3.93.59 \(\int \frac {-25 e^{-7+x} x+e^{-7+x} (50+75 x+25 x^2) \log (2+x)+(50+25 x) \log ^2(2+x)}{(2+x) \log ^2(2+x)} \, dx\)

Optimal. Leaf size=18 \[ 25 \left (4+x+\frac {e^{-7+x} x}{\log (2+x)}\right ) \]

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Rubi [F]  time = 1.19, 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 {-25 e^{-7+x} x+e^{-7+x} \left (50+75 x+25 x^2\right ) \log (2+x)+(50+25 x) \log ^2(2+x)}{(2+x) \log ^2(2+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25*E^(-7 + x)*x + E^(-7 + x)*(50 + 75*x + 25*x^2)*Log[2 + x] + (50 + 25*x)*Log[2 + x]^2)/((2 + x)*Log[2
+ x]^2),x]

[Out]

25*x - 25*Defer[Int][E^(-7 + x)/Log[2 + x]^2, x] + 50*Defer[Int][E^(-7 + x)/((2 + x)*Log[2 + x]^2), x] - 25*De
fer[Int][E^(-7 + x)/Log[2 + x], x] + 25*Defer[Int][(E^(-7 + x)*(2 + x))/Log[2 + x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (25+\frac {25 e^{-7+x} \left (-x+2 \log (2+x)+3 x \log (2+x)+x^2 \log (2+x)\right )}{(2+x) \log ^2(2+x)}\right ) \, dx\\ &=25 x+25 \int \frac {e^{-7+x} \left (-x+2 \log (2+x)+3 x \log (2+x)+x^2 \log (2+x)\right )}{(2+x) \log ^2(2+x)} \, dx\\ &=25 x+25 \int \frac {e^{-7+x} \left (-x+\left (2+3 x+x^2\right ) \log (2+x)\right )}{(2+x) \log ^2(2+x)} \, dx\\ &=25 x+25 \int \left (-\frac {e^{-7+x} x}{(2+x) \log ^2(2+x)}+\frac {e^{-7+x} (1+x)}{\log (2+x)}\right ) \, dx\\ &=25 x-25 \int \frac {e^{-7+x} x}{(2+x) \log ^2(2+x)} \, dx+25 \int \frac {e^{-7+x} (1+x)}{\log (2+x)} \, dx\\ &=25 x-25 \int \left (\frac {e^{-7+x}}{\log ^2(2+x)}-\frac {2 e^{-7+x}}{(2+x) \log ^2(2+x)}\right ) \, dx+25 \int \left (-\frac {e^{-7+x}}{\log (2+x)}+\frac {e^{-7+x} (2+x)}{\log (2+x)}\right ) \, dx\\ &=25 x-25 \int \frac {e^{-7+x}}{\log ^2(2+x)} \, dx-25 \int \frac {e^{-7+x}}{\log (2+x)} \, dx+25 \int \frac {e^{-7+x} (2+x)}{\log (2+x)} \, dx+50 \int \frac {e^{-7+x}}{(2+x) \log ^2(2+x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 22, normalized size = 1.22 \begin {gather*} \frac {25 \left (e^7 x+\frac {e^x x}{\log (2+x)}\right )}{e^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*E^(-7 + x)*x + E^(-7 + x)*(50 + 75*x + 25*x^2)*Log[2 + x] + (50 + 25*x)*Log[2 + x]^2)/((2 + x)*
Log[2 + x]^2),x]

[Out]

(25*(E^7*x + (E^x*x)/Log[2 + x]))/E^7

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fricas [A]  time = 0.66, size = 21, normalized size = 1.17 \begin {gather*} \frac {25 \, {\left (x e^{\left (x - 7\right )} + x \log \left (x + 2\right )\right )}}{\log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x+50)*log(2+x)^2+(25*x^2+75*x+50)*exp(x-7)*log(2+x)-25*x*exp(x-7))/(2+x)/log(2+x)^2,x, algorith
m="fricas")

[Out]

25*(x*e^(x - 7) + x*log(x + 2))/log(x + 2)

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giac [A]  time = 0.25, size = 23, normalized size = 1.28 \begin {gather*} \frac {25 \, {\left (x e^{7} \log \left (x + 2\right ) + x e^{x}\right )} e^{\left (-7\right )}}{\log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x+50)*log(2+x)^2+(25*x^2+75*x+50)*exp(x-7)*log(2+x)-25*x*exp(x-7))/(2+x)/log(2+x)^2,x, algorith
m="giac")

[Out]

25*(x*e^7*log(x + 2) + x*e^x)*e^(-7)/log(x + 2)

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maple [A]  time = 0.52, size = 18, normalized size = 1.00

method	result	size
default	\(25 x +\frac {25 x \,{\mathrm e}^{x -7}}{\ln \left (2+x \right )}\)	\(18\)
risch	\(25 x +\frac {25 x \,{\mathrm e}^{x -7}}{\ln \left (2+x \right )}\)	\(18\)
norman	\(\frac {25 x \,{\mathrm e}^{x -7}+25 x \ln \left (2+x \right )}{\ln \left (2+x \right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x+50)*ln(2+x)^2+(25*x^2+75*x+50)*exp(x-7)*ln(2+x)-25*x*exp(x-7))/(2+x)/ln(2+x)^2,x,method=_RETURNVERB
OSE)

[Out]

25*x+25*x/ln(2+x)*exp(x-7)

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maxima [A]  time = 0.40, size = 23, normalized size = 1.28 \begin {gather*} \frac {25 \, {\left (x e^{7} \log \left (x + 2\right ) + x e^{x}\right )} e^{\left (-7\right )}}{\log \left (x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x+50)*log(2+x)^2+(25*x^2+75*x+50)*exp(x-7)*log(2+x)-25*x*exp(x-7))/(2+x)/log(2+x)^2,x, algorith
m="maxima")

[Out]

25*(x*e^7*log(x + 2) + x*e^x)*e^(-7)/log(x + 2)

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mupad [B]  time = 7.75, size = 17, normalized size = 0.94 \begin {gather*} 25\,x+\frac {25\,x\,{\mathrm {e}}^{-7}\,{\mathrm {e}}^x}{\ln \left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 2)^2*(25*x + 50) - 25*x*exp(x - 7) + log(x + 2)*exp(x - 7)*(75*x + 25*x^2 + 50))/(log(x + 2)^2*(x
+ 2)),x)

[Out]

25*x + (25*x*exp(-7)*exp(x))/log(x + 2)

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sympy [A]  time = 0.30, size = 15, normalized size = 0.83 \begin {gather*} \frac {25 x e^{x - 7}}{\log {\left (x + 2 \right )}} + 25 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x+50)*ln(2+x)**2+(25*x**2+75*x+50)*exp(x-7)*ln(2+x)-25*x*exp(x-7))/(2+x)/ln(2+x)**2,x)

[Out]

25*x*exp(x - 7)/log(x + 2) + 25*x

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