\frac{2 - x}{1 + \log (- x + \frac{x^{2}}{1 - \frac{x}{i π + \log (\frac{5}{4})}})}$

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Rubi [F] time = 13.69, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{number of rules}{integrand size}$ = 0.000, Rules used = {} $\begin{matrix} \int \frac{2 x^{2} + (- 4 x + 2 x^{2}) (i π + \log (\frac{5}{4})) + (2 - 4 x + x^{2}) {(i π + \log (\frac{5}{4}))}^{2} + (x^{3} + (- 2 x^{2} + x^{3}) (i π + \log (\frac{5}{4})) + (x - x^{2}) {(i π + \log (\frac{5}{4}))}^{2}) \log (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})})}{- x^{3} + (2 x^{2} - x^{3}) (i π + \log (\frac{5}{4})) + (- x + x^{2}) {(i π + \log (\frac{5}{4}))}^{2} + (- 2 x^{3} + (4 x^{2} - 2 x^{3}) (i π + \log (\frac{5}{4})) + (- 2 x + 2 x^{2}) {(i π + \log (\frac{5}{4}))}^{2}) \log (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})}) + (- x^{3} + (2 x^{2} - x^{3}) (i π + \log (\frac{5}{4})) + (- x + x^{2}) {(i π + \log (\frac{5}{4}))}^{2}) \log^{2} (\frac{x^{2} + (- x + x^{2}) (i π + \log (\frac{5}{4}))}{i π - x + \log (\frac{5}{4})})} d x \end{matrix}$

Veriﬁcation is not applicable to the result.

[In]

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

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

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

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

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

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

[Out]

Defer[Int][(-1 - Log[(x*(x + (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^(-1), x] + Defer[Int][(1 + L

og[(x*(x + (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^(-2), x] - 2*Defer[Int][1/(x*(1 + Log[(x*(x +

(-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^2), x] - (2*I)*Defer[Int][1/((-Pi - I*x + I*Log[5/4])*(1

+ Log[(x*(x + (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^2), x] - I*(Pi - I*Log[5/4])^2*(1 + ((Pi -

I*Log[5/4])*(2 + I*Pi + Log[5/4]))/Sqrt[-(Pi - I*Log[5/4])^4])*Defer[Int][1/((-Sqrt[-(Pi - I*Log[5/4])^4] + (2

*I - Pi + I*Log[5/4])*(I*Pi + Log[5/4]) + 2*x*(Pi - I*(1 + Log[5/4])))*(1 + Log[(x*(x + (-1 + x)*(I*Pi + Log[5

/4])))/(I*Pi - x + Log[5/4])])^2), x] - I*(Pi - I*Log[5/4])^2*(1 - ((Pi - I*Log[5/4])*(2 + I*Pi + Log[5/4]))/S

qrt[-(Pi - I*Log[5/4])^4])*Defer[Int][1/((Sqrt[-(Pi - I*Log[5/4])^4] + (2*I - Pi + I*Log[5/4])*(I*Pi + Log[5/4

]) + 2*x*(Pi - I*(1 + Log[5/4])))*(1 + Log[(x*(x + (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^2), x]

- 4*(Pi - I*(1 + Log[5/4]))*Defer[Int][1/((-2*Pi + 2*x*(Pi - I*(1 + Log[5/4])) + I*Log[25/16])*(1 + Log[(x*(x

+ (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])])^2), x]

Rubi steps

$\begin{matrix} \begin{aligned} integral & = \int \frac{2 x^{2} + 2 (- 2 + x) x (i π + \log (\frac{5}{4})) + (2 - 4 x + x^{2}) {(i π + \log (\frac{5}{4}))}^{2} + x (x^{2} + (- 1 + x) {(π - i \log (\frac{5}{4}))}^{2} + (- 2 + x) x (i π + \log (\frac{5}{4}))) \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})}{x (π^{2} + x (π - i \log (\frac{5}{4})) (2 i - π + i \log (\frac{5}{4})) - \log^{2} (\frac{5}{4}) - x^{2} (1 + i π + \log (\frac{5}{4})) - i π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x \\ = \int (\frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} + \frac{(2 - x) (- i π^{2} + i \log^{2} (\frac{5}{4}) - x^{2} (π - i (1 + \log (\frac{5}{4}))) - π \log (\frac{25}{16}) + 2 x (i π^{2} - i \log (\frac{5}{4}) (1 + \log (\frac{5}{4})) + π (1 + \log (\frac{25}{16}))))}{x (i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}}) d x \\ = \int \frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} d x + \int \frac{(2 - x) (- i π^{2} + i \log^{2} (\frac{5}{4}) - x^{2} (π - i (1 + \log (\frac{5}{4}))) - π \log (\frac{25}{16}) + 2 x (i π^{2} - i \log (\frac{5}{4}) (1 + \log (\frac{5}{4})) + π (1 + \log (\frac{25}{16}))))}{x (i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x \\ = \int \frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} d x + \int (\frac{1}{{(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} - \frac{2}{x {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} - \frac{i (- 2 + x) {(π - i \log (\frac{5}{4}))}^{2}}{(i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}}) d x \\ = - (2 \int \frac{1}{x {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x) - (i {(π - i \log (\frac{5}{4}))}^{2}) \int \frac{- 2 + x}{(i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x + \int \frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} d x + \int \frac{1}{{(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x \\ = - (2 \int \frac{1}{x {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x) - (i {(π - i \log (\frac{5}{4}))}^{2}) \int (- \frac{2 i}{(- π^{2} - x (π - i \log (\frac{5}{4})) (2 i - π + i \log (\frac{5}{4})) + \log^{2} (\frac{5}{4}) + x^{2} (1 + i π + \log (\frac{5}{4})) + i π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} + \frac{x}{(i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}}) d x + \int \frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} d x + \int \frac{1}{{(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x \\ = - (2 \int \frac{1}{x {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x) - (i {(π - i \log (\frac{5}{4}))}^{2}) \int \frac{x}{(i π^{2} - i \log^{2} (\frac{5}{4}) + x (2 i - π + i \log (\frac{5}{4})) (i π + \log (\frac{5}{4})) + x^{2} (π - i (1 + \log (\frac{5}{4}))) + π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x - (2 {(π - i \log (\frac{5}{4}))}^{2}) \int \frac{1}{(- π^{2} - x (π - i \log (\frac{5}{4})) (2 i - π + i \log (\frac{5}{4})) + \log^{2} (\frac{5}{4}) + x^{2} (1 + i π + \log (\frac{5}{4})) + i π \log (\frac{25}{16})) {(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x + \int \frac{1}{- 1 - \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})})} d x + \int \frac{1}{{(1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))}^{2}} d x \\ = Rest of rules removed due to large latex content \end{aligned} \end{matrix}$

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Mathematica [B] time = 0.47, size = 143, normalized size = 3.76 $\begin{matrix} - \frac{i (- 2 + x) (- i π + x - \log (\frac{5}{4})) (π (- 1 + x) - i (x - \log (\frac{5}{4}) + x \log (\frac{5}{4})))}{(π^{2} (- 1 + x) + \log^{2} (\frac{5}{4}) + x^{2} (1 + \log (\frac{5}{4})) - x \log (\frac{5}{4}) (2 + \log (\frac{5}{4})) + i π (x^{2} - 2 x (1 + \log (\frac{5}{4})) + \log (\frac{25}{16}))) (1 + \log (\frac{x (x + (- 1 + x) (i π + \log (\frac{5}{4})))}{i π - x + \log (\frac{5}{4})}))} \end{matrix}$

Antiderivative was successfully veriﬁed.

[In]

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

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

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

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

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

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

[Out]

((-I)*(-2 + x)*((-I)*Pi + x - Log[5/4])*(Pi*(-1 + x) - I*(x - Log[5/4] + x*Log[5/4])))/((Pi^2*(-1 + x) + Log[5

/4]^2 + x^2*(1 + Log[5/4]) - x*Log[5/4]*(2 + Log[5/4]) + I*Pi*(x^2 - 2*x*(1 + Log[5/4]) + Log[25/16]))*(1 + Lo

g[(x*(x + (-1 + x)*(I*Pi + Log[5/4])))/(I*Pi - x + Log[5/4])]))

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fricas [A] time = 1.08, size = 46, normalized size = 1.21 $\begin{matrix} - \frac{x - 2}{\log (\frac{(i π + 1) x^{2} - i π x + (x^{2} - x) \log (\frac{5}{4})}{i π - x + \log (\frac{5}{4})}) + 1} \end{matrix}$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

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

g(5/4)+I*pi)^2+(-x^3+2*x^2)*(log(5/4)+I*pi)-x^3),x, algorithm="fricas")

[Out]

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

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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 $\begin{matrix} Timed out \end{matrix}$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

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

g(5/4)+I*pi)^2+(-x^3+2*x^2)*(log(5/4)+I*pi)-x^3),x, algorithm="giac")

[Out]

Timed out

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maple [A] time = 110.66, size = 51, normalized size = 1.34


method	result	size

risch	$- \frac{x - 2}{\ln (\frac{(x^{2} - x) (\ln (5) - 2 \ln (2) + i π) + x^{2}}{\ln (5) - 2 \ln (2) + i π - x}) + 1}$	$51$
norman	$\frac{\frac{4 i π \ln (5) + 2 π^{2} + 2 i π \ln (5)^{2} - 8 i \ln (2) π \ln (5) - 8 i π \ln (2) - 4 \ln (2) π^{2} + 2 π^{2} \ln (5) + 2 i π^{3} + 8 i \ln (2)^{2} π - 16 \ln (2)^{3} + 24 \ln (2)^{2} \ln (5) - 12 \ln (2) \ln (5)^{2} + 2 \ln (5)^{3} + 2 i π + 24 \ln (2)^{2} - 24 \ln (2) \ln (5) + 6 \ln (5)^{2} - 12 \ln (2) + 6 \ln (5) + 2}{2 i π \ln (5) + i π \ln (5)^{2} - 4 i \ln (2) π \ln (5) - 4 i π \ln (2) - 2 \ln (2) π^{2} + π^{2} \ln (5) + i π^{3} + 4 i \ln (2)^{2} π - 8 \ln (2)^{3} + 12 \ln (2)^{2} \ln (5) - 6 \ln (2) \ln (5)^{2} + \ln (5)^{3} + π^{2} + i π + 12 \ln (2)^{2} - 12 \ln (2) \ln (5) + 3 \ln (5)^{2} - 6 \ln (2) + 3 \ln (5) + 1} - x}{\ln (\frac{(x^{2} - x) (\ln (\frac{5}{4}) + i π) + x^{2}}{\ln (\frac{5}{4}) + i π - x}) + 1}$	$283$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

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

^2)*(ln(5/4)+I*Pi)-x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A] time = 18.75, size = 54, normalized size = 1.42 $\begin{matrix} \frac{- i x + 2 i}{\log (i π + (- i π - \log (5) + 2 \log (2) - 1) x + \log (5) - 2 \log (2)) - \log (- i π + x - \log (5) + 2 \log (2)) + \log (x) + 1} \end{matrix}$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

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

g(5/4)+I*pi)^2+(-x^3+2*x^2)*(log(5/4)+I*pi)-x^3),x, algorithm="maxima")

[Out]

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

og(2)) + log(x) + 1)

________________________________________________________________________________________

mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.03 $\begin{matrix} Hanged \end{matrix}$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x^2 - (x - x^2)*(Pi*1i + log(5/4)))/(Pi*1i - x + log(5/4)))*((x - x^2)*(Pi*1i + log(5/4))^2 - (Pi*1

i + log(5/4))*(2*x^2 - x^3) + x^3) - (4*x - 2*x^2)*(Pi*1i + log(5/4)) + (Pi*1i + log(5/4))^2*(x^2 - 4*x + 2) +

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

(x - x^2)*(Pi*1i + log(5/4))^2 - (Pi*1i + log(5/4))*(2*x^2 - x^3) + x^3) - (Pi*1i + log(5/4))*(2*x^2 - x^3) +

x^3 + log((x^2 - (x - x^2)*(Pi*1i + log(5/4)))/(Pi*1i - x + log(5/4)))*((2*x - 2*x^2)*(Pi*1i + log(5/4))^2 - (

Pi*1i + log(5/4))*(4*x^2 - 2*x^3) + 2*x^3)),x)

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 $\begin{matrix} Timed out \end{matrix}$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2+x)*(ln(5/4)+I*pi)**2+(x**3-2*x**2)*(ln(5/4)+I*pi)+x**3)*ln(((x**2-x)*(ln(5/4)+I*pi)+x**2)/(

ln(5/4)+I*pi-x))+(x**2-4*x+2)*(ln(5/4)+I*pi)**2+(2*x**2-4*x)*(ln(5/4)+I*pi)+2*x**2)/(((x**2-x)*(ln(5/4)+I*pi)*

*2+(-x**3+2*x**2)*(ln(5/4)+I*pi)-x**3)*ln(((x**2-x)*(ln(5/4)+I*pi)+x**2)/(ln(5/4)+I*pi-x))**2+((2*x**2-2*x)*(l

n(5/4)+I*pi)**2+(-2*x**3+4*x**2)*(ln(5/4)+I*pi)-2*x**3)*ln(((x**2-x)*(ln(5/4)+I*pi)+x**2)/(ln(5/4)+I*pi-x))+(x

**2-x)*(ln(5/4)+I*pi)**2+(-x**3+2*x**2)*(ln(5/4)+I*pi)-x**3),x)

[Out]

Timed out

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