3.17.23 \(\int \frac {10-2 e^x+(-5+e^x (1+x)) \log (4 x^2)}{\log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=15 \[ \frac {\left (-5+e^x\right ) x}{\log \left (4 x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 25, normalized size of antiderivative = 1.67, number of steps used = 10, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6742, 2360, 2297, 2300, 2178, 2288} \begin {gather*} \frac {e^x x}{\log \left (4 x^2\right )}-\frac {5 x}{\log \left (4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 - 2*E^x + (-5 + E^x*(1 + x))*Log[4*x^2])/Log[4*x^2]^2,x]

[Out]

(-5*x)/Log[4*x^2] + (E^x*x)/Log[4*x^2]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2297

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

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5 \left (-2+\log \left (4 x^2\right )\right )}{\log ^2\left (4 x^2\right )}+\frac {e^x \left (-2+\log \left (4 x^2\right )+x \log \left (4 x^2\right )\right )}{\log ^2\left (4 x^2\right )}\right ) \, dx\\ &=-\left (5 \int \frac {-2+\log \left (4 x^2\right )}{\log ^2\left (4 x^2\right )} \, dx\right )+\int \frac {e^x \left (-2+\log \left (4 x^2\right )+x \log \left (4 x^2\right )\right )}{\log ^2\left (4 x^2\right )} \, dx\\ &=\frac {e^x x}{\log \left (4 x^2\right )}-5 \int \left (-\frac {2}{\log ^2\left (4 x^2\right )}+\frac {1}{\log \left (4 x^2\right )}\right ) \, dx\\ &=\frac {e^x x}{\log \left (4 x^2\right )}-5 \int \frac {1}{\log \left (4 x^2\right )} \, dx+10 \int \frac {1}{\log ^2\left (4 x^2\right )} \, dx\\ &=-\frac {5 x}{\log \left (4 x^2\right )}+\frac {e^x x}{\log \left (4 x^2\right )}+5 \int \frac {1}{\log \left (4 x^2\right )} \, dx-\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}}\\ &=-\frac {5 x \text {Ei}\left (\frac {1}{2} \log \left (4 x^2\right )\right )}{4 \sqrt {x^2}}-\frac {5 x}{\log \left (4 x^2\right )}+\frac {e^x x}{\log \left (4 x^2\right )}+\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}}\\ &=-\frac {5 x}{\log \left (4 x^2\right )}+\frac {e^x x}{\log \left (4 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 15, normalized size = 1.00 \begin {gather*} \frac {\left (-5+e^x\right ) x}{\log \left (4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 - 2*E^x + (-5 + E^x*(1 + x))*Log[4*x^2])/Log[4*x^2]^2,x]

[Out]

((-5 + E^x)*x)/Log[4*x^2]

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fricas [A]  time = 0.75, size = 17, normalized size = 1.13 \begin {gather*} \frac {x e^{x} - 5 \, x}{\log \left (4 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="fricas")

[Out]

(x*e^x - 5*x)/log(4*x^2)

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giac [A]  time = 0.27, size = 17, normalized size = 1.13 \begin {gather*} \frac {x e^{x} - 5 \, x}{\log \left (4 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="giac")

[Out]

(x*e^x - 5*x)/log(4*x^2)

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maple [A]  time = 0.38, size = 22, normalized size = 1.47

method	result	size
norman	\(\frac {{\mathrm e}^{x} x -5 x}{\ln \left (4 x^{2}\right )}\)	\(22\)
default	\(-\frac {5 x}{\ln \left (4 x^{2}\right )}+\frac {x \,{\mathrm e}^{x}}{\ln \left (4 x^{2}\right )}\)	\(29\)
risch	\(\frac {2 i x \left ({\mathrm e}^{x}-5\right )}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (2)+4 i \ln \relax (x )}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(1/2*x)^2-5)*ln(4*x^2)-2*exp(1/2*x)^2+10)/ln(4*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

(x*exp(1/2*x)^2-5*x)/ln(4*x^2)

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maxima [A]  time = 0.55, size = 17, normalized size = 1.13 \begin {gather*} \frac {x e^{x} - 5 \, x}{2 \, {\left (\log \relax (2) + \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="maxima")

[Out]

1/2*(x*e^x - 5*x)/(log(2) + log(x))

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mupad [B]  time = 1.08, size = 14, normalized size = 0.93 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^x-5\right )}{\ln \left (4\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*x^2)*(exp(x)*(x + 1) - 5) - 2*exp(x) + 10)/log(4*x^2)^2,x)

[Out]

(x*(exp(x) - 5))/log(4*x^2)

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sympy [A]  time = 0.28, size = 20, normalized size = 1.33 \begin {gather*} \frac {x e^{x}}{\log {\left (4 x^{2} \right )}} - \frac {5 x}{\log {\left (4 x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)**2-5)*ln(4*x**2)-2*exp(1/2*x)**2+10)/ln(4*x**2)**2,x)

[Out]

x*exp(x)/log(4*x**2) - 5*x/log(4*x**2)

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