3.26.74 \(\int \frac {e^{3+2 x} (-2 x+16 x^2)+e^{3+2 x} (2 x-6 x^2-8 x^3) \log (x-4 x^2)+(-1+4 x+e^{3+2 x} (9-38 x+8 x^2)) \log ^3(x-4 x^2)}{(-1+4 x) \log ^3(x-4 x^2)} \, dx\)

Optimal. Leaf size=31 \[ 4+x-e^{3+2 x} \left (5-x+\frac {x^2}{\log ^2\left (x-4 x^2\right )}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.39, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 + 2*x)*(-2*x + 16*x^2) + E^(3 + 2*x)*(2*x - 6*x^2 - 8*x^3)*Log[x - 4*x^2] + (-1 + 4*x + E^(3 + 2*x)*
(9 - 38*x + 8*x^2))*Log[x - 4*x^2]^3)/((-1 + 4*x)*Log[x - 4*x^2]^3),x]

[Out]

-1/2*E^(3 + 2*x) - (E^(3 + 2*x)*(9 - 2*x))/2 + x + Defer[Int][E^(3 + 2*x)/Log[(1 - 4*x)*x]^3, x]/2 + 4*Defer[I
nt][(E^(3 + 2*x)*x)/Log[(1 - 4*x)*x]^3, x] + Defer[Int][E^(3 + 2*x)/((-1 + 4*x)*Log[(1 - 4*x)*x]^3), x]/2 - 2*
Defer[Int][(E^(3 + 2*x)*x)/Log[(1 - 4*x)*x]^2, x] - 2*Defer[Int][(E^(3 + 2*x)*x^2)/Log[(1 - 4*x)*x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+e^{3+2 x} (-9+2 x)+\frac {2 e^{3+2 x} x (-1+8 x)}{(-1+4 x) \log ^3((1-4 x) x)}-\frac {2 e^{3+2 x} x (1+x)}{\log ^2((1-4 x) x)}\right ) \, dx\\ &=x+2 \int \frac {e^{3+2 x} x (-1+8 x)}{(-1+4 x) \log ^3((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x (1+x)}{\log ^2((1-4 x) x)} \, dx+\int e^{3+2 x} (-9+2 x) \, dx\\ &=-\frac {1}{2} e^{3+2 x} (9-2 x)+x+2 \int \left (\frac {e^{3+2 x}}{4 \log ^3((1-4 x) x)}+\frac {2 e^{3+2 x} x}{\log ^3((1-4 x) x)}+\frac {e^{3+2 x}}{4 (-1+4 x) \log ^3((1-4 x) x)}\right ) \, dx-2 \int \left (\frac {e^{3+2 x} x}{\log ^2((1-4 x) x)}+\frac {e^{3+2 x} x^2}{\log ^2((1-4 x) x)}\right ) \, dx-\int e^{3+2 x} \, dx\\ &=-\frac {1}{2} e^{3+2 x}-\frac {1}{2} e^{3+2 x} (9-2 x)+x+\frac {1}{2} \int \frac {e^{3+2 x}}{\log ^3((1-4 x) x)} \, dx+\frac {1}{2} \int \frac {e^{3+2 x}}{(-1+4 x) \log ^3((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x}{\log ^2((1-4 x) x)} \, dx-2 \int \frac {e^{3+2 x} x^2}{\log ^2((1-4 x) x)} \, dx+4 \int \frac {e^{3+2 x} x}{\log ^3((1-4 x) x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.35, size = 35, normalized size = 1.13 \begin {gather*} e^{3+2 x} (-5+x)+x-\frac {e^{3+2 x} x^2}{\log ^2\left (x-4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + 2*x)*(-2*x + 16*x^2) + E^(3 + 2*x)*(2*x - 6*x^2 - 8*x^3)*Log[x - 4*x^2] + (-1 + 4*x + E^(3 +
2*x)*(9 - 38*x + 8*x^2))*Log[x - 4*x^2]^3)/((-1 + 4*x)*Log[x - 4*x^2]^3),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.95, size = 47, normalized size = 1.52 \begin {gather*} -\frac {x^{2} e^{\left (2 \, x + 3\right )} - {\left ({\left (x - 5\right )} e^{\left (2 \, x + 3\right )} + x\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-38*x+9)*exp(2*x+3)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2+2*x)*exp(2*x+3)*log(-4*x^2+x)+(16*x
^2-2*x)*exp(2*x+3))/(4*x-1)/log(-4*x^2+x)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.22, size = 71, normalized size = 2.29 \begin {gather*} \frac {x e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2} - x^{2} e^{\left (2 \, x + 3\right )} + x \log \left (-4 \, x^{2} + x\right )^{2} - 5 \, e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-38*x+9)*exp(2*x+3)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2+2*x)*exp(2*x+3)*log(-4*x^2+x)+(16*x
^2-2*x)*exp(2*x+3))/(4*x-1)/log(-4*x^2+x)^3,x, algorithm="giac")

[Out]

(x*e^(2*x + 3)*log(-4*x^2 + x)^2 - x^2*e^(2*x + 3) + x*log(-4*x^2 + x)^2 - 5*e^(2*x + 3)*log(-4*x^2 + x)^2)/lo
g(-4*x^2 + x)^2

________________________________________________________________________________________

maple [C]  time = 0.66, size = 134, normalized size = 4.32

method	result	size
risch	\({\mathrm e}^{2 x +3} x +x -5 \,{\mathrm e}^{2 x +3}+\frac {4 x^{2} {\mathrm e}^{2 x +3}}{\left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}-\pi \mathrm {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{3}-2 \pi +2 i \ln \relax (x )+2 i \ln \left (x -\frac {1}{4}\right )\right )^{2}}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2-38*x+9)*exp(2*x+3)+4*x-1)*ln(-4*x^2+x)^3+(-8*x^3-6*x^2+2*x)*exp(2*x+3)*ln(-4*x^2+x)+(16*x^2-2*x)*
exp(2*x+3))/(4*x-1)/ln(-4*x^2+x)^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.72, size = 115, normalized size = 3.71 \begin {gather*} \frac {x \log \relax (x)^{2} + {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} + x\right )} \log \left (-4 \, x + 1\right )^{2} - {\left (x^{2} e^{3} - {\left (x e^{3} - 5 \, e^{3}\right )} \log \relax (x)^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} \log \relax (x) + x \log \relax (x)\right )} \log \left (-4 \, x + 1\right )}{\log \relax (x)^{2} + 2 \, \log \relax (x) \log \left (-4 \, x + 1\right ) + \log \left (-4 \, x + 1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2-38*x+9)*exp(2*x+3)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2+2*x)*exp(2*x+3)*log(-4*x^2+x)+(16*x
^2-2*x)*exp(2*x+3))/(4*x-1)/log(-4*x^2+x)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.65, size = 216, normalized size = 6.97 \begin {gather*} x+\frac {\frac {\left (x-4\,x^2\right )\,\left (x\,{\mathrm {e}}^{2\,x+3}+x^2\,{\mathrm {e}}^{2\,x+3}\right )}{8\,x-1}-\frac {{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (64\,x^5+136\,x^4+10\,x^3-15\,x^2+2\,x\right )}{{\left (8\,x-1\right )}^3}}{\ln \left (x-4\,x^2\right )}-\frac {x^2\,{\mathrm {e}}^{2\,x+3}+\frac {x\,{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (x+1\right )}{8\,x-1}}{{\ln \left (x-4\,x^2\right )}^2}-\frac {{\mathrm {e}}^{2\,x+3}\,\left (\frac {x^7}{2}+\frac {15\,x^6}{16}-\frac {3\,x^5}{16}-\frac {291\,x^4}{256}+\frac {2775\,x^3}{512}-\frac {493\,x^2}{256}+\frac {121\,x}{512}-\frac {5}{512}\right )}{x^3-\frac {3\,x^2}{8}+\frac {3\,x}{64}-\frac {1}{512}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x + 3)*(2*x - 16*x^2) - log(x - 4*x^2)^3*(4*x + exp(2*x + 3)*(8*x^2 - 38*x + 9) - 1) + exp(2*x + 3
)*log(x - 4*x^2)*(6*x^2 - 2*x + 8*x^3))/(log(x - 4*x^2)^3*(4*x - 1)),x)

[Out]

x + (((x - 4*x^2)*(x*exp(2*x + 3) + x^2*exp(2*x + 3)))/(8*x - 1) - (exp(2*x + 3)*log(x - 4*x^2)*(x - 4*x^2)*(2
*x - 15*x^2 + 10*x^3 + 136*x^4 + 64*x^5))/(8*x - 1)^3)/log(x - 4*x^2) - (x^2*exp(2*x + 3) + (x*exp(2*x + 3)*lo
g(x - 4*x^2)*(x - 4*x^2)*(x + 1))/(8*x - 1))/log(x - 4*x^2)^2 - (exp(2*x + 3)*((121*x)/512 - (493*x^2)/256 + (
2775*x^3)/512 - (291*x^4)/256 - (3*x^5)/16 + (15*x^6)/16 + x^7/2 - 5/512))/((3*x)/64 - (3*x^2)/8 + x^3 - 1/512
)

________________________________________________________________________________________

sympy [A]  time = 0.41, size = 44, normalized size = 1.42 \begin {gather*} x + \frac {\left (- x^{2} + x \log {\left (- 4 x^{2} + x \right )}^{2} - 5 \log {\left (- 4 x^{2} + x \right )}^{2}\right ) e^{2 x + 3}}{\log {\left (- 4 x^{2} + x \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2-38*x+9)*exp(2*x+3)+4*x-1)*ln(-4*x**2+x)**3+(-8*x**3-6*x**2+2*x)*exp(2*x+3)*ln(-4*x**2+x)+(
16*x**2-2*x)*exp(2*x+3))/(4*x-1)/ln(-4*x**2+x)**3,x)

[Out]

x + (-x**2 + x*log(-4*x**2 + x)**2 - 5*log(-4*x**2 + x)**2)*exp(2*x + 3)/log(-4*x**2 + x)**2

________________________________________________________________________________________