3.36.95 \(\int \frac {-64-648 x-1522 x^2-216 x^3+e^x (40 x^3-40 x^4-310 x^5-40 x^6+(-80 x^2-230 x^3+270 x^4+40 x^5) \log (4))+e^{2 x} (50 x^6+50 x^7+(-50 x^5-100 x^6) \log (4)+50 x^5 \log ^2(4))}{x^5} \, dx\)

Optimal. Leaf size=30 \[ \frac {\left (-2-\frac {4}{x}+x-5 \left (5+x+e^x x (-x+\log (4))\right )\right )^2}{x^2} \]

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Rubi [C]  time = 0.34, antiderivative size = 174, normalized size of antiderivative = 5.80, number of steps used = 23, number of rules used = 7, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {14, 2199, 2176, 2194, 2177, 2178, 2196} \begin {gather*} -40 \log (4) \text {Ei}(x)+10 (4-23 \log (4)) \text {Ei}(x)-10 (4-27 \log (4)) \text {Ei}(x)+\frac {16}{x^4}+\frac {216}{x^3}+25 e^{2 x} x^2+\frac {761}{x^2}+\frac {40 e^x \log (4)}{x^2}-40 e^x x-25 e^{2 x} x+40 e^x+\frac {25 e^{2 x}}{2}+\frac {216}{x}+25 e^{2 x} x (1-2 \log (4))-10 e^x (31-\log (256))-\frac {25}{2} e^{2 x} (1-\log (16))-25 e^{2 x} (1-\log (4)) \log (4)+\frac {40 e^x \log (4)}{x}-\frac {10 e^x (4-23 \log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 648*x - 1522*x^2 - 216*x^3 + E^x*(40*x^3 - 40*x^4 - 310*x^5 - 40*x^6 + (-80*x^2 - 230*x^3 + 270*x^4
+ 40*x^5)*Log[4]) + E^(2*x)*(50*x^6 + 50*x^7 + (-50*x^5 - 100*x^6)*Log[4] + 50*x^5*Log[4]^2))/x^5,x]

[Out]

40*E^x + (25*E^(2*x))/2 + 16/x^4 + 216/x^3 + 761/x^2 + 216/x - 40*E^x*x - 25*E^(2*x)*x + 25*E^(2*x)*x^2 - 10*E
xpIntegralEi[x]*(4 - 27*Log[4]) - (10*E^x*(4 - 23*Log[4]))/x + 10*ExpIntegralEi[x]*(4 - 23*Log[4]) + 25*E^(2*x
)*x*(1 - 2*Log[4]) + (40*E^x*Log[4])/x^2 + (40*E^x*Log[4])/x - 40*ExpIntegralEi[x]*Log[4] - 25*E^(2*x)*(1 - Lo
g[4])*Log[4] - (25*E^(2*x)*(1 - Log[16]))/2 - 10*E^x*(31 - Log[256])

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 2196

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

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 {2 \left (32+324 x+761 x^2+108 x^3\right )}{x^5}+\frac {10 e^x \left (-4 x^4-x^2 (4-27 \log (4))+x (4-23 \log (4))-x^3 (31-4 \log (4))-8 \log (4)\right )}{x^3}+50 e^{2 x} (x-\log (4)) (1+x-\log (4))\right ) \, dx\\ &=-\left (2 \int \frac {32+324 x+761 x^2+108 x^3}{x^5} \, dx\right )+10 \int \frac {e^x \left (-4 x^4-x^2 (4-27 \log (4))+x (4-23 \log (4))-x^3 (31-4 \log (4))-8 \log (4)\right )}{x^3} \, dx+50 \int e^{2 x} (x-\log (4)) (1+x-\log (4)) \, dx\\ &=-\left (2 \int \left (\frac {32}{x^5}+\frac {324}{x^4}+\frac {761}{x^3}+\frac {108}{x^2}\right ) \, dx\right )+10 \int \left (-4 e^x x-31 e^x \left (1-\frac {8 \log (2)}{31}\right )+\frac {e^x (4-23 \log (4))}{x^2}-\frac {8 e^x \log (4)}{x^3}+\frac {e^x (-4+27 \log (4))}{x}\right ) \, dx+50 \int \left (e^{2 x} x^2+e^{2 x} x (1-2 \log (4))+e^{2 x} (-1+\log (4)) \log (4)\right ) \, dx\\ &=\frac {16}{x^4}+\frac {216}{x^3}+\frac {761}{x^2}+\frac {216}{x}-40 \int e^x x \, dx+50 \int e^{2 x} x^2 \, dx-(10 (31-8 \log (2))) \int e^x \, dx-(10 (4-27 \log (4))) \int \frac {e^x}{x} \, dx+(10 (4-23 \log (4))) \int \frac {e^x}{x^2} \, dx+(50 (1-2 \log (4))) \int e^{2 x} x \, dx-(80 \log (4)) \int \frac {e^x}{x^3} \, dx-(50 (1-\log (4)) \log (4)) \int e^{2 x} \, dx\\ &=\frac {16}{x^4}+\frac {216}{x^3}+\frac {761}{x^2}+\frac {216}{x}-40 e^x x+25 e^{2 x} x^2-10 e^x (31-8 \log (2))-10 \text {Ei}(x) (4-27 \log (4))-\frac {10 e^x (4-23 \log (4))}{x}+25 e^{2 x} x (1-2 \log (4))+\frac {40 e^x \log (4)}{x^2}-25 e^{2 x} (1-\log (4)) \log (4)+40 \int e^x \, dx-50 \int e^{2 x} x \, dx+(10 (4-23 \log (4))) \int \frac {e^x}{x} \, dx-(25 (1-2 \log (4))) \int e^{2 x} \, dx-(40 \log (4)) \int \frac {e^x}{x^2} \, dx\\ &=40 e^x+\frac {16}{x^4}+\frac {216}{x^3}+\frac {761}{x^2}+\frac {216}{x}-40 e^x x-25 e^{2 x} x+25 e^{2 x} x^2-10 e^x (31-8 \log (2))-10 \text {Ei}(x) (4-27 \log (4))-\frac {10 e^x (4-23 \log (4))}{x}+10 \text {Ei}(x) (4-23 \log (4))-\frac {25}{2} e^{2 x} (1-2 \log (4))+25 e^{2 x} x (1-2 \log (4))+\frac {40 e^x \log (4)}{x^2}+\frac {40 e^x \log (4)}{x}-25 e^{2 x} (1-\log (4)) \log (4)+25 \int e^{2 x} \, dx-(40 \log (4)) \int \frac {e^x}{x} \, dx\\ &=40 e^x+\frac {25 e^{2 x}}{2}+\frac {16}{x^4}+\frac {216}{x^3}+\frac {761}{x^2}+\frac {216}{x}-40 e^x x-25 e^{2 x} x+25 e^{2 x} x^2-10 e^x (31-8 \log (2))-10 \text {Ei}(x) (4-27 \log (4))-\frac {10 e^x (4-23 \log (4))}{x}+10 \text {Ei}(x) (4-23 \log (4))-\frac {25}{2} e^{2 x} (1-2 \log (4))+25 e^{2 x} x (1-2 \log (4))+\frac {40 e^x \log (4)}{x^2}+\frac {40 e^x \log (4)}{x}-40 \text {Ei}(x) \log (4)-25 e^{2 x} (1-\log (4)) \log (4)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 95, normalized size = 3.17 \begin {gather*} 2 \left (\frac {8}{x^4}+\frac {108}{x^3}+\frac {761}{2 x^2}+\frac {108}{x}+e^{2 x} \left (\frac {25 x^2}{2}-\frac {25}{2} x \log (16)+\frac {25}{4} \left (-2 \log (4)+2 \log ^2(4)+\log (16)\right )\right )+e^x \left (-20 x+\frac {20 \log (4)}{x^2}+\frac {5 (-4+27 \log (4))}{x}+5 (-27+\log (256))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 648*x - 1522*x^2 - 216*x^3 + E^x*(40*x^3 - 40*x^4 - 310*x^5 - 40*x^6 + (-80*x^2 - 230*x^3 + 2
70*x^4 + 40*x^5)*Log[4]) + E^(2*x)*(50*x^6 + 50*x^7 + (-50*x^5 - 100*x^6)*Log[4] + 50*x^5*Log[4]^2))/x^5,x]

[Out]

2*(8/x^4 + 108/x^3 + 761/(2*x^2) + 108/x + E^(2*x)*((25*x^2)/2 - (25*x*Log[16])/2 + (25*(-2*Log[4] + 2*Log[4]^
2 + Log[16]))/4) + E^x*(-20*x + (20*Log[4])/x^2 + (5*(-4 + 27*Log[4]))/x + 5*(-27 + Log[256])))

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fricas [B]  time = 0.83, size = 85, normalized size = 2.83 \begin {gather*} \frac {216 \, x^{3} + 761 \, x^{2} + 25 \, {\left (x^{6} - 4 \, x^{5} \log \relax (2) + 4 \, x^{4} \log \relax (2)^{2}\right )} e^{\left (2 \, x\right )} - 10 \, {\left (4 \, x^{5} + 27 \, x^{4} + 4 \, x^{3} - 2 \, {\left (4 \, x^{4} + 27 \, x^{3} + 4 \, x^{2}\right )} \log \relax (2)\right )} e^{x} + 216 \, x + 16}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5*log(2)^2+2*(-100*x^6-50*x^5)*log(2)+50*x^7+50*x^6)*exp(x)^2+(2*(40*x^5+270*x^4-230*x^3-80*
x^2)*log(2)-40*x^6-310*x^5-40*x^4+40*x^3)*exp(x)-216*x^3-1522*x^2-648*x-64)/x^5,x, algorithm="fricas")

[Out]

(216*x^3 + 761*x^2 + 25*(x^6 - 4*x^5*log(2) + 4*x^4*log(2)^2)*e^(2*x) - 10*(4*x^5 + 27*x^4 + 4*x^3 - 2*(4*x^4
+ 27*x^3 + 4*x^2)*log(2))*e^x + 216*x + 16)/x^4

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giac [B]  time = 0.16, size = 100, normalized size = 3.33 \begin {gather*} \frac {25 \, x^{6} e^{\left (2 \, x\right )} - 100 \, x^{5} e^{\left (2 \, x\right )} \log \relax (2) + 100 \, x^{4} e^{\left (2 \, x\right )} \log \relax (2)^{2} - 40 \, x^{5} e^{x} + 80 \, x^{4} e^{x} \log \relax (2) - 270 \, x^{4} e^{x} + 540 \, x^{3} e^{x} \log \relax (2) - 40 \, x^{3} e^{x} + 80 \, x^{2} e^{x} \log \relax (2) + 216 \, x^{3} + 761 \, x^{2} + 216 \, x + 16}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5*log(2)^2+2*(-100*x^6-50*x^5)*log(2)+50*x^7+50*x^6)*exp(x)^2+(2*(40*x^5+270*x^4-230*x^3-80*
x^2)*log(2)-40*x^6-310*x^5-40*x^4+40*x^3)*exp(x)-216*x^3-1522*x^2-648*x-64)/x^5,x, algorithm="giac")

[Out]

(25*x^6*e^(2*x) - 100*x^5*e^(2*x)*log(2) + 100*x^4*e^(2*x)*log(2)^2 - 40*x^5*e^x + 80*x^4*e^x*log(2) - 270*x^4
*e^x + 540*x^3*e^x*log(2) - 40*x^3*e^x + 80*x^2*e^x*log(2) + 216*x^3 + 761*x^2 + 216*x + 16)/x^4

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maple [B]  time = 0.06, size = 80, normalized size = 2.67

method	result	size
risch	\(\frac {216 x^{3}+761 x^{2}+216 x +16}{x^{4}}+\left (100 \ln \relax (2)^{2}-100 x \ln \relax (2)+25 x^{2}\right ) {\mathrm e}^{2 x}+\frac {10 \left (8 x^{2} \ln \relax (2)-4 x^{3}+54 x \ln \relax (2)-27 x^{2}+8 \ln \relax (2)-4 x \right ) {\mathrm e}^{x}}{x^{2}}\)	\(80\)
default	\(\frac {16}{x^{4}}+\frac {216}{x^{3}}+\frac {761}{x^{2}}+\frac {216}{x}-\frac {40 \,{\mathrm e}^{x}}{x}-40 \,{\mathrm e}^{x} x -270 \,{\mathrm e}^{x}+80 \,{\mathrm e}^{x} \ln \relax (2)+25 \,{\mathrm e}^{2 x} x^{2}+100 \ln \relax (2)^{2} {\mathrm e}^{2 x}+\frac {80 \ln \relax (2) {\mathrm e}^{x}}{x^{2}}+\frac {540 \ln \relax (2) {\mathrm e}^{x}}{x}-100 x \ln \relax (2) {\mathrm e}^{2 x}\)	\(90\)
norman	\(\frac {16+\left (-270+80 \ln \relax (2)\right ) x^{4} {\mathrm e}^{x}+\left (540 \ln \relax (2)-40\right ) x^{3} {\mathrm e}^{x}+216 x +761 x^{2}+216 x^{3}-40 x^{5} {\mathrm e}^{x}+25 x^{6} {\mathrm e}^{2 x}+80 x^{2} \ln \relax (2) {\mathrm e}^{x}+100 x^{4} \ln \relax (2)^{2} {\mathrm e}^{2 x}-100 \,{\mathrm e}^{2 x} \ln \relax (2) x^{5}}{x^{4}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((200*x^5*ln(2)^2+2*(-100*x^6-50*x^5)*ln(2)+50*x^7+50*x^6)*exp(x)^2+(2*(40*x^5+270*x^4-230*x^3-80*x^2)*ln(
2)-40*x^6-310*x^5-40*x^4+40*x^3)*exp(x)-216*x^3-1522*x^2-648*x-64)/x^5,x,method=_RETURNVERBOSE)

[Out]

(216*x^3+761*x^2+216*x+16)/x^4+(100*ln(2)^2-100*x*ln(2)+25*x^2)*exp(2*x)+10*(8*x^2*ln(2)-4*x^3+54*x*ln(2)-27*x
^2+8*ln(2)-4*x)/x^2*exp(x)

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maxima [C]  time = 0.47, size = 131, normalized size = 4.37 \begin {gather*} -50 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \relax (2) + 100 \, e^{\left (2 \, x\right )} \log \relax (2)^{2} + \frac {25}{2} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {25}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 40 \, {\left (x - 1\right )} e^{x} + 540 \, {\rm Ei}\relax (x) \log \relax (2) - 50 \, e^{\left (2 \, x\right )} \log \relax (2) + 80 \, e^{x} \log \relax (2) - 460 \, \Gamma \left (-1, -x\right ) \log \relax (2) + 160 \, \Gamma \left (-2, -x\right ) \log \relax (2) + \frac {216}{x} + \frac {761}{x^{2}} + \frac {216}{x^{3}} + \frac {16}{x^{4}} - 40 \, {\rm Ei}\relax (x) - 310 \, e^{x} + 40 \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x^5*log(2)^2+2*(-100*x^6-50*x^5)*log(2)+50*x^7+50*x^6)*exp(x)^2+(2*(40*x^5+270*x^4-230*x^3-80*
x^2)*log(2)-40*x^6-310*x^5-40*x^4+40*x^3)*exp(x)-216*x^3-1522*x^2-648*x-64)/x^5,x, algorithm="maxima")

[Out]

-50*(2*x - 1)*e^(2*x)*log(2) + 100*e^(2*x)*log(2)^2 + 25/2*(2*x^2 - 2*x + 1)*e^(2*x) + 25/2*(2*x - 1)*e^(2*x)
- 40*(x - 1)*e^x + 540*Ei(x)*log(2) - 50*e^(2*x)*log(2) + 80*e^x*log(2) - 460*gamma(-1, -x)*log(2) + 160*gamma
(-2, -x)*log(2) + 216/x + 761/x^2 + 216/x^3 + 16/x^4 - 40*Ei(x) - 310*e^x + 40*gamma(-1, -x)

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mupad [B]  time = 2.16, size = 81, normalized size = 2.70 \begin {gather*} {\mathrm {e}}^x\,\left (80\,\ln \relax (2)-270\right )+100\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2+25\,x^2\,{\mathrm {e}}^{2\,x}-x\,\left (40\,{\mathrm {e}}^x+100\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\right )+\frac {216\,x+x^2\,\left (80\,{\mathrm {e}}^x\,\ln \relax (2)+761\right )+x^3\,\left ({\mathrm {e}}^x\,\left (540\,\ln \relax (2)-40\right )+216\right )+16}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(648*x - exp(2*x)*(200*x^5*log(2)^2 - 2*log(2)*(50*x^5 + 100*x^6) + 50*x^6 + 50*x^7) + exp(x)*(2*log(2)*(
80*x^2 + 230*x^3 - 270*x^4 - 40*x^5) - 40*x^3 + 40*x^4 + 310*x^5 + 40*x^6) + 1522*x^2 + 216*x^3 + 64)/x^5,x)

[Out]

exp(x)*(80*log(2) - 270) + 100*exp(2*x)*log(2)^2 + 25*x^2*exp(2*x) - x*(40*exp(x) + 100*exp(2*x)*log(2)) + (21
6*x + x^2*(80*exp(x)*log(2) + 761) + x^3*(exp(x)*(540*log(2) - 40) + 216) + 16)/x^4

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sympy [B]  time = 0.25, size = 88, normalized size = 2.93 \begin {gather*} \frac {\left (25 x^{4} - 100 x^{3} \log {\relax (2 )} + 100 x^{2} \log {\relax (2 )}^{2}\right ) e^{2 x} + \left (- 40 x^{3} - 270 x^{2} + 80 x^{2} \log {\relax (2 )} - 40 x + 540 x \log {\relax (2 )} + 80 \log {\relax (2 )}\right ) e^{x}}{x^{2}} - \frac {- 216 x^{3} - 761 x^{2} - 216 x - 16}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((200*x**5*ln(2)**2+2*(-100*x**6-50*x**5)*ln(2)+50*x**7+50*x**6)*exp(x)**2+(2*(40*x**5+270*x**4-230*
x**3-80*x**2)*ln(2)-40*x**6-310*x**5-40*x**4+40*x**3)*exp(x)-216*x**3-1522*x**2-648*x-64)/x**5,x)

[Out]

((25*x**4 - 100*x**3*log(2) + 100*x**2*log(2)**2)*exp(2*x) + (-40*x**3 - 270*x**2 + 80*x**2*log(2) - 40*x + 54
0*x*log(2) + 80*log(2))*exp(x))/x**2 - (-216*x**3 - 761*x**2 - 216*x - 16)/x**4

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