3.84.81 \(\int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} (-4+2 x+x^2)+e^{1+x} (40 x-20 x^2-10 x^3)+3^x ((-25 x^2-50 x^3-25 x^4) \log (3)+e^{1+x} (5+5 x-5 x^2-5 x^3+(5 x+10 x^2+5 x^3) \log (3)))}{125 x^2+250 x^3+125 x^4+e^{2+2 x} (5+10 x+5 x^2)+e^{1+x} (-50 x-100 x^2-50 x^3)} \, dx\)

Optimal. Leaf size=30 \[ \frac {x}{5}+\frac {1}{1+x}-\frac {3^x x}{-e^{1+x}+5 x} \]

________________________________________________________________________________________

Rubi [F]  time = 2.76, 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 {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-100*x^2 + 50*x^3 + 25*x^4 + E^(2 + 2*x)*(-4 + 2*x + x^2) + E^(1 + x)*(40*x - 20*x^2 - 10*x^3) + 3^x*((-2
5*x^2 - 50*x^3 - 25*x^4)*Log[3] + E^(1 + x)*(5 + 5*x - 5*x^2 - 5*x^3 + (5*x + 10*x^2 + 5*x^3)*Log[3])))/(125*x
^2 + 250*x^3 + 125*x^4 + E^(2 + 2*x)*(5 + 10*x + 5*x^2) + E^(1 + x)*(-50*x - 100*x^2 - 50*x^3)),x]

[Out]

x/5 + (1 + x)^(-1) + Defer[Int][3^x/(E^(1 + x) - 5*x), x] + 5*Defer[Int][(3^x*x)/(E^(1 + x) - 5*x)^2, x] - (1
- Log[3])*Defer[Int][(3^x*x)/(E^(1 + x) - 5*x), x] - 5*Defer[Int][(3^x*x^2)/(E^(1 + x) - 5*x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{5 \left (e^{1+x}-5 x\right )^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{\left (e^{1+x}-5 x\right )^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {25\ 3^x (-1+x) x}{\left (e^{1+x}-5 x\right )^2}+\frac {-4+2 x+x^2}{(1+x)^2}+\frac {5\ 3^x (1-x (1-\log (3)))}{e^{1+x}-5 x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-4+2 x+x^2}{(1+x)^2} \, dx-5 \int \frac {3^x (-1+x) x}{\left (e^{1+x}-5 x\right )^2} \, dx+\int \frac {3^x (1-x (1-\log (3)))}{e^{1+x}-5 x} \, dx\\ &=\frac {1}{5} \int \left (1-\frac {5}{(1+x)^2}\right ) \, dx-5 \int \left (-\frac {3^x x}{\left (e^{1+x}-5 x\right )^2}+\frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2}\right ) \, dx+\int \frac {3^x (1+x (-1+\log (3)))}{e^{1+x}-5 x} \, dx\\ &=\frac {x}{5}+\frac {1}{1+x}+5 \int \frac {3^x x}{\left (e^{1+x}-5 x\right )^2} \, dx-5 \int \frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2} \, dx+\int \left (\frac {3^x}{e^{1+x}-5 x}+\frac {3^x x (-1+\log (3))}{e^{1+x}-5 x}\right ) \, dx\\ &=\frac {x}{5}+\frac {1}{1+x}+5 \int \frac {3^x x}{\left (e^{1+x}-5 x\right )^2} \, dx-5 \int \frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2} \, dx+(-1+\log (3)) \int \frac {3^x x}{e^{1+x}-5 x} \, dx+\int \frac {3^x}{e^{1+x}-5 x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 30, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (x+\frac {5\ 3^x x}{e^{1+x}-5 x}+\frac {5}{1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100*x^2 + 50*x^3 + 25*x^4 + E^(2 + 2*x)*(-4 + 2*x + x^2) + E^(1 + x)*(40*x - 20*x^2 - 10*x^3) + 3^
x*((-25*x^2 - 50*x^3 - 25*x^4)*Log[3] + E^(1 + x)*(5 + 5*x - 5*x^2 - 5*x^3 + (5*x + 10*x^2 + 5*x^3)*Log[3])))/
(125*x^2 + 250*x^3 + 125*x^4 + E^(2 + 2*x)*(5 + 10*x + 5*x^2) + E^(1 + x)*(-50*x - 100*x^2 - 50*x^3)),x]

[Out]

(x + (5*3^x*x)/(E^(1 + x) - 5*x) + 5/(1 + x))/5

________________________________________________________________________________________

fricas [B]  time = 0.89, size = 58, normalized size = 1.93 \begin {gather*} \frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} + x + 5\right )} e^{\left (x + 1\right )} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x + 1\right )} e^{\left (x + 1\right )} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(x+1)+(-25*x^4-50*x^3-25*x^2)*log(3))*exp(x*log(3
))+(x^2+2*x-4)*exp(x+1)^2+(-10*x^3-20*x^2+40*x)*exp(x+1)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(x+1)^2+(-5
0*x^3-100*x^2-50*x)*exp(x+1)+125*x^4+250*x^3+125*x^2),x, algorithm="fricas")

[Out]

1/5*(5*x^3 - 5*(x^2 + x)*3^x + 5*x^2 - (x^2 + x + 5)*e^(x + 1) + 25*x)/(5*x^2 - (x + 1)*e^(x + 1) + 5*x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {25 \, x^{4} + 50 \, x^{3} - 5 \, {\left ({\left (x^{3} + x^{2} - {\left (x^{3} + 2 \, x^{2} + x\right )} \log \relax (3) - x - 1\right )} e^{\left (x + 1\right )} + 5 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \relax (3)\right )} 3^{x} - 100 \, x^{2} + {\left (x^{2} + 2 \, x - 4\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} - 4 \, x\right )} e^{\left (x + 1\right )}}{5 \, {\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{\left (x + 1\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(x+1)+(-25*x^4-50*x^3-25*x^2)*log(3))*exp(x*log(3
))+(x^2+2*x-4)*exp(x+1)^2+(-10*x^3-20*x^2+40*x)*exp(x+1)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(x+1)^2+(-5
0*x^3-100*x^2-50*x)*exp(x+1)+125*x^4+250*x^3+125*x^2),x, algorithm="giac")

[Out]

integrate(1/5*(25*x^4 + 50*x^3 - 5*((x^3 + x^2 - (x^3 + 2*x^2 + x)*log(3) - x - 1)*e^(x + 1) + 5*(x^4 + 2*x^3
+ x^2)*log(3))*3^x - 100*x^2 + (x^2 + 2*x - 4)*e^(2*x + 2) - 10*(x^3 + 2*x^2 - 4*x)*e^(x + 1))/(25*x^4 + 50*x^
3 + 25*x^2 + (x^2 + 2*x + 1)*e^(2*x + 2) - 10*(x^3 + 2*x^2 + x)*e^(x + 1)), x)

________________________________________________________________________________________

maple [B]  time = 0.45, size = 66, normalized size = 2.20

method	result	size
norman	\(\frac {x^{3}+4 x -\frac {4 \,{\mathrm e}^{x +1}}{5}-x \,{\mathrm e}^{x \ln \relax (3)}-x^{2} {\mathrm e}^{x \ln \relax (3)}-\frac {x^{2} {\mathrm e}^{x +1}}{5}}{5 x^{2}-x \,{\mathrm e}^{x +1}+5 x -{\mathrm e}^{x +1}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((5*x^3+10*x^2+5*x)*ln(3)-5*x^3-5*x^2+5*x+5)*exp(x+1)+(-25*x^4-50*x^3-25*x^2)*ln(3))*exp(x*ln(3))+(x^2+2
*x-4)*exp(x+1)^2+(-10*x^3-20*x^2+40*x)*exp(x+1)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(x+1)^2+(-50*x^3-100
*x^2-50*x)*exp(x+1)+125*x^4+250*x^3+125*x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.52, size = 67, normalized size = 2.23 \begin {gather*} \frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} e + x e + 5 \, e\right )} e^{x} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x e + e\right )} e^{x} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((5*x^3+10*x^2+5*x)*log(3)-5*x^3-5*x^2+5*x+5)*exp(x+1)+(-25*x^4-50*x^3-25*x^2)*log(3))*exp(x*log(3
))+(x^2+2*x-4)*exp(x+1)^2+(-10*x^3-20*x^2+40*x)*exp(x+1)+25*x^4+50*x^3-100*x^2)/((5*x^2+10*x+5)*exp(x+1)^2+(-5
0*x^3-100*x^2-50*x)*exp(x+1)+125*x^4+250*x^3+125*x^2),x, algorithm="maxima")

[Out]

1/5*(5*x^3 - 5*(x^2 + x)*3^x + 5*x^2 - (x^2*e + x*e + 5*e)*e^x + 25*x)/(5*x^2 - (x*e + e)*e^x + 5*x)

________________________________________________________________________________________

mupad [B]  time = 5.48, size = 27, normalized size = 0.90 \begin {gather*} \frac {x}{5}+\frac {1}{x+1}-\frac {3^x\,x}{5\,\left (x-\frac {{\mathrm {e}}^{x+1}}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x + 2)*(2*x + x^2 - 4) - exp(x + 1)*(20*x^2 - 40*x + 10*x^3) - 100*x^2 + 50*x^3 + 25*x^4 + exp(x*lo
g(3))*(exp(x + 1)*(5*x + log(3)*(5*x + 10*x^2 + 5*x^3) - 5*x^2 - 5*x^3 + 5) - log(3)*(25*x^2 + 50*x^3 + 25*x^4
)))/(exp(2*x + 2)*(10*x + 5*x^2 + 5) - exp(x + 1)*(50*x + 100*x^2 + 50*x^3) + 125*x^2 + 250*x^3 + 125*x^4),x)

[Out]

x/5 + 1/(x + 1) - (3^x*x)/(5*(x - exp(x + 1)/5))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((5*x**3+10*x**2+5*x)*ln(3)-5*x**3-5*x**2+5*x+5)*exp(x+1)+(-25*x**4-50*x**3-25*x**2)*ln(3))*exp(x*
ln(3))+(x**2+2*x-4)*exp(x+1)**2+(-10*x**3-20*x**2+40*x)*exp(x+1)+25*x**4+50*x**3-100*x**2)/((5*x**2+10*x+5)*ex
p(x+1)**2+(-50*x**3-100*x**2-50*x)*exp(x+1)+125*x**4+250*x**3+125*x**2),x)

[Out]

Timed out

________________________________________________________________________________________