3.13.7 \(\int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} (1-10 x+25 x^2+e^{16} (1-25 x+27 x^2-20 x^3+50 x^4))+e^{\frac {3-x^2+5 x^3}{-1+5 x}} (1-25 x+27 x^2-20 x^3+50 x^4) \log (x)}{5-50 x+125 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{5} e^{x \left (x-\frac {3}{x-5 x^2}\right )} x \left (e^{16}+\log (x)\right ) \]

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Rubi [F]  time = 5.62, 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^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 10*x + 25*x^2 + E^16*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)) + E^((3
- x^2 + 5*x^3)/(-1 + 5*x))*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)*Log[x])/(5 - 50*x + 125*x^2),x]

[Out]

((25 + 27*E^16)*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x)), x])/125 + (Log[x]*Defer[Int][E^((3 - x^2 + 5*x^3)
/(-1 + 5*x)), x])/5 - (2*Defer[Int][E^((-13 + 80*x - x^2 + 5*x^3)/(-1 + 5*x)), x])/125 + ((1 + E^16)*Defer[Int
][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(1 - 5*x)^2, x])/5 + (Log[x]*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(1
- 5*x)^2, x])/5 + (2*Log[x]*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*x^2, x])/5 + (2*Defer[Int][E^((-13 +
80*x - x^2 + 5*x^3)/(-1 + 5*x))*x^2, x])/5 - ((2 + 5*E^16)*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5
*x)^2, x])/5 + ((25 + 27*E^16)*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x)^2, x])/125 - (4*Log[x]*D
efer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x)^2, x])/5 - (2*Defer[Int][E^((-13 + 80*x - x^2 + 5*x^3)/(
-1 + 5*x))/(-1 + 5*x)^2, x])/125 - ((2 + 5*E^16)*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x), x])/5
+ (2*(25 + 27*E^16)*Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x), x])/125 - (3*Log[x]*Defer[Int][E^
((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x), x])/5 - (4*Defer[Int][E^((-13 + 80*x - x^2 + 5*x^3)/(-1 + 5*x))/(-1
+ 5*x), x])/125 - Defer[Int][Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x)), x]/x, x]/5 + (3*Defer[Int][Defer[In
t][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(1 - 5*x)^2, x]/x, x])/5 - (2*Defer[Int][Defer[Int][E^((3 - x^2 + 5*x^3)/(
-1 + 5*x))*x^2, x]/x, x])/5 + (3*Defer[Int][Defer[Int][E^((3 - x^2 + 5*x^3)/(-1 + 5*x))/(-1 + 5*x), x]/x, x])/
5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5 (-1+5 x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{(-1+5 x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1+e^{16}\right )}{(1-5 x)^2}-\frac {10 e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1+\frac {5 e^{16}}{2}\right ) x}{(1-5 x)^2}+\frac {25 e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1+\frac {27 e^{16}}{25}\right ) x^2}{(1-5 x)^2}-\frac {20 e^{\frac {-13+80 x-x^2+5 x^3}{-1+5 x}} x^3}{(1-5 x)^2}+\frac {50 e^{\frac {-13+80 x-x^2+5 x^3}{-1+5 x}} x^4}{(1-5 x)^2}+\frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \log (x)}{(1-5 x)^2}-\frac {25 e^{\frac {3-x^2+5 x^3}{-1+5 x}} x \log (x)}{(1-5 x)^2}+\frac {27 e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^2 \log (x)}{(1-5 x)^2}-\frac {20 e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^3 \log (x)}{(1-5 x)^2}+\frac {50 e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^4 \log (x)}{(1-5 x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \log (x)}{(1-5 x)^2} \, dx-4 \int \frac {e^{\frac {-13+80 x-x^2+5 x^3}{-1+5 x}} x^3}{(1-5 x)^2} \, dx-4 \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^3 \log (x)}{(1-5 x)^2} \, dx-5 \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x \log (x)}{(1-5 x)^2} \, dx+\frac {27}{5} \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^2 \log (x)}{(1-5 x)^2} \, dx+10 \int \frac {e^{\frac {-13+80 x-x^2+5 x^3}{-1+5 x}} x^4}{(1-5 x)^2} \, dx+10 \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^4 \log (x)}{(1-5 x)^2} \, dx+\left (-2-5 e^{16}\right ) \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x}{(1-5 x)^2} \, dx+\frac {1}{5} \left (1+e^{16}\right ) \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}}}{(1-5 x)^2} \, dx+\frac {1}{5} \left (25+27 e^{16}\right ) \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} x^2}{(1-5 x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{5} e^{x^2+\frac {3}{-1+5 x}} x \left (e^{16}+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 10*x + 25*x^2 + E^16*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)) +
E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)*Log[x])/(5 - 50*x + 125*x^2),x]

[Out]

(E^(x^2 + 3/(-1 + 5*x))*x*(E^16 + Log[x]))/5

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fricas [B]  time = 0.95, size = 53, normalized size = 1.89 \begin {gather*} \frac {1}{5} \, x e^{\left (\frac {5 \, x^{3} - x^{2} + 3}{5 \, x - 1}\right )} \log \relax (x) + \frac {1}{5} \, x e^{\left (\frac {5 \, x^{3} - x^{2} + 3}{5 \, x - 1} + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x)+((50*x^4-20*x^3+27*x^2-25*x+1)*exp(
16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/(5*x-1)))/(125*x^2-50*x+5),x, algorithm="fricas")

[Out]

1/5*x*e^((5*x^3 - x^2 + 3)/(5*x - 1))*log(x) + 1/5*x*e^((5*x^3 - x^2 + 3)/(5*x - 1) + 16)

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giac [B]  time = 0.93, size = 59, normalized size = 2.11 \begin {gather*} \frac {1}{5} \, {\left (x e^{\left (\frac {5 \, x^{3} - x^{2} + 15 \, x}{5 \, x - 1}\right )} \log \relax (x) + x e^{\left (\frac {5 \, x^{3} - x^{2} + 15 \, x}{5 \, x - 1} + 16\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x)+((50*x^4-20*x^3+27*x^2-25*x+1)*exp(
16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/(5*x-1)))/(125*x^2-50*x+5),x, algorithm="giac")

[Out]

1/5*(x*e^((5*x^3 - x^2 + 15*x)/(5*x - 1))*log(x) + x*e^((5*x^3 - x^2 + 15*x)/(5*x - 1) + 16))*e^(-3)

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maple [B]  time = 0.22, size = 55, normalized size = 1.96

method	result	size
risch	\(\frac {x \,{\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} \ln \relax (x )}{5}+\frac {x \,{\mathrm e}^{\frac {5 x^{3}-x^{2}+80 x -13}{5 x -1}}}{5}\)	\(55\)
norman	\(\frac {x^{2} {\mathrm e}^{16} {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}}+x^{2} {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} \ln \relax (x )-\frac {x \,{\mathrm e}^{16} {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}}}{5}-\frac {x \,{\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} \ln \relax (x )}{5}}{5 x -1}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*ln(x)+((50*x^4-20*x^3+27*x^2-25*x+1)*exp(16)+25*
x^2-10*x+1)*exp((5*x^3-x^2+3)/(5*x-1)))/(125*x^2-50*x+5),x,method=_RETURNVERBOSE)

[Out]

1/5*x*exp((5*x^3-x^2+3)/(5*x-1))*ln(x)+1/5*x*exp((5*x^3-x^2+80*x-13)/(5*x-1))

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maxima [A]  time = 0.52, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{5} \, {\left (x e^{16} + x \log \relax (x)\right )} e^{\left (x^{2} + \frac {3}{5 \, x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x)+((50*x^4-20*x^3+27*x^2-25*x+1)*exp(
16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/(5*x-1)))/(125*x^2-50*x+5),x, algorithm="maxima")

[Out]

1/5*(x*e^16 + x*log(x))*e^(x^2 + 3/(5*x - 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {5\,x^3-x^2+3}{5\,x-1}}\,\left ({\mathrm {e}}^{16}\,\left (50\,x^4-20\,x^3+27\,x^2-25\,x+1\right )-10\,x+25\,x^2+1\right )+{\mathrm {e}}^{\frac {5\,x^3-x^2+3}{5\,x-1}}\,\ln \relax (x)\,\left (50\,x^4-20\,x^3+27\,x^2-25\,x+1\right )}{125\,x^2-50\,x+5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((5*x^3 - x^2 + 3)/(5*x - 1))*(exp(16)*(27*x^2 - 25*x - 20*x^3 + 50*x^4 + 1) - 10*x + 25*x^2 + 1) + ex
p((5*x^3 - x^2 + 3)/(5*x - 1))*log(x)*(27*x^2 - 25*x - 20*x^3 + 50*x^4 + 1))/(125*x^2 - 50*x + 5),x)

[Out]

int((exp((5*x^3 - x^2 + 3)/(5*x - 1))*(exp(16)*(27*x^2 - 25*x - 20*x^3 + 50*x^4 + 1) - 10*x + 25*x^2 + 1) + ex
p((5*x^3 - x^2 + 3)/(5*x - 1))*log(x)*(27*x^2 - 25*x - 20*x^3 + 50*x^4 + 1))/(125*x^2 - 50*x + 5), x)

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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(((50*x**4-20*x**3+27*x**2-25*x+1)*exp((5*x**3-x**2+3)/(5*x-1))*ln(x)+((50*x**4-20*x**3+27*x**2-25*x+
1)*exp(16)+25*x**2-10*x+1)*exp((5*x**3-x**2+3)/(5*x-1)))/(125*x**2-50*x+5),x)

[Out]

Timed out

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