3.5.17 $\int \frac{816+512e+80e^2+(128+40e)\log (x)+5{\log }^2(x)+(168+104e+16e^2+(26+8e)\log (x)+{\log }^2(x))\log (80x)}{9+6e+e^2}dx$

Optimal. Leaf size=23 $$x{(4-\frac{\log (x)}{-3-e})}^2(4+\log (80x))$$

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Rubi [B]  time = 0.42, antiderivative size = 207, normalized size of antiderivative = 9.00, number of steps used = 17, number of rules used = 6, integrand size = 62, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.097, Rules used = {12, 2295, 2296, 6741, 6742, 2361} $$\begin{array}{c}\frac{16(17+5e)x}{3+e}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{4(13+4e)x}{(3+e{)}^2}-\frac{8(7+2e)x}{3+e}+\frac{4x}{(3+e{)}^2}+\frac{4x{\log }^2(x)}{(3+e{)}^2}+\frac{x\log (80x){\log }^2(x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}-\frac{2(13+4e)x\log (x)}{(3+e{)}^2}-\frac{6x\log (x)}{(3+e{)}^2}+\frac{2(13+4e)x\log (80x)\log (x)}{(3+e{)}^2}-\frac{2x\log (80x)\log (x)}{(3+e{)}^2}-\frac{2(13+4e)x\log (80x)}{(3+e{)}^2}+\frac{8(7+2e)x\log (80x)}{3+e}+\frac{2x\log (80x)}{(3+e{)}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(816 + 512*E + 80*E^2 + (128 + 40*E)*Log[x] + 5*Log[x]^2 + (168 + 104*E + 16*E^2 + (26 + 8*E)*Log[x] + Log
[x]^2)*Log[80*x])/(9 + 6*E + E^2),x]

[Out]

(4*x)/(3 + E)^2 - (8*(7 + 2*E)*x)/(3 + E) + (4*(13 + 4*E)*x)/(3 + E)^2 - (8*(16 + 5*E)*x)/(3 + E)^2 + (16*(17
+ 5*E)*x)/(3 + E) - (6*x*Log[x])/(3 + E)^2 - (2*(13 + 4*E)*x*Log[x])/(3 + E)^2 + (8*(16 + 5*E)*x*Log[x])/(3 +
E)^2 + (4*x*Log[x]^2)/(3 + E)^2 + (2*x*Log[80*x])/(3 + E)^2 + (8*(7 + 2*E)*x*Log[80*x])/(3 + E) - (2*(13 + 4*E
)*x*Log[80*x])/(3 + E)^2 - (2*x*Log[x]*Log[80*x])/(3 + E)^2 + (2*(13 + 4*E)*x*Log[x]*Log[80*x])/(3 + E)^2 + (x
*Log[x]^2*Log[80*x])/(3 + E)^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int (816+512e+80e^2+(128+40e)\log (x)+5{\log }^2(x)+(168+104e+16e^2+(26+8e)\log (x)+{\log }^2(x))\log (80x))dx}{(3+e{)}^2}\\ & =\frac{16(17+5e)x}{3+e}+\frac{\int (168+104e+16e^2+(26+8e)\log (x)+{\log }^2(x))\log (80x)dx}{(3+e{)}^2}+\frac{5\int {\log }^2(x)dx}{(3+e{)}^2}+\frac{(8(16+5e))\int \log (x)dx}{(3+e{)}^2}\\ & =-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{5x{\log }^2(x)}{(3+e{)}^2}+\frac{\int (168(1+\frac{1}{21}e(13+2e))+(26+8e)\log (x)+{\log }^2(x))\log (80x)dx}{(3+e{)}^2}-\frac{10\int \log (x)dx}{(3+e{)}^2}\\ & =\frac{10x}{(3+e{)}^2}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{10x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{5x{\log }^2(x)}{(3+e{)}^2}+\frac{\int (8(3+e)(7+2e)\log (80x)+2(13+4e)\log (x)\log (80x)+{\log }^2(x)\log (80x))dx}{(3+e{)}^2}\\ & =\frac{10x}{(3+e{)}^2}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{10x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{5x{\log }^2(x)}{(3+e{)}^2}+\frac{\int {\log }^2(x)\log (80x)dx}{(3+e{)}^2}+\frac{(8(7+2e))\int \log (80x)dx}{3+e}+\frac{(2(13+4e))\int \log (x)\log (80x)dx}{(3+e{)}^2}\\ & =\frac{10x}{(3+e{)}^2}-\frac{8(7+2e)x}{3+e}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{10x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{5x{\log }^2(x)}{(3+e{)}^2}+\frac{2x\log (80x)}{(3+e{)}^2}+\frac{8(7+2e)x\log (80x)}{3+e}-\frac{2(13+4e)x\log (80x)}{(3+e{)}^2}-\frac{2x\log (x)\log (80x)}{(3+e{)}^2}+\frac{2(13+4e)x\log (x)\log (80x)}{(3+e{)}^2}+\frac{x{\log }^2(x)\log (80x)}{(3+e{)}^2}-\frac{\int (2-2\log (x)+{\log }^2(x))dx}{(3+e{)}^2}-\frac{(2(13+4e))\int (-1+\log (x))dx}{(3+e{)}^2}\\ & =\frac{8x}{(3+e{)}^2}-\frac{8(7+2e)x}{3+e}+\frac{2(13+4e)x}{(3+e{)}^2}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{10x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{5x{\log }^2(x)}{(3+e{)}^2}+\frac{2x\log (80x)}{(3+e{)}^2}+\frac{8(7+2e)x\log (80x)}{3+e}-\frac{2(13+4e)x\log (80x)}{(3+e{)}^2}-\frac{2x\log (x)\log (80x)}{(3+e{)}^2}+\frac{2(13+4e)x\log (x)\log (80x)}{(3+e{)}^2}+\frac{x{\log }^2(x)\log (80x)}{(3+e{)}^2}-\frac{\int {\log }^2(x)dx}{(3+e{)}^2}+\frac{2\int \log (x)dx}{(3+e{)}^2}-\frac{(2(13+4e))\int \log (x)dx}{(3+e{)}^2}\\ & =\frac{6x}{(3+e{)}^2}-\frac{8(7+2e)x}{3+e}+\frac{4(13+4e)x}{(3+e{)}^2}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{8x\log (x)}{(3+e{)}^2}-\frac{2(13+4e)x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{4x{\log }^2(x)}{(3+e{)}^2}+\frac{2x\log (80x)}{(3+e{)}^2}+\frac{8(7+2e)x\log (80x)}{3+e}-\frac{2(13+4e)x\log (80x)}{(3+e{)}^2}-\frac{2x\log (x)\log (80x)}{(3+e{)}^2}+\frac{2(13+4e)x\log (x)\log (80x)}{(3+e{)}^2}+\frac{x{\log }^2(x)\log (80x)}{(3+e{)}^2}+\frac{2\int \log (x)dx}{(3+e{)}^2}\\ & =\frac{4x}{(3+e{)}^2}-\frac{8(7+2e)x}{3+e}+\frac{4(13+4e)x}{(3+e{)}^2}-\frac{8(16+5e)x}{(3+e{)}^2}+\frac{16(17+5e)x}{3+e}-\frac{6x\log (x)}{(3+e{)}^2}-\frac{2(13+4e)x\log (x)}{(3+e{)}^2}+\frac{8(16+5e)x\log (x)}{(3+e{)}^2}+\frac{4x{\log }^2(x)}{(3+e{)}^2}+\frac{2x\log (80x)}{(3+e{)}^2}+\frac{8(7+2e)x\log (80x)}{3+e}-\frac{2(13+4e)x\log (80x)}{(3+e{)}^2}-\frac{2x\log (x)\log (80x)}{(3+e{)}^2}+\frac{2(13+4e)x\log (x)\log (80x)}{(3+e{)}^2}+\frac{x{\log }^2(x)\log (80x)}{(3+e{)}^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.12, size = 23, normalized size = 1.00 $$\begin{array}{c}\frac{x(4(3+e)+\log (x){)}^2(4+\log (80x))}{(3+e{)}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(816 + 512*E + 80*E^2 + (128 + 40*E)*Log[x] + 5*Log[x]^2 + (168 + 104*E + 16*E^2 + (26 + 8*E)*Log[x]
+ Log[x]^2)*Log[80*x])/(9 + 6*E + E^2),x]

[Out]

(x*(4*(3 + E) + Log[x])^2*(4 + Log[80*x]))/(3 + E)^2

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fricas [B]  time = 0.67, size = 95, normalized size = 4.13 $$\begin{array}{c}\frac{x\log (x{)}^3+(8xe+x\log \left(80\right)+28x)\log (x{)}^2+64x{e}^2+384xe+16(x{e}^2+6xe+9x)\log \left(80\right)+8(2x{e}^2+16xe+(xe+3x)\log \left(80\right)+30x)\log (x)+576x}{e^2+6e+9}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)^2+(8*exp(1)+26)*log(x)+16*exp(1)^2+104*exp(1)+168)*log(80*x)+5*log(x)^2+(40*exp(1)+128)*log
(x)+80*exp(1)^2+512*exp(1)+816)/(exp(1)^2+6*exp(1)+9),x, algorithm="fricas")

[Out]

(x*log(x)^3 + (8*x*e + x*log(80) + 28*x)*log(x)^2 + 64*x*e^2 + 384*x*e + 16*(x*e^2 + 6*x*e + 9*x)*log(80) + 8*
(2*x*e^2 + 16*x*e + (x*e + 3*x)*log(80) + 30*x)*log(x) + 576*x)/(e^2 + 6*e + 9)

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giac [B]  time = 0.21, size = 125, normalized size = 5.43 $$\begin{array}{c}\frac{8xe\log \left(80\right)\log (x)+8xe\log (x{)}^2+x\log \left(80\right)\log (x{)}^2+x\log (x{)}^3+16x{e}^2\log \left(80\right)+96xe\log \left(80\right)+16x{e}^2\log (x)+88xe\log (x)+24x\log \left(80\right)\log (x)+28x\log (x{)}^2+8(x\log (x)-x)(5e+16)+64x{e}^2+424xe+144x\log \left(80\right)+112x\log (x)+704x}{e^2+6e+9}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)^2+(8*exp(1)+26)*log(x)+16*exp(1)^2+104*exp(1)+168)*log(80*x)+5*log(x)^2+(40*exp(1)+128)*log
(x)+80*exp(1)^2+512*exp(1)+816)/(exp(1)^2+6*exp(1)+9),x, algorithm="giac")

[Out]

(8*x*e*log(80)*log(x) + 8*x*e*log(x)^2 + x*log(80)*log(x)^2 + x*log(x)^3 + 16*x*e^2*log(80) + 96*x*e*log(80) +
16*x*e^2*log(x) + 88*x*e*log(x) + 24*x*log(80)*log(x) + 28*x*log(x)^2 + 8*(x*log(x) - x)*(5*e + 16) + 64*x*e^
2 + 424*x*e + 144*x*log(80) + 112*x*log(x) + 704*x)/(e^2 + 6*e + 9)

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maple [B]  time = 0.12, size = 72, normalized size = 3.13

method	result	size
norman	$\frac{(64\mathrm{e}+192)x+(16\mathrm{e}+48)x\ln \left(80x\right)+\frac{x\ln (x{)}^2\ln \left(80x\right)}{3+\mathrm{e}}+32x\ln (x)+8x\ln (x)\ln \left(80x\right)+\frac{4x\ln (x{)}^2}{3+\mathrm{e}}}{3+\mathrm{e}}$	$72$
default	$\frac{576x+128x\mathrm{e}\ln (x)+384x\mathrm{e}+240x\ln (x)+16{\mathrm{e}}^2\ln \left(80\right)x+16x{\mathrm{e}}^2\ln (x)+64{\mathrm{e}}^{2}x+8\ln \left(80\right)\mathrm{e}\ln (x)x+96\mathrm{e}\ln \left(80\right)x+8\mathrm{e}\ln (x{)}^{2}x+\ln \left(80\right)\ln (x{)}^{2}x+24\ln \left(80\right)\ln (x)x+144\ln \left(80\right)x+x\ln (x{)}^3+28x\ln (x{)}^2}{{\mathrm{e}}^2+6\mathrm{e}+9}$	$118$
risch	$\frac{x\ln (x{)}^3}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{(56+16\mathrm{e}+8\ln (2)+2\ln (5))x\ln (x{)}^2}{2{\mathrm{e}}^2+12\mathrm{e}+18}+\frac{4(60+4{\mathrm{e}}^2+8\mathrm{e}\ln (2)+2\mathrm{e}\ln (5)+32\mathrm{e}+24\ln (2)+6\ln (5))x\ln (x)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{64x{\mathrm{e}}^2\ln (2)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{16x{\mathrm{e}}^2\ln (5)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{384x\mathrm{e}\ln (2)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{96x\mathrm{e}\ln (5)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{64x{\mathrm{e}}^2}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{384x\mathrm{e}}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{576x\ln (2)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{144x\ln (5)}{{\mathrm{e}}^2+6\mathrm{e}+9}+\frac{576x}{{\mathrm{e}}^2+6\mathrm{e}+9}$	$235$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)^2+(8*exp(1)+26)*ln(x)+16*exp(1)^2+104*exp(1)+168)*ln(80*x)+5*ln(x)^2+(40*exp(1)+128)*ln(x)+80*exp(
1)^2+512*exp(1)+816)/(exp(1)^2+6*exp(1)+9),x,method=_RETURNVERBOSE)

[Out]

((64*exp(1)+192)*x+(16*exp(1)+48)*x*ln(80*x)+1/(3+exp(1))*x*ln(x)^2*ln(80*x)+32*x*ln(x)+8*x*ln(x)*ln(80*x)+4/(
3+exp(1))*x*ln(x)^2)/(3+exp(1))

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maxima [B]  time = 0.42, size = 133, normalized size = 5.78 $$\begin{array}{c}-\frac{2x(4e+11)\log (x)+x\log (x{)}^2-5(\log (x{)}^2-2\log (x)+2)x+2x(8e^2+44e+61)-8(x\log (x)-x)(5e+16)-80x{e}^2-512xe-((\log (x{)}^2-2\log (x)+2)x+2(x\log (x)-x)(4e+13)+16x{e}^2+104xe+168x)\log \left(80x\right)-816x}{e^2+6e+9}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)^2+(8*exp(1)+26)*log(x)+16*exp(1)^2+104*exp(1)+168)*log(80*x)+5*log(x)^2+(40*exp(1)+128)*log
(x)+80*exp(1)^2+512*exp(1)+816)/(exp(1)^2+6*exp(1)+9),x, algorithm="maxima")

[Out]

-(2*x*(4*e + 11)*log(x) + x*log(x)^2 - 5*(log(x)^2 - 2*log(x) + 2)*x + 2*x*(8*e^2 + 44*e + 61) - 8*(x*log(x) -
x)*(5*e + 16) - 80*x*e^2 - 512*x*e - ((log(x)^2 - 2*log(x) + 2)*x + 2*(x*log(x) - x)*(4*e + 13) + 16*x*e^2 +
104*x*e + 168*x)*log(80*x) - 816*x)/(e^2 + 6*e + 9)

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mupad [B]  time = 0.55, size = 76, normalized size = 3.30 $$\begin{array}{c}\frac{x({\ln (x)}^3+{\ln (x)}^2(\ln \left(80x\right)+8\mathrm{e}-\ln (x)+28)+16{(\mathrm{e}+3)}^2(\ln \left(80x\right)-\ln (x)+4)+8\ln (x)(\mathrm{e}+3)(\ln \left(80x\right)+2\mathrm{e}-\ln (x)+10))}{6\mathrm{e}+{\mathrm{e}}^2+9}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((512*exp(1) + 80*exp(2) + log(80*x)*(104*exp(1) + 16*exp(2) + log(x)^2 + log(x)*(8*exp(1) + 26) + 168) + 5
*log(x)^2 + log(x)*(40*exp(1) + 128) + 816)/(6*exp(1) + exp(2) + 9),x)

[Out]

(x*(log(x)^3 + log(x)^2*(log(80*x) + 8*exp(1) - log(x) + 28) + 16*(exp(1) + 3)^2*(log(80*x) - log(x) + 4) + 8*
log(x)*(exp(1) + 3)*(log(80*x) + 2*exp(1) - log(x) + 10)))/(6*exp(1) + exp(2) + 9)

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sympy [B]  time = 0.40, size = 80, normalized size = 3.48 $$\begin{array}{c}\frac{x\log {(x)}^3}{e^2+9+6e}+x(64+16\log \left(80\right))+\frac{(x\log \left(80\right)+8ex+28x)\log {(x)}^2}{e^2+9+6e}+\frac{(8x\log \left(80\right)+16ex+80x)\log (x)}{e+3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)**2+(8*exp(1)+26)*ln(x)+16*exp(1)**2+104*exp(1)+168)*ln(80*x)+5*ln(x)**2+(40*exp(1)+128)*ln(x
)+80*exp(1)**2+512*exp(1)+816)/(exp(1)**2+6*exp(1)+9),x)

[Out]

x*log(x)**3/(exp(2) + 9 + 6*E) + x*(64 + 16*log(80)) + (x*log(80) + 8*E*x + 28*x)*log(x)**2/(exp(2) + 9 + 6*E)
+ (8*x*log(80) + 16*E*x + 80*x)*log(x)/(E + 3)

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