3.81.39 \(\int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} (32+24 x+4 x^2) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{2+2 x} \log ^2(4+x)}{5+2 x} \]

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Rubi [F]  time = 1.99, 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^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2 + 2*x)*(10 + 4*x)*Log[4 + x] + E^(2 + 2*x)*(32 + 24*x + 4*x^2)*Log[4 + x]^2)/(100 + 105*x + 36*x^2 +
4*x^3),x]

[Out]

(-2*ExpIntegralEi[2*(4 + x)]*Log[4 + x])/(3*E^6) + (2*ExpIntegralEi[5 + 2*x]*Log[4 + x])/(3*E^3) - (2*Defer[In
t][ExpIntegralEi[5 + 2*x]/(4 + x), x])/(3*E^3) - (2*Defer[Int][ExpIntegralEi[8 + 2*x]/(-4 - x), x])/(3*E^6) -
2*Defer[Int][(E^(2 + 2*x)*Log[4 + x]^2)/(5 + 2*x)^2, x] + 2*Defer[Int][(E^(2 + 2*x)*Log[4 + x]^2)/(5 + 2*x), x
]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 22, normalized size = 1.05 \begin {gather*} \frac {2 e^{2+2 x} \log ^2(4+x)}{10+4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + 2*x)*(10 + 4*x)*Log[4 + x] + E^(2 + 2*x)*(32 + 24*x + 4*x^2)*Log[4 + x]^2)/(100 + 105*x + 36
*x^2 + 4*x^3),x]

[Out]

(2*E^(2 + 2*x)*Log[4 + x]^2)/(10 + 4*x)

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fricas [A]  time = 1.20, size = 20, normalized size = 0.95 \begin {gather*} \frac {e^{\left (2 \, x + 2\right )} \log \left (x + 4\right )^{2}}{2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x+32)*exp(x+1)^2*log(4+x)^2+(4*x+10)*exp(x+1)^2*log(4+x))/(4*x^3+36*x^2+105*x+100),x, alg
orithm="fricas")

[Out]

e^(2*x + 2)*log(x + 4)^2/(2*x + 5)

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giac [A]  time = 0.36, size = 20, normalized size = 0.95 \begin {gather*} \frac {e^{\left (2 \, x + 2\right )} \log \left (x + 4\right )^{2}}{2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x+32)*exp(x+1)^2*log(4+x)^2+(4*x+10)*exp(x+1)^2*log(4+x))/(4*x^3+36*x^2+105*x+100),x, alg
orithm="giac")

[Out]

e^(2*x + 2)*log(x + 4)^2/(2*x + 5)

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maple [A]  time = 0.03, size = 21, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{2 x +2} \ln \left (4+x \right )^{2}}{5+2 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+24*x+32)*exp(x+1)^2*ln(4+x)^2+(4*x+10)*exp(x+1)^2*ln(4+x))/(4*x^3+36*x^2+105*x+100),x,method=_RETU
RNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 20, normalized size = 0.95 \begin {gather*} \frac {e^{\left (2 \, x + 2\right )} \log \left (x + 4\right )^{2}}{2 \, x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+24*x+32)*exp(x+1)^2*log(4+x)^2+(4*x+10)*exp(x+1)^2*log(4+x))/(4*x^3+36*x^2+105*x+100),x, alg
orithm="maxima")

[Out]

e^(2*x + 2)*log(x + 4)^2/(2*x + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,x+2}\,\left (4\,x^2+24\,x+32\right )\,{\ln \left (x+4\right )}^2+{\mathrm {e}}^{2\,x+2}\,\left (4\,x+10\right )\,\ln \left (x+4\right )}{4\,x^3+36\,x^2+105\,x+100} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 4)^2*exp(2*x + 2)*(24*x + 4*x^2 + 32) + log(x + 4)*exp(2*x + 2)*(4*x + 10))/(105*x + 36*x^2 + 4*x
^3 + 100),x)

[Out]

int((log(x + 4)^2*exp(2*x + 2)*(24*x + 4*x^2 + 32) + log(x + 4)*exp(2*x + 2)*(4*x + 10))/(105*x + 36*x^2 + 4*x
^3 + 100), x)

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sympy [A]  time = 0.31, size = 17, normalized size = 0.81 \begin {gather*} \frac {e^{2 x + 2} \log {\left (x + 4 \right )}^{2}}{2 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+24*x+32)*exp(x+1)**2*ln(4+x)**2+(4*x+10)*exp(x+1)**2*ln(4+x))/(4*x**3+36*x**2+105*x+100),x)

[Out]

exp(2*x + 2)*log(x + 4)**2/(2*x + 5)

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