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

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

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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 = {} $$\begin{array}{c}\int \frac{-30+54x+e^{2-x}(15-12x-12x^2-3x^3)+(-120x-12x^2+e^{2-x}(60x+36x^2+6x^3))\log (5+x)+(60x+12x^2+e^{2-x}(-30x-21x^2-3x^3)){\log }^2(5+x)}{20+e^{2-x}(-20-4x)+4x+e^{4-2x}(5+x)}dx\end{array}$$

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

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

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 $$\begin{array}{c}-\frac{3(x^2\log {(x+5)}^2-2x^2\log (x+5)+x^2-x)}{e^{(-x+2)}-2}\end{array}$$

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 $$\begin{array}{c}-\frac{3(x^2{e}^x\log {(x+5)}^2-2x^2{e}^x\log (x+5)+x^2{e}^x-x{e}^x)}{e^2-2e^x}\end{array}$$

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	$-\frac{3x^2\ln {(5+x)}^2}{{\mathrm{e}}^{2-x}-2}+\frac{6x^2\ln (5+x)}{{\mathrm{e}}^{2-x}-2}-\frac{3x(x-1)}{{\mathrm{e}}^{2-x}-2}$	$58$

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 $$\begin{array}{c}-\frac{3(x^2{e}^x\log {(x+5)}^2-2x^2{e}^x\log (x+5)+(x^2-x)e^x)}{e^2-2e^x}\end{array}$$

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 $$\begin{array}{c}\frac{3x{\mathrm{e}}^{x-2}(x{\ln (x+5)}^2-2x\ln (x+5)+x-1)}{2{\mathrm{e}}^{x-2}-1}\end{array}$$

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 $$\begin{array}{c}\frac{-3x^2\log {(x+5)}^2+6x^2\log (x+5)-3x^2+3x}{e^{2-x}-2}\end{array}$$

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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