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

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

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Rubi [A]  time = 0.23, 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, 2197, 6712, 2288} \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 2197

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 2288

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 6712

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]] /; FreeQ
[{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 \operatorname {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 [A]  time = 0.17, size = 47, normalized size = 1.74 \begin {gather*} \frac {4 \left (625^x \left (625^x-6 x\right ) \log (5)+\left (-4 625^x \log (25)+6 x \log (625)\right ) \log (x)+4 \log (625) \log ^2(x)\right )}{x^2 \log (625)} \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]

(4*(625^x*(625^x - 6*x)*Log[5] + (-4*625^x*Log[25] + 6*x*Log[625])*Log[x] + 4*Log[625]*Log[x]^2))/(x^2*Log[625
])

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fricas [A]  time = 1.15, size = 40, normalized size = 1.48 \begin {gather*} -\frac {6 \cdot 5^{4 \, x} x + 8 \, {\left (5^{4 \, x} - 3 \, x\right )} \log \relax (x) - 16 \, \log \relax (x)^{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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left ({\left (4 \, x \log \relax (5) - 1\right )} 5^{8 \, x} - {\left (12 \, x^{2} \log \relax (5) - 3 \, x + 4\right )} 5^{4 \, x} - 4 \, {\left (2 \, {\left (2 \, x \log \relax (5) - 1\right )} 5^{4 \, x} + 3 \, x - 4\right )} \log \relax (x) - 16 \, \log \relax (x)^{2} + 12 \, x\right )}}{x^{3}}\,{d x} \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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maple [A]  time = 0.18, size = 44, normalized size = 1.63

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

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]

16*ln(x)^2/x^2+8*(3*x-625^x)/x^2*ln(x)-625^x*(6*x-625^x)/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 64 \, \Gamma \left (-1, -8 \, x \log \relax (5)\right ) \log \relax (5)^{2} + 128 \, \Gamma \left (-2, -4 \, x \log \relax (5)\right ) \log \relax (5)^{2} + 128 \, \Gamma \left (-2, -8 \, x \log \relax (5)\right ) \log \relax (5)^{2} - 24 \, {\rm Ei}\left (4 \, x \log \relax (5)\right ) \log \relax (5) + 24 \, \Gamma \left (-1, -4 \, x \log \relax (5)\right ) \log \relax (5) + \frac {24 \, \log \relax (x)}{x} - \frac {8 \, {\left (5^{4 \, x} \log \relax (x) - 2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) - 1\right )}}{x^{2}} - \frac {16 \, \log \relax (x)}{x^{2}} - \frac {8}{x^{2}} + 8 \, \int \frac {5^{4 \, x}}{x^{3}}\,{d x} \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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mupad [B]  time = 7.12, size = 31, normalized size = 1.15 \begin {gather*} \frac {\left (4\,\ln \relax (x)-5^{4\,x}\right )\,\left (6\,x+4\,\ln \relax (x)-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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sympy [B]  time = 0.43, size = 54, normalized size = 2.00 \begin {gather*} \frac {24 \log {\relax (x )}}{x} + \frac {16 \log {\relax (x )}^{2}}{x^{2}} + \frac {x^{2} e^{8 x \log {\relax (5 )}} + \left (- 6 x^{3} - 8 x^{2} \log {\relax (x )}\right ) e^{4 x \log {\relax (5 )}}}{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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