3.73.75 \(\int \frac {e^{1-x} (4+4 x+7 x^2-9 x^3-36 x^4+(-4 x^2+4 x^3+9 x^4-9 x^5) \log (3))-36 e^{1-x} x^3 \log (4-9 x^2)+e^{1-x} (-4 x^2+9 x^4) \log ^2(4-9 x^2)}{-4 x^2+9 x^4} \, dx\)

Optimal. Leaf size=29 \[ e^{1-x} \left (4+\frac {1}{x}+x \log (3)-\log ^2\left (4-9 x^2\right )\right ) \]

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Rubi [F]  time = 1.92, 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^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(1 - x)*(4 + 4*x + 7*x^2 - 9*x^3 - 36*x^4 + (-4*x^2 + 4*x^3 + 9*x^4 - 9*x^5)*Log[3]) - 36*E^(1 - x)*x^3
*Log[4 - 9*x^2] + E^(1 - x)*(-4*x^2 + 9*x^4)*Log[4 - 9*x^2]^2)/(-4*x^2 + 9*x^4),x]

[Out]

E^(1 - x)/x + E^(1 - x)*(4 - Log[3]) + E^(1 - x)*Log[3] + E^(1 - x)*x*Log[3] - 2*E^(5/3)*ExpIntegralEi[(-2 - 3
*x)/3]*Log[4 - 9*x^2] - 2*E^(1/3)*ExpIntegralEi[(2 - 3*x)/3]*Log[4 - 9*x^2] + 6*E^(5/3)*Defer[Int][ExpIntegral
Ei[-2/3 - x]/(-2 + 3*x), x] + 6*E^(5/3)*Defer[Int][ExpIntegralEi[-2/3 - x]/(2 + 3*x), x] + 6*E^(1/3)*Defer[Int
][ExpIntegralEi[2/3 - x]/(-2 + 3*x), x] + 6*E^(1/3)*Defer[Int][ExpIntegralEi[2/3 - x]/(2 + 3*x), x] + Defer[In
t][E^(1 - x)*Log[4 - 9*x^2]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{x^2 \left (-4+9 x^2\right )} \, dx\\ &=\int \frac {e^{1-x} \left (\left (-4+9 x^2\right ) \left (1+x-x^2 (-4+\log (3))+x^3 \log (3)\right )+36 x^3 \log \left (4-9 x^2\right )+\left (4 x^2-9 x^4\right ) \log ^2\left (4-9 x^2\right )\right )}{x^2 \left (4-9 x^2\right )} \, dx\\ &=\int \left (\frac {e^{1-x} \left (-1-x-x^2 (4-\log (3))-x^3 \log (3)\right )}{x^2}-\frac {36 e^{1-x} x \log \left (4-9 x^2\right )}{-4+9 x^2}+e^{1-x} \log ^2\left (4-9 x^2\right )\right ) \, dx\\ &=-\left (36 \int \frac {e^{1-x} x \log \left (4-9 x^2\right )}{-4+9 x^2} \, dx\right )+\int \frac {e^{1-x} \left (-1-x-x^2 (4-\log (3))-x^3 \log (3)\right )}{x^2} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+36 \int \frac {\sqrt [3]{e} x \left (-e^{4/3} \text {Ei}\left (-\frac {2}{3}-x\right )-\text {Ei}\left (\frac {2}{3}-x\right )\right )}{4-9 x^2} \, dx+\int \left (-\frac {e^{1-x}}{x^2}-\frac {e^{1-x}}{x}-4 e^{1-x} \left (1-\frac {\log (3)}{4}\right )-e^{1-x} x \log (3)\right ) \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \frac {x \left (-e^{4/3} \text {Ei}\left (-\frac {2}{3}-x\right )-\text {Ei}\left (\frac {2}{3}-x\right )\right )}{4-9 x^2} \, dx-(4-\log (3)) \int e^{1-x} \, dx-\log (3) \int e^{1-x} x \, dx-\int \frac {e^{1-x}}{x^2} \, dx-\int \frac {e^{1-x}}{x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=\frac {e^{1-x}}{x}-e \text {Ei}(-x)+e^{1-x} (4-\log (3))+e^{1-x} x \log (3)-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \left (\frac {e^{4/3} x \text {Ei}\left (-\frac {2}{3}-x\right )}{-4+9 x^2}+\frac {x \text {Ei}\left (\frac {2}{3}-x\right )}{-4+9 x^2}\right ) \, dx-\log (3) \int e^{1-x} \, dx+\int \frac {e^{1-x}}{x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=\frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \frac {x \text {Ei}\left (\frac {2}{3}-x\right )}{-4+9 x^2} \, dx+\left (36 e^{5/3}\right ) \int \frac {x \text {Ei}\left (-\frac {2}{3}-x\right )}{-4+9 x^2} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=\frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (36 \sqrt [3]{e}\right ) \int \left (\frac {\text {Ei}\left (\frac {2}{3}-x\right )}{6 (-2+3 x)}+\frac {\text {Ei}\left (\frac {2}{3}-x\right )}{6 (2+3 x)}\right ) \, dx+\left (36 e^{5/3}\right ) \int \left (\frac {\text {Ei}\left (-\frac {2}{3}-x\right )}{6 (-2+3 x)}+\frac {\text {Ei}\left (-\frac {2}{3}-x\right )}{6 (2+3 x)}\right ) \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ &=\frac {e^{1-x}}{x}+e^{1-x} (4-\log (3))+e^{1-x} \log (3)+e^{1-x} x \log (3)-2 e^{5/3} \text {Ei}\left (\frac {1}{3} (-2-3 x)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \text {Ei}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )+\left (6 \sqrt [3]{e}\right ) \int \frac {\text {Ei}\left (\frac {2}{3}-x\right )}{-2+3 x} \, dx+\left (6 \sqrt [3]{e}\right ) \int \frac {\text {Ei}\left (\frac {2}{3}-x\right )}{2+3 x} \, dx+\left (6 e^{5/3}\right ) \int \frac {\text {Ei}\left (-\frac {2}{3}-x\right )}{-2+3 x} \, dx+\left (6 e^{5/3}\right ) \int \frac {\text {Ei}\left (-\frac {2}{3}-x\right )}{2+3 x} \, dx+\int e^{1-x} \log ^2\left (4-9 x^2\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.06, size = 43, normalized size = 1.48 \begin {gather*} -e^{-x} \left (-4 e-\frac {e}{x}-e x \log (3)\right )-e^{1-x} \log ^2\left (4-9 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - x)*(4 + 4*x + 7*x^2 - 9*x^3 - 36*x^4 + (-4*x^2 + 4*x^3 + 9*x^4 - 9*x^5)*Log[3]) - 36*E^(1 -
x)*x^3*Log[4 - 9*x^2] + E^(1 - x)*(-4*x^2 + 9*x^4)*Log[4 - 9*x^2]^2)/(-4*x^2 + 9*x^4),x]

[Out]

-((-4*E - E/x - E*x*Log[3])/E^x) - E^(1 - x)*Log[4 - 9*x^2]^2

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fricas [A]  time = 0.66, size = 43, normalized size = 1.48 \begin {gather*} -\frac {x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} - {\left (x^{2} \log \relax (3) + 4 \, x + 1\right )} e^{\left (-x + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^4-4*x^2)*exp(-x+1)*log(-9*x^2+4)^2-36*x^3*exp(-x+1)*log(-9*x^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*
log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(-x+1))/(9*x^4-4*x^2),x, algorithm="fricas")

[Out]

-(x*e^(-x + 1)*log(-9*x^2 + 4)^2 - (x^2*log(3) + 4*x + 1)*e^(-x + 1))/x

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giac [A]  time = 0.28, size = 51, normalized size = 1.76 \begin {gather*} \frac {x^{2} e^{\left (-x + 1\right )} \log \relax (3) - x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} + 4 \, x e^{\left (-x + 1\right )} + e^{\left (-x + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^4-4*x^2)*exp(-x+1)*log(-9*x^2+4)^2-36*x^3*exp(-x+1)*log(-9*x^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*
log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(-x+1))/(9*x^4-4*x^2),x, algorithm="giac")

[Out]

(x^2*e^(-x + 1)*log(3) - x*e^(-x + 1)*log(-9*x^2 + 4)^2 + 4*x*e^(-x + 1) + e^(-x + 1))/x

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maple [A]  time = 0.15, size = 41, normalized size = 1.41

method	result	size
risch	\(-{\mathrm e}^{1-x} \ln \left (-9 x^{2}+4\right )^{2}+\frac {\left (x^{2} \ln \relax (3)+4 x +1\right ) {\mathrm e}^{1-x}}{x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*x^4-4*x^2)*exp(1-x)*ln(-9*x^2+4)^2-36*x^3*exp(1-x)*ln(-9*x^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*ln(3)-36*x
^4-9*x^3+7*x^2+4*x+4)*exp(1-x))/(9*x^4-4*x^2),x,method=_RETURNVERBOSE)

[Out]

-exp(1-x)*ln(-9*x^2+4)^2+(x^2*ln(3)+4*x+1)/x*exp(1-x)

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maxima [B]  time = 0.51, size = 67, normalized size = 2.31 \begin {gather*} \frac {{\left (x^{2} e \log \relax (3) - x e \log \left (3 \, x + 2\right )^{2} - 2 \, x e \log \left (3 \, x + 2\right ) \log \left (-3 \, x + 2\right ) - x e \log \left (-3 \, x + 2\right )^{2} + 4 \, x e + e\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x^4-4*x^2)*exp(-x+1)*log(-9*x^2+4)^2-36*x^3*exp(-x+1)*log(-9*x^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*
log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(-x+1))/(9*x^4-4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.48, size = 40, normalized size = 1.38 \begin {gather*} \frac {{\mathrm {e}}^{1-x}\,\left (\ln \relax (3)\,x^2+4\,x+1\right )}{x}-{\ln \left (4-9\,x^2\right )}^2\,{\mathrm {e}}^{1-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4 - 9*x^2)^2*exp(1 - x)*(4*x^2 - 9*x^4) - exp(1 - x)*(4*x - log(3)*(4*x^2 - 4*x^3 - 9*x^4 + 9*x^5) +
7*x^2 - 9*x^3 - 36*x^4 + 4) + 36*x^3*log(4 - 9*x^2)*exp(1 - x))/(4*x^2 - 9*x^4),x)

[Out]

(exp(1 - x)*(4*x + x^2*log(3) + 1))/x - log(4 - 9*x^2)^2*exp(1 - x)

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sympy [A]  time = 0.40, size = 29, normalized size = 1.00 \begin {gather*} \frac {\left (x^{2} \log {\relax (3 )} - x \log {\left (4 - 9 x^{2} \right )}^{2} + 4 x + 1\right ) e^{1 - x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x**4-4*x**2)*exp(-x+1)*ln(-9*x**2+4)**2-36*x**3*exp(-x+1)*ln(-9*x**2+4)+((-9*x**5+9*x**4+4*x**3-
4*x**2)*ln(3)-36*x**4-9*x**3+7*x**2+4*x+4)*exp(-x+1))/(9*x**4-4*x**2),x)

[Out]

(x**2*log(3) - x*log(4 - 9*x**2)**2 + 4*x + 1)*exp(1 - x)/x

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