3.5.19 \(\int \frac {-30+54 x+e^{2-x} (15-12 x-12 x^2-3 x^3)+(-120 x-12 x^2+e^{2-x} (60 x+36 x^2+6 x^3)) \log (5+x)+(60 x+12 x^2+e^{2-x} (-30 x-21 x^2-3 x^3)) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx\)

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

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Rubi [F]  time = 48.98, 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 {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 12*x^2 + E^(2 - x)*(60*x + 36*x^2 + 6*x^3
))*Log[5 + x] + (60*x + 12*x^2 + E^(2 - x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20 - 4*x)
+ 4*x + E^(4 - 2*x)*(5 + x)),x]

[Out]

33*E^(-2 + x) - (3*x)/2 + 3*E^(-2 + x)*x + (3*E^2*x)/(2*(E^2 - 2*E^x)) + (3*x^2)/2 - 3*E^(-2 + x)*x^2 - (3*E^2
*x^2)/(2*(E^2 - 2*E^x)) - (66*ExpIntegralEi[5 + x])/E^7 - (3*x*Log[1 - 2*E^(-2 + x)])/2 - (3*x^2*Log[1 - 2*E^(
-2 + x)])/2 + 15*Log[E^2 - 2*E^x] + 36*E^(-2 + x)*Log[5 + x] - 6*E^(-2 + x)*x*Log[5 + x] + 6*E^(-2 + x)*x^2*Lo
g[5 + x] - (150*ExpIntegralEi[5 + x]*Log[5 + x])/E^7 - (3*PolyLog[2, 2*E^(-2 + x)])/2 - 3*x*PolyLog[2, 2*E^(-2
+ x)] + 3*PolyLog[3, 2*E^(-2 + x)] - 6*Defer[Int][(E^(-2 + 2*x)*x)/(-E^2 + 2*E^x), x] - 12*Log[5 + x]*Defer[I
nt][(E^(-2 + 2*x)*x)/(-E^2 + 2*E^x), x] + 6*Defer[Int][(E^(-2 + 2*x)*x^2)/(-E^2 + 2*E^x), x] - 12*Log[5 + x]*D
efer[Int][(E^(-2 + 2*x)*x^2)/(-E^2 + 2*E^x), x] - 300*Defer[Int][E^(-2 + 2*x)/((-E^2 + 2*E^x)*(5 + x)), x] + 3
00*Log[5 + x]*Defer[Int][E^(-2 + 2*x)/((-E^2 + 2*E^x)*(5 + x)), x] + (150*Defer[Int][ExpIntegralEi[5 + x]/(5 +
x), x])/E^7 - 60*Defer[Int][(E^(-2 + 2*x)*Log[5 + x])/(-E^2 + 2*E^x), x] + 12*Defer[Int][(E^(2*x)*x^2*Log[5 +
x])/(-E^2 + 2*E^x)^2, x] - 45*Defer[Int][E^(-2 + x)*Log[5 + x]^2, x] + 12*Defer[Int][(E^(-2 + 2*x)*x*Log[5 +
x]^2)/(-E^2 + 2*E^x), x] - 6*Defer[Int][(E^(2*x)*x^2*Log[5 + x]^2)/(-E^2 + 2*E^x)^2, x] + 6*Defer[Int][(E^(-2
+ 2*x)*x^2*Log[5 + x]^2)/(-E^2 + 2*E^x), x] + 24*Defer[Int][E^(-2 + x)*(5 + x)*Log[5 + x]^2, x] - 3*Defer[Int]
[E^(-2 + x)*(5 + x)^2*Log[5 + x]^2, x] + 12*Defer[Int][Defer[Int][(E^(-2 + 2*x)*x)/(-E^2 + 2*E^x), x]/(5 + x),
x] + 12*Defer[Int][Defer[Int][(E^(-2 + 2*x)*x^2)/(-E^2 + 2*E^x), x]/(5 + x), x] - 300*Defer[Int][Defer[Int][E
^(-2 + 2*x)/((-E^2 + 2*E^x)*(5 + x)), x]/(5 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)\right )}{\left (e^2-2 e^x\right )^2 (5+x)} \, dx\\ &=\int \left (-\frac {6 e^{2 x} x \left (-1+x-2 x \log (5+x)+x \log ^2(5+x)\right )}{\left (-e^2+2 e^x\right )^2}-\frac {3 e^{-2+x} \left (-5+4 x+4 x^2+x^3-20 x \log (5+x)-12 x^2 \log (5+x)-2 x^3 \log (5+x)+10 x \log ^2(5+x)+7 x^2 \log ^2(5+x)+x^3 \log ^2(5+x)\right )}{5+x}-\frac {6 e^{-2+2 x} \left (-5+4 x+4 x^2+x^3-20 x \log (5+x)-12 x^2 \log (5+x)-2 x^3 \log (5+x)+10 x \log ^2(5+x)+7 x^2 \log ^2(5+x)+x^3 \log ^2(5+x)\right )}{\left (e^2-2 e^x\right ) (5+x)}\right ) \, dx\\ &=-\left (3 \int \frac {e^{-2+x} \left (-5+4 x+4 x^2+x^3-20 x \log (5+x)-12 x^2 \log (5+x)-2 x^3 \log (5+x)+10 x \log ^2(5+x)+7 x^2 \log ^2(5+x)+x^3 \log ^2(5+x)\right )}{5+x} \, dx\right )-6 \int \frac {e^{2 x} x \left (-1+x-2 x \log (5+x)+x \log ^2(5+x)\right )}{\left (-e^2+2 e^x\right )^2} \, dx-6 \int \frac {e^{-2+2 x} \left (-5+4 x+4 x^2+x^3-20 x \log (5+x)-12 x^2 \log (5+x)-2 x^3 \log (5+x)+10 x \log ^2(5+x)+7 x^2 \log ^2(5+x)+x^3 \log ^2(5+x)\right )}{\left (e^2-2 e^x\right ) (5+x)} \, dx\\ &=-\left (3 \int \frac {e^{-2+x} \left (-5+4 x+4 x^2+x^3-2 x \left (10+6 x+x^2\right ) \log (5+x)+x \left (10+7 x+x^2\right ) \log ^2(5+x)\right )}{5+x} \, dx\right )-6 \int \left (-\frac {e^{2 x} x}{\left (-e^2+2 e^x\right )^2}+\frac {e^{2 x} x^2}{\left (-e^2+2 e^x\right )^2}-\frac {2 e^{2 x} x^2 \log (5+x)}{\left (-e^2+2 e^x\right )^2}+\frac {e^{2 x} x^2 \log ^2(5+x)}{\left (-e^2+2 e^x\right )^2}\right ) \, dx-6 \int \frac {e^{-2+2 x} \left (-5+4 x+4 x^2+x^3-2 x \left (10+6 x+x^2\right ) \log (5+x)+x \left (10+7 x+x^2\right ) \log ^2(5+x)\right )}{\left (e^2-2 e^x\right ) (5+x)} \, dx\\ &=-\left (3 \int \left (\frac {e^{-2+x} \left (-5+4 x+4 x^2+x^3\right )}{5+x}-\frac {2 e^{-2+x} x \left (10+6 x+x^2\right ) \log (5+x)}{5+x}+e^{-2+x} x (2+x) \log ^2(5+x)\right ) \, dx\right )+6 \int \frac {e^{2 x} x}{\left (-e^2+2 e^x\right )^2} \, dx-6 \int \frac {e^{2 x} x^2}{\left (-e^2+2 e^x\right )^2} \, dx-6 \int \frac {e^{2 x} x^2 \log ^2(5+x)}{\left (-e^2+2 e^x\right )^2} \, dx-6 \int \left (\frac {5 e^{-2+2 x}}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {4 e^{-2+2 x} x}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {4 e^{-2+2 x} x^2}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {e^{-2+2 x} x^3}{\left (-e^2+2 e^x\right ) (5+x)}+\frac {20 e^{-2+2 x} x \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)}+\frac {12 e^{-2+2 x} x^2 \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)}+\frac {2 e^{-2+2 x} x^3 \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {10 e^{-2+2 x} x \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {7 e^{-2+2 x} x^2 \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)}-\frac {e^{-2+2 x} x^3 \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)}\right ) \, dx+12 \int \frac {e^{2 x} x^2 \log (5+x)}{\left (-e^2+2 e^x\right )^2} \, dx\\ &=-\left (3 \int \frac {e^{-2+x} \left (-5+4 x+4 x^2+x^3\right )}{5+x} \, dx\right )-3 \int e^{-2+x} x (2+x) \log ^2(5+x) \, dx+6 \int \frac {e^{-2+2 x} x^3}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+6 \int \left (\frac {x}{4}+\frac {e^4 x}{4 \left (e^2-2 e^x\right )^2}-\frac {e^2 x}{2 \left (e^2-2 e^x\right )}\right ) \, dx-6 \int \left (\frac {x^2}{4}+\frac {e^4 x^2}{4 \left (e^2-2 e^x\right )^2}-\frac {e^2 x^2}{2 \left (e^2-2 e^x\right )}\right ) \, dx+6 \int \frac {e^{-2+x} x \left (10+6 x+x^2\right ) \log (5+x)}{5+x} \, dx-6 \int \frac {e^{2 x} x^2 \log ^2(5+x)}{\left (-e^2+2 e^x\right )^2} \, dx+6 \int \frac {e^{-2+2 x} x^3 \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+12 \int \frac {e^{2 x} x^2 \log (5+x)}{\left (-e^2+2 e^x\right )^2} \, dx-12 \int \frac {e^{-2+2 x} x^3 \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+24 \int \frac {e^{-2+2 x} x}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+24 \int \frac {e^{-2+2 x} x^2}{\left (-e^2+2 e^x\right ) (5+x)} \, dx-30 \int \frac {e^{-2+2 x}}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+42 \int \frac {e^{-2+2 x} x^2 \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx+60 \int \frac {e^{-2+2 x} x \log ^2(5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx-72 \int \frac {e^{-2+2 x} x^2 \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx-120 \int \frac {e^{-2+2 x} x \log (5+x)}{\left (-e^2+2 e^x\right ) (5+x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 37, normalized size = 1.32 \begin {gather*} \frac {3 e^x x \left (-1+x-2 x \log (5+x)+x \log ^2(5+x)\right )}{-e^2+2 e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 12*x^2 + E^(2 - x)*(60*x + 36*x^2 +
6*x^3))*Log[5 + x] + (60*x + 12*x^2 + E^(2 - x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20
- 4*x) + 4*x + E^(4 - 2*x)*(5 + x)),x]

[Out]

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

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fricas [A]  time = 1.47, size = 38, normalized size = 1.36 \begin {gather*} -\frac {3 \, {\left (x^{2} \log \left (x + 5\right )^{2} - 2 \, x^{2} \log \left (x + 5\right ) + x^{2} - x\right )}}{e^{\left (-x + 2\right )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*
log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="
fricas")

[Out]

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

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giac [A]  time = 0.49, size = 46, normalized size = 1.64 \begin {gather*} -\frac {3 \, {\left (x^{2} e^{x} \log \left (x + 5\right )^{2} - 2 \, x^{2} e^{x} \log \left (x + 5\right ) + x^{2} e^{x} - x e^{x}\right )}}{e^{2} - 2 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*
log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="
giac")

[Out]

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

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maple [B]  time = 0.05, size = 58, normalized size = 2.07

method	result	size
risch	\(-\frac {3 x^{2} \ln \left (5+x \right )^{2}}{{\mathrm e}^{2-x}-2}+\frac {6 x^{2} \ln \left (5+x \right )}{{\mathrm e}^{2-x}-2}-\frac {3 x \left (x -1\right )}{{\mathrm e}^{2-x}-2}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*ln(5+x)^2+((6*x^3+36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*ln(5+x)
+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x,method=_RETURNVERBOS
E)

[Out]

-3*x^2/(exp(2-x)-2)*ln(5+x)^2+6*x^2/(exp(2-x)-2)*ln(5+x)-3*x*(x-1)/(exp(2-x)-2)

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maxima [A]  time = 0.76, size = 45, normalized size = 1.61 \begin {gather*} -\frac {3 \, {\left (x^{2} e^{x} \log \left (x + 5\right )^{2} - 2 \, x^{2} e^{x} \log \left (x + 5\right ) + {\left (x^{2} - x\right )} e^{x}\right )}}{e^{2} - 2 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*
log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="
maxima")

[Out]

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

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mupad [B]  time = 0.28, size = 35, normalized size = 1.25 \begin {gather*} \frac {3\,x\,{\mathrm {e}}^{x-2}\,\left (x\,{\ln \left (x+5\right )}^2-2\,x\,\ln \left (x+5\right )+x-1\right )}{2\,{\mathrm {e}}^{x-2}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2 - x)*(12*x + 12*x^2 + 3*x^3 - 15) - log(x + 5)^2*(60*x - exp(2 - x)*(30*x + 21*x^2 + 3*x^3) + 12*x
^2) - 54*x + log(x + 5)*(120*x - exp(2 - x)*(60*x + 36*x^2 + 6*x^3) + 12*x^2) + 30)/(4*x - exp(2 - x)*(4*x + 2
0) + exp(4 - 2*x)*(x + 5) + 20),x)

[Out]

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

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sympy [A]  time = 0.34, size = 36, normalized size = 1.29 \begin {gather*} \frac {- 3 x^{2} \log {\left (x + 5 \right )}^{2} + 6 x^{2} \log {\left (x + 5 \right )} - 3 x^{2} + 3 x}{e^{2 - x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x**3-21*x**2-30*x)*exp(2-x)+12*x**2+60*x)*ln(5+x)**2+((6*x**3+36*x**2+60*x)*exp(2-x)-12*x**2-1
20*x)*ln(5+x)+(-3*x**3-12*x**2-12*x+15)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)**2+(-4*x-20)*exp(2-x)+20+4*x),x)

[Out]

(-3*x**2*log(x + 5)**2 + 6*x**2*log(x + 5) - 3*x**2 + 3*x)/(exp(2 - x) - 2)

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