3.65.99 \(\int \frac {e^{e^{4 x}} (30-6 x+(-6+6 x+e^{4 x} (-120+144 x-24 x^2)) \log (-1+x))}{-25+35 x-11 x^2+x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {6 e^{e^{4 x}} \log (-1+x)}{5-x} \]

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Rubi [F]  time = 2.24, 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^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^(4*x)*(30 - 6*x + (-6 + 6*x + E^(4*x)*(-120 + 144*x - 24*x^2))*Log[-1 + x]))/(-25 + 35*x - 11*x^2 + x
^3),x]

[Out]

6*Log[-1 + x]*Defer[Int][E^E^(4*x)/(-5 + x)^2, x] - (3*Defer[Int][E^E^(4*x)/(-5 + x), x])/2 - 24*Log[-1 + x]*D
efer[Int][E^(E^(4*x) + 4*x)/(-5 + x), x] + (3*Defer[Int][E^E^(4*x)/(-1 + x), x])/2 - 6*Defer[Int][Defer[Int][E
^E^(4*x)/(-5 + x)^2, x]/(-1 + x), x] + 24*Defer[Int][Defer[Int][E^(E^(4*x) + 4*x)/(-5 + x), x]/(-1 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {30 e^{e^{4 x}}}{(-5+x)^2 (-1+x)}-\frac {6 e^{e^{4 x}} x}{(-5+x)^2 (-1+x)}-\frac {24 e^{e^{4 x}+4 x} \log (-1+x)}{-5+x}-\frac {6 e^{e^{4 x}} \log (-1+x)}{(-5+x)^2 (-1+x)}+\frac {6 e^{e^{4 x}} x \log (-1+x)}{(-5+x)^2 (-1+x)}\right ) \, dx\\ &=-\left (6 \int \frac {e^{e^{4 x}} x}{(-5+x)^2 (-1+x)} \, dx\right )-6 \int \frac {e^{e^{4 x}} \log (-1+x)}{(-5+x)^2 (-1+x)} \, dx+6 \int \frac {e^{e^{4 x}} x \log (-1+x)}{(-5+x)^2 (-1+x)} \, dx-24 \int \frac {e^{e^{4 x}+4 x} \log (-1+x)}{-5+x} \, dx+30 \int \frac {e^{e^{4 x}}}{(-5+x)^2 (-1+x)} \, dx\\ &=-\left (6 \int \left (\frac {5 e^{e^{4 x}}}{4 (-5+x)^2}-\frac {e^{e^{4 x}}}{16 (-5+x)}+\frac {e^{e^{4 x}}}{16 (-1+x)}\right ) \, dx\right )-6 \int \frac {-20 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\int \frac {e^{e^{4 x}}}{-5+x} \, dx-\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{16 (1-x)} \, dx+6 \int \frac {-4 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\int \frac {e^{e^{4 x}}}{-5+x} \, dx-\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{16 (1-x)} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx+30 \int \left (\frac {e^{e^{4 x}}}{4 (-5+x)^2}-\frac {e^{e^{4 x}}}{16 (-5+x)}+\frac {e^{e^{4 x}}}{16 (-1+x)}\right ) \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ &=\frac {3}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx-\frac {3}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx-\frac {3}{8} \int \frac {-20 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\int \frac {e^{e^{4 x}}}{-5+x} \, dx-\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{1-x} \, dx+\frac {3}{8} \int \frac {-4 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\int \frac {e^{e^{4 x}}}{-5+x} \, dx-\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{1-x} \, dx-\frac {15}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx+\frac {15}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ &=\frac {3}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx-\frac {3}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+\frac {3}{8} \int \left (\frac {4 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x}+\frac {\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{-1+x}\right ) \, dx-\frac {3}{8} \int \left (\frac {20 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x}+\frac {\int \frac {e^{e^{4 x}}}{-1+x} \, dx}{-1+x}\right ) \, dx-\frac {15}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx+\frac {15}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ &=\frac {3}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx-\frac {3}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+\frac {3}{8} \int \frac {4 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x} \, dx-\frac {3}{8} \int \frac {20 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x} \, dx-\frac {15}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx+\frac {15}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ &=\frac {3}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx-\frac {3}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+\frac {3}{8} \int \left (\frac {4 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx}{-1+x}-\frac {\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x}\right ) \, dx-\frac {3}{8} \int \left (\frac {20 \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx}{-1+x}-\frac {\int \frac {e^{e^{4 x}}}{-5+x} \, dx}{-1+x}\right ) \, dx-\frac {15}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx+\frac {15}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ &=\frac {3}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx-\frac {3}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx+\frac {3}{2} \int \frac {\int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx}{-1+x} \, dx-\frac {15}{8} \int \frac {e^{e^{4 x}}}{-5+x} \, dx+\frac {15}{8} \int \frac {e^{e^{4 x}}}{-1+x} \, dx-\frac {15}{2} \int \frac {\int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx}{-1+x} \, dx+24 \int \frac {\int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx}{-1+x} \, dx-\frac {1}{2} (3 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx+\frac {1}{2} (15 \log (-1+x)) \int \frac {e^{e^{4 x}}}{(-5+x)^2} \, dx-(24 \log (-1+x)) \int \frac {e^{e^{4 x}+4 x}}{-5+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 18, normalized size = 0.90 \begin {gather*} -\frac {6 e^{e^{4 x}} \log (-1+x)}{-5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(4*x)*(30 - 6*x + (-6 + 6*x + E^(4*x)*(-120 + 144*x - 24*x^2))*Log[-1 + x]))/(-25 + 35*x - 11*x
^2 + x^3),x]

[Out]

(-6*E^E^(4*x)*Log[-1 + x])/(-5 + x)

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fricas [A]  time = 0.76, size = 16, normalized size = 0.80 \begin {gather*} -\frac {6 \, e^{\left (e^{\left (4 \, x\right )}\right )} \log \left (x - 1\right )}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(x-1)-6*x+30)*exp(exp(x)^4)/(x^3-11*x^2+35*x-25),x, algorit
hm="fricas")

[Out]

-6*e^(e^(4*x))*log(x - 1)/(x - 5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {6 \, {\left ({\left (4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (4 \, x\right )} - x + 1\right )} \log \left (x - 1\right ) + x - 5\right )} e^{\left (e^{\left (4 \, x\right )}\right )}}{x^{3} - 11 \, x^{2} + 35 \, x - 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(x-1)-6*x+30)*exp(exp(x)^4)/(x^3-11*x^2+35*x-25),x, algorit
hm="giac")

[Out]

integrate(-6*((4*(x^2 - 6*x + 5)*e^(4*x) - x + 1)*log(x - 1) + x - 5)*e^(e^(4*x))/(x^3 - 11*x^2 + 35*x - 25),
x)

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maple [A]  time = 0.06, size = 17, normalized size = 0.85

method	result	size
risch	\(-\frac {6 \ln \left (x -1\right ) {\mathrm e}^{{\mathrm e}^{4 x}}}{x -5}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*ln(x-1)-6*x+30)*exp(exp(x)^4)/(x^3-11*x^2+35*x-25),x,method=_RETURNV
ERBOSE)

[Out]

-6/(x-5)*ln(x-1)*exp(exp(4*x))

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maxima [A]  time = 0.42, size = 16, normalized size = 0.80 \begin {gather*} -\frac {6 \, e^{\left (e^{\left (4 \, x\right )}\right )} \log \left (x - 1\right )}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(x-1)-6*x+30)*exp(exp(x)^4)/(x^3-11*x^2+35*x-25),x, algorit
hm="maxima")

[Out]

-6*e^(e^(4*x))*log(x - 1)/(x - 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (6\,x+\ln \left (x-1\right )\,\left ({\mathrm {e}}^{4\,x}\,\left (24\,x^2-144\,x+120\right )-6\,x+6\right )-30\right )}{x^3-11\,x^2+35\,x-25} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(4*x))*(6*x + log(x - 1)*(exp(4*x)*(24*x^2 - 144*x + 120) - 6*x + 6) - 30))/(35*x - 11*x^2 + x^3
- 25),x)

[Out]

int(-(exp(exp(4*x))*(6*x + log(x - 1)*(exp(4*x)*(24*x^2 - 144*x + 120) - 6*x + 6) - 30))/(35*x - 11*x^2 + x^3
- 25), x)

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sympy [A]  time = 0.34, size = 17, normalized size = 0.85 \begin {gather*} - \frac {6 e^{e^{4 x}} \log {\left (x - 1 \right )}}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-24*x**2+144*x-120)*exp(x)**4+6*x-6)*ln(x-1)-6*x+30)*exp(exp(x)**4)/(x**3-11*x**2+35*x-25),x)

[Out]

-6*exp(exp(4*x))*log(x - 1)/(x - 5)

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