3.92.78 \(\int \frac {30 x^2-20 x^5-3 x^6+e^x (3 x^2-3 x^3-2 x^5-x^6)+(48 x-8 x^4) \log (2)+9 \log ^2(2)}{x^4} \, dx\)

Optimal. Leaf size=36 \[ \left (\frac {3}{x}+x^2\right ) \left (6-e^x-x-\left (-5+\frac {x-\log (2)}{x}\right )^2\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 53, normalized size of antiderivative = 1.47, number of steps used = 14, number of rules used = 6, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2199, 2177, 2178, 2176, 2194} \begin {gather*} -x^3-\frac {3 \log ^2(2)}{x^3}-e^x x^2-10 x^2-\frac {24 \log (2)}{x^2}-\frac {3 e^x}{x}-\frac {30}{x}-8 x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30*x^2 - 20*x^5 - 3*x^6 + E^x*(3*x^2 - 3*x^3 - 2*x^5 - x^6) + (48*x - 8*x^4)*Log[2] + 9*Log[2]^2)/x^4,x]

[Out]

-30/x - (3*E^x)/x - 10*x^2 - E^x*x^2 - x^3 - (24*Log[2])/x^2 - 8*x*Log[2] - (3*Log[2]^2)/x^3

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^x \left (-3+3 x+2 x^3+x^4\right )}{x^2}+\frac {30 x^2-20 x^5-3 x^6+48 x \log (2)-8 x^4 \log (2)+9 \log ^2(2)}{x^4}\right ) \, dx\\ &=-\int \frac {e^x \left (-3+3 x+2 x^3+x^4\right )}{x^2} \, dx+\int \frac {30 x^2-20 x^5-3 x^6+48 x \log (2)-8 x^4 \log (2)+9 \log ^2(2)}{x^4} \, dx\\ &=-\int \left (-\frac {3 e^x}{x^2}+\frac {3 e^x}{x}+2 e^x x+e^x x^2\right ) \, dx+\int \left (\frac {30}{x^2}-20 x-3 x^2-8 \log (2)+\frac {48 \log (2)}{x^3}+\frac {9 \log ^2(2)}{x^4}\right ) \, dx\\ &=-\frac {30}{x}-10 x^2-x^3-\frac {24 \log (2)}{x^2}-8 x \log (2)-\frac {3 \log ^2(2)}{x^3}-2 \int e^x x \, dx+3 \int \frac {e^x}{x^2} \, dx-3 \int \frac {e^x}{x} \, dx-\int e^x x^2 \, dx\\ &=-\frac {30}{x}-\frac {3 e^x}{x}-2 e^x x-10 x^2-e^x x^2-x^3-3 \text {Ei}(x)-\frac {24 \log (2)}{x^2}-8 x \log (2)-\frac {3 \log ^2(2)}{x^3}+2 \int e^x \, dx+2 \int e^x x \, dx+3 \int \frac {e^x}{x} \, dx\\ &=2 e^x-\frac {30}{x}-\frac {3 e^x}{x}-10 x^2-e^x x^2-x^3-\frac {24 \log (2)}{x^2}-8 x \log (2)-\frac {3 \log ^2(2)}{x^3}-2 \int e^x \, dx\\ &=-\frac {30}{x}-\frac {3 e^x}{x}-10 x^2-e^x x^2-x^3-\frac {24 \log (2)}{x^2}-8 x \log (2)-\frac {3 \log ^2(2)}{x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 46, normalized size = 1.28 \begin {gather*} -\frac {3 \left (10+e^x\right ) x^2+\left (10+e^x\right ) x^5+x^6+24 x \log (2)+8 x^4 \log (2)+3 \log ^2(2)}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30*x^2 - 20*x^5 - 3*x^6 + E^x*(3*x^2 - 3*x^3 - 2*x^5 - x^6) + (48*x - 8*x^4)*Log[2] + 9*Log[2]^2)/x
^4,x]

[Out]

-((3*(10 + E^x)*x^2 + (10 + E^x)*x^5 + x^6 + 24*x*Log[2] + 8*x^4*Log[2] + 3*Log[2]^2)/x^3)

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fricas [A]  time = 0.55, size = 48, normalized size = 1.33 \begin {gather*} -\frac {x^{6} + 10 \, x^{5} + 30 \, x^{2} + {\left (x^{5} + 3 \, x^{2}\right )} e^{x} + 8 \, {\left (x^{4} + 3 \, x\right )} \log \relax (2) + 3 \, \log \relax (2)^{2}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^6-2*x^5-3*x^3+3*x^2)*exp(x)+9*log(2)^2+(-8*x^4+48*x)*log(2)-3*x^6-20*x^5+30*x^2)/x^4,x, algorit
hm="fricas")

[Out]

-(x^6 + 10*x^5 + 30*x^2 + (x^5 + 3*x^2)*e^x + 8*(x^4 + 3*x)*log(2) + 3*log(2)^2)/x^3

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giac [A]  time = 0.20, size = 50, normalized size = 1.39 \begin {gather*} -\frac {x^{6} + x^{5} e^{x} + 10 \, x^{5} + 8 \, x^{4} \log \relax (2) + 3 \, x^{2} e^{x} + 30 \, x^{2} + 24 \, x \log \relax (2) + 3 \, \log \relax (2)^{2}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^6-2*x^5-3*x^3+3*x^2)*exp(x)+9*log(2)^2+(-8*x^4+48*x)*log(2)-3*x^6-20*x^5+30*x^2)/x^4,x, algorit
hm="giac")

[Out]

-(x^6 + x^5*e^x + 10*x^5 + 8*x^4*log(2) + 3*x^2*e^x + 30*x^2 + 24*x*log(2) + 3*log(2)^2)/x^3

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maple [A]  time = 0.09, size = 50, normalized size = 1.39

method	result	size
risch	\(-x^{3}-8 x \ln \relax (2)-10 x^{2}+\frac {-3 \ln \relax (2)^{2}-24 x \ln \relax (2)-30 x^{2}}{x^{3}}-\frac {\left (x^{3}+3\right ) {\mathrm e}^{x}}{x}\)	\(50\)
default	\(-10 x^{2}-\frac {30}{x}-x^{3}-\frac {24 \ln \relax (2)}{x^{2}}-\frac {3 \,{\mathrm e}^{x}}{x}-{\mathrm e}^{x} x^{2}-\frac {3 \ln \relax (2)^{2}}{x^{3}}-8 x \ln \relax (2)\)	\(52\)
norman	\(\frac {-30 x^{2}-10 x^{5}-x^{6}-3 \ln \relax (2)^{2}-24 x \ln \relax (2)-8 x^{4} \ln \relax (2)-x^{5} {\mathrm e}^{x}-3 \,{\mathrm e}^{x} x^{2}}{x^{3}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^6-2*x^5-3*x^3+3*x^2)*exp(x)+9*ln(2)^2+(-8*x^4+48*x)*ln(2)-3*x^6-20*x^5+30*x^2)/x^4,x,method=_RETURNVE
RBOSE)

[Out]

-x^3-8*x*ln(2)-10*x^2+(-3*ln(2)^2-24*x*ln(2)-30*x^2)/x^3-(x^3+3)/x*exp(x)

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maxima [C]  time = 0.39, size = 67, normalized size = 1.86 \begin {gather*} -x^{3} - 10 \, x^{2} - {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 2 \, {\left (x - 1\right )} e^{x} - 8 \, x \log \relax (2) - \frac {30}{x} - \frac {24 \, \log \relax (2)}{x^{2}} - \frac {3 \, \log \relax (2)^{2}}{x^{3}} - 3 \, {\rm Ei}\relax (x) + 3 \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^6-2*x^5-3*x^3+3*x^2)*exp(x)+9*log(2)^2+(-8*x^4+48*x)*log(2)-3*x^6-20*x^5+30*x^2)/x^4,x, algorit
hm="maxima")

[Out]

-x^3 - 10*x^2 - (x^2 - 2*x + 2)*e^x - 2*(x - 1)*e^x - 8*x*log(2) - 30/x - 24*log(2)/x^2 - 3*log(2)^2/x^3 - 3*E
i(x) + 3*gamma(-1, -x)

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mupad [B]  time = 0.17, size = 47, normalized size = 1.31 \begin {gather*} -x\,\ln \left (256\right )-x^2\,\left ({\mathrm {e}}^x+10\right )-x^3-\frac {24\,x\,\ln \relax (2)+3\,{\ln \relax (2)}^2+x^2\,\left (3\,{\mathrm {e}}^x+30\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(3*x^3 - 3*x^2 + 2*x^5 + x^6) - log(2)*(48*x - 8*x^4) - 9*log(2)^2 - 30*x^2 + 20*x^5 + 3*x^6)/x^4
,x)

[Out]

- x*log(256) - x^2*(exp(x) + 10) - x^3 - (24*x*log(2) + 3*log(2)^2 + x^2*(3*exp(x) + 30))/x^3

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sympy [B]  time = 0.23, size = 48, normalized size = 1.33 \begin {gather*} - x^{3} - 10 x^{2} - 8 x \log {\relax (2 )} + \frac {\left (- x^{3} - 3\right ) e^{x}}{x} - \frac {30 x^{2} + 24 x \log {\relax (2 )} + 3 \log {\relax (2 )}^{2}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**6-2*x**5-3*x**3+3*x**2)*exp(x)+9*ln(2)**2+(-8*x**4+48*x)*ln(2)-3*x**6-20*x**5+30*x**2)/x**4,x)

[Out]

-x**3 - 10*x**2 - 8*x*log(2) + (-x**3 - 3)*exp(x)/x - (30*x**2 + 24*x*log(2) + 3*log(2)**2)/x**3

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