3.92.45 \(\int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} (-8+6 x-24 x^2 \log (5))+(32-24 x+5^{4 x} (16-32 x \log (5))) \log (x)-32 \log ^2(x)}{x^3} \, dx\) [9145]

Optimal. Leaf size=27 \[ \left (-3+\frac {5^{4 x}-x}{x}+\frac {x-4 \log (x)}{x}\right )^2 \]

[Out]

((exp(4*x*ln(5))-x)/x-3+(x-4*ln(x))/x)^2

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Rubi [A]
time = 0.22, antiderivative size = 53, normalized size of antiderivative = 1.96, number of steps used = 6, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {14, 2228, 6844, 2326} \begin {gather*} \frac {5^{8 x}}{x^2}+\frac {16 \log ^2(x)}{x^2}-\frac {8\ 625^x \left (3 x^2 \log (5)+2 x \log (25) \log (x)\right )}{x^3 \log (625)}+\frac {24 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24*x + 5^(8*x)*(-2 + 8*x*Log[5]) + 5^(4*x)*(-8 + 6*x - 24*x^2*Log[5]) + (32 - 24*x + 5^(4*x)*(16 - 32*x*L
og[5]))*Log[x] - 32*Log[x]^2)/x^3,x]

[Out]

5^(8*x)/x^2 + (24*Log[x])/x + (16*Log[x]^2)/x^2 - (8*625^x*(3*x^2*Log[5] + 2*x*Log[25]*Log[x]))/(x^3*Log[625])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6844

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, Dist[(-c)*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; Fre
eQ[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2\ 5^{8 x} (-1+4 x \log (5))}{x^3}-\frac {8 (-1+\log (x)) (3 x+4 \log (x))}{x^3}-\frac {2\ 625^x \left (4-3 x+12 x^2 \log (5)-8 \log (x)+8 x \log (25) \log (x)\right )}{x^3}\right ) \, dx\\ &=2 \int \frac {5^{8 x} (-1+4 x \log (5))}{x^3} \, dx-2 \int \frac {625^x \left (4-3 x+12 x^2 \log (5)-8 \log (x)+8 x \log (25) \log (x)\right )}{x^3} \, dx-8 \int \frac {(-1+\log (x)) (3 x+4 \log (x))}{x^3} \, dx\\ &=\frac {5^{8 x}}{x^2}-\frac {8\ 625^x \left (3 x^2 \log (5)+2 x \log (25) \log (x)\right )}{x^3 \log (625)}+8 \text {Subst}\left (\int (3+4 x) \, dx,x,\frac {\log (x)}{x}\right )\\ &=\frac {5^{8 x}}{x^2}+\frac {24 \log (x)}{x}+\frac {16 \log ^2(x)}{x^2}-\frac {8\ 625^x \left (3 x^2 \log (5)+2 x \log (25) \log (x)\right )}{x^3 \log (625)}\\ \end {aligned} \end {gather*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 1.14, size = 231, normalized size = 8.56 \begin {gather*} 2 \left (\frac {625^{2 x}}{2 x^2}-\frac {3\ 625^x}{x}+7 \log ^2(625)-6 \gamma \log ^2(625)-\log (625) \log (152587890625)+\gamma \log (625) \log (152587890625)+3 \log ^2(625) \log \left (\frac {1}{x}\right )-\log (625) \log (390625) \log \left (\frac {1}{x}\right )-\frac {4\ 625^x \log (x)}{x^2}+\frac {12 \log (x)}{x}+3 \log ^2(625) \log (x)-4 \gamma \log ^2(625) \log (x)+\log (625) \log (390625) \log (x)-\log (625) \log (152587890625) \log (x)+\gamma \log (625) \log (152587890625) \log (x)+\text {ExpIntegralEi}(x \log (625)) \log (625) \log (152587890625) \log (x)+\text {Gamma}(0,-x \log (625)) \log (625) \log (152587890625) \log (x)+2 \log ^2(625) \log \left (\frac {1}{x}\right ) \log (x)+\log (625) \log (152587890625) \log (-x) \log (x)+\frac {8 \log ^2(x)}{x^2}-\log (625) \log (390625) \log ^2(x)-6 \log ^2(625) \log (\log (625))+\log ^2(390625) \log (\log (625))-4 \log ^2(625) \log (x) \log (\log (625))+\log (625) \log (152587890625) \log (x) \log (\log (625))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*x + 5^(8*x)*(-2 + 8*x*Log[5]) + 5^(4*x)*(-8 + 6*x - 24*x^2*Log[5]) + (32 - 24*x + 5^(4*x)*(16 -
32*x*Log[5]))*Log[x] - 32*Log[x]^2)/x^3,x]

[Out]

2*(625^(2*x)/(2*x^2) - (3*625^x)/x + 7*Log[625]^2 - 6*EulerGamma*Log[625]^2 - Log[625]*Log[152587890625] + Eul
erGamma*Log[625]*Log[152587890625] + 3*Log[625]^2*Log[x^(-1)] - Log[625]*Log[390625]*Log[x^(-1)] - (4*625^x*Lo
g[x])/x^2 + (12*Log[x])/x + 3*Log[625]^2*Log[x] - 4*EulerGamma*Log[625]^2*Log[x] + Log[625]*Log[390625]*Log[x]
- Log[625]*Log[152587890625]*Log[x] + EulerGamma*Log[625]*Log[152587890625]*Log[x] + ExpIntegralEi[x*Log[625]
]*Log[625]*Log[152587890625]*Log[x] + Gamma[0, -(x*Log[625])]*Log[625]*Log[152587890625]*Log[x] + 2*Log[625]^2
*Log[x^(-1)]*Log[x] + Log[625]*Log[152587890625]*Log[-x]*Log[x] + (8*Log[x]^2)/x^2 - Log[625]*Log[390625]*Log[
x]^2 - 6*Log[625]^2*Log[Log[625]] + Log[390625]^2*Log[Log[625]] - 4*Log[625]^2*Log[x]*Log[Log[625]] + Log[625]
*Log[152587890625]*Log[x]*Log[Log[625]])

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Maple [A]
time = 2.07, size = 54, normalized size = 2.00

method	result	size
risch	\(\frac {16 \ln \left (x \right )^{2}}{x^{2}}+\frac {8 \left (3 x -625^{x}\right ) \ln \left (x \right )}{x^{2}}-\frac {625^{x} \left (6 x -625^{x}\right )}{x^{2}}\)	\(44\)
default	\(\frac {-6 \,{\mathrm e}^{4 x \ln \left (5\right )} x -8 \ln \left (x \right ) {\mathrm e}^{4 x \ln \left (5\right )}}{x^{2}}+\frac {16 \ln \left (x \right )^{2}}{x^{2}}+\frac {24 \ln \left (x \right )}{x}+\frac {{\mathrm e}^{8 x \ln \left (5\right )}}{x^{2}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(x)^2+((-32*x*ln(5)+16)*exp(4*x*ln(5))-24*x+32)*ln(x)+(8*x*ln(5)-2)*exp(4*x*ln(5))^2+(-24*x^2*ln(5)
+6*x-8)*exp(4*x*ln(5))+24*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

(-6*exp(4*x*ln(5))*x-8*ln(x)*exp(4*x*ln(5)))/x^2+16*ln(x)^2/x^2+24*ln(x)/x+1/x^2*exp(4*x*ln(5))^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x)+(8*x*log(5)-2)*exp(4*x*log(5))^2+(-
24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+24*x)/x^3,x, algorithm="maxima")

[Out]

64*gamma(-1, -8*x*log(5))*log(5)^2 + 128*gamma(-2, -4*x*log(5))*log(5)^2 + 128*gamma(-2, -8*x*log(5))*log(5)^2
- 24*Ei(4*x*log(5))*log(5) + 24*gamma(-1, -4*x*log(5))*log(5) + 24*log(x)/x - 8*(5^(4*x)*log(x) - 2*log(x)^2
- 2*log(x) - 1)/x^2 - 16*log(x)/x^2 - 8/x^2 + 8*integrate(5^(4*x)/x^3, x)

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Fricas [A]
time = 0.41, size = 40, normalized size = 1.48 \begin {gather*} -\frac {6 \cdot 5^{4 \, x} x + 8 \, {\left (5^{4 \, x} - 3 \, x\right )} \log \left (x\right ) - 16 \, \log \left (x\right )^{2} - 5^{8 \, x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x)+(8*x*log(5)-2)*exp(4*x*log(5))^2+(-
24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+24*x)/x^3,x, algorithm="fricas")

[Out]

-(6*5^(4*x)*x + 8*(5^(4*x) - 3*x)*log(x) - 16*log(x)^2 - 5^(8*x))/x^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
time = 0.16, size = 54, normalized size = 2.00 \begin {gather*} \frac {24 \log {\left (x \right )}}{x} + \frac {16 \log {\left (x \right )}^{2}}{x^{2}} + \frac {x^{2} e^{8 x \log {\left (5 \right )}} + \left (- 6 x^{3} - 8 x^{2} \log {\left (x \right )}\right ) e^{4 x \log {\left (5 \right )}}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(x)**2+((-32*x*ln(5)+16)*exp(4*x*ln(5))-24*x+32)*ln(x)+(8*x*ln(5)-2)*exp(4*x*ln(5))**2+(-24*x
**2*ln(5)+6*x-8)*exp(4*x*ln(5))+24*x)/x**3,x)

[Out]

24*log(x)/x + 16*log(x)**2/x**2 + (x**2*exp(8*x*log(5)) + (-6*x**3 - 8*x**2*log(x))*exp(4*x*log(5)))/x**4

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x)+(8*x*log(5)-2)*exp(4*x*log(5))^2+(-
24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+24*x)/x^3,x, algorithm="giac")

[Out]

integrate(2*((4*x*log(5) - 1)*5^(8*x) - (12*x^2*log(5) - 3*x + 4)*5^(4*x) - 4*(2*(2*x*log(5) - 1)*5^(4*x) + 3*
x - 4)*log(x) - 16*log(x)^2 + 12*x)/x^3, x)

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Mupad [B]
time = 7.12, size = 31, normalized size = 1.15 \begin {gather*} \frac {\left (4\,\ln \left (x\right )-5^{4\,x}\right )\,\left (6\,x+4\,\ln \left (x\right )-5^{4\,x}\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(24*x + exp(4*x*log(5))*(32*x*log(5) - 16) - 32) - 24*x + 32*log(x)^2 - exp(8*x*log(5))*(8*x*log(
5) - 2) + exp(4*x*log(5))*(24*x^2*log(5) - 6*x + 8))/x^3,x)

[Out]

((4*log(x) - 5^(4*x))*(6*x + 4*log(x) - 5^(4*x)))/x^2

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