3.8.64 \(\int \frac {360+600 x+262 x^2+48 x^3-34 x^4-30 x^5+e^4 (-900-900 x-15 x^2+360 x^3+114 x^4-36 x^5-15 x^6)}{36 x^2+60 x^3+25 x^4+e^4 (-360 x-600 x^2-178 x^3+120 x^4+50 x^5)+e^8 (900+1500 x+265 x^2-600 x^3-214 x^4+60 x^5+25 x^6)} \, dx\)

Optimal. Leaf size=33 \[ \frac {2-x+\frac {2 x}{5+\frac {6}{x}}}{e^4+\frac {x}{-5+x^2}} \]

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Rubi [B]  time = 0.33, antiderivative size = 90, normalized size of antiderivative = 2.73, number of steps used = 7, number of rules used = 4, integrand size = 140, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2074, 638, 618, 206} \begin {gather*} -\frac {15 e^4 \left (6+13 e^4\right )-\left (18+99 e^4+118 e^8\right ) x}{e^8 \left (30+89 e^4\right ) \left (-e^4 x^2-x+5 e^4\right )}-\frac {3 x}{5 e^4}+\frac {6408}{25 \left (30+89 e^4\right ) (5 x+6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(360 + 600*x + 262*x^2 + 48*x^3 - 34*x^4 - 30*x^5 + E^4*(-900 - 900*x - 15*x^2 + 360*x^3 + 114*x^4 - 36*x^
5 - 15*x^6))/(36*x^2 + 60*x^3 + 25*x^4 + E^4*(-360*x - 600*x^2 - 178*x^3 + 120*x^4 + 50*x^5) + E^8*(900 + 1500
*x + 265*x^2 - 600*x^3 - 214*x^4 + 60*x^5 + 25*x^6)),x]

[Out]

(-3*x)/(5*E^4) + 6408/(25*(30 + 89*E^4)*(6 + 5*x)) - (15*E^4*(6 + 13*E^4) - (18 + 99*E^4 + 118*E^8)*x)/(E^8*(3
0 + 89*E^4)*(5*E^4 - x - E^4*x^2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {3}{5 e^4}-\frac {6408}{5 \left (30+89 e^4\right ) (6+5 x)^2}+\frac {5 e^4 \left (18+159 e^4+236 e^8\right )-\left (18+99 e^4+298 e^8+390 e^{12}\right ) x}{e^8 \left (30+89 e^4\right ) \left (5 e^4-x-e^4 x^2\right )^2}+\frac {18+99 e^4+118 e^8}{e^8 \left (30+89 e^4\right ) \left (-5 e^4+x+e^4 x^2\right )}\right ) \, dx\\ &=-\frac {3 x}{5 e^4}+\frac {6408}{25 \left (30+89 e^4\right ) (6+5 x)}+\frac {\int \frac {5 e^4 \left (18+159 e^4+236 e^8\right )-\left (18+99 e^4+298 e^8+390 e^{12}\right ) x}{\left (5 e^4-x-e^4 x^2\right )^2} \, dx}{e^8 \left (30+89 e^4\right )}+\frac {\left (18+99 e^4+118 e^8\right ) \int \frac {1}{-5 e^4+x+e^4 x^2} \, dx}{e^8 \left (30+89 e^4\right )}\\ &=-\frac {3 x}{5 e^4}+\frac {6408}{25 \left (30+89 e^4\right ) (6+5 x)}-\frac {15 e^4 \left (6+13 e^4\right )-\left (18+99 e^4+118 e^8\right ) x}{e^8 \left (30+89 e^4\right ) \left (5 e^4-x-e^4 x^2\right )}+\frac {\left (18+99 e^4+118 e^8\right ) \int \frac {1}{5 e^4-x-e^4 x^2} \, dx}{e^8 \left (30+89 e^4\right )}-\frac {\left (2 \left (18+99 e^4+118 e^8\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+20 e^8-x^2} \, dx,x,1+2 e^4 x\right )}{e^8 \left (30+89 e^4\right )}\\ &=-\frac {3 x}{5 e^4}+\frac {6408}{25 \left (30+89 e^4\right ) (6+5 x)}-\frac {15 e^4 \left (6+13 e^4\right )-\left (18+99 e^4+118 e^8\right ) x}{e^8 \left (30+89 e^4\right ) \left (5 e^4-x-e^4 x^2\right )}-\frac {2 \left (18+99 e^4+118 e^8\right ) \tanh ^{-1}\left (\frac {1+2 e^4 x}{\sqrt {1+20 e^8}}\right )}{e^8 \left (30+89 e^4\right ) \sqrt {1+20 e^8}}-\frac {\left (2 \left (18+99 e^4+118 e^8\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+20 e^8-x^2} \, dx,x,-1-2 e^4 x\right )}{e^8 \left (30+89 e^4\right )}\\ &=-\frac {3 x}{5 e^4}+\frac {6408}{25 \left (30+89 e^4\right ) (6+5 x)}-\frac {15 e^4 \left (6+13 e^4\right )-\left (18+99 e^4+118 e^8\right ) x}{e^8 \left (30+89 e^4\right ) \left (5 e^4-x-e^4 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 88, normalized size = 2.67 \begin {gather*} \frac {-15 x (6+5 x)+e^4 \left (450+147 x-280 x^2-75 x^3\right )+e^8 \left (-360+450 x+447 x^2-90 x^3-75 x^4\right )}{25 e^8 \left (x (6+5 x)+e^4 \left (-30-25 x+6 x^2+5 x^3\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(360 + 600*x + 262*x^2 + 48*x^3 - 34*x^4 - 30*x^5 + E^4*(-900 - 900*x - 15*x^2 + 360*x^3 + 114*x^4 -
36*x^5 - 15*x^6))/(36*x^2 + 60*x^3 + 25*x^4 + E^4*(-360*x - 600*x^2 - 178*x^3 + 120*x^4 + 50*x^5) + E^8*(900
+ 1500*x + 265*x^2 - 600*x^3 - 214*x^4 + 60*x^5 + 25*x^6)),x]

[Out]

(-15*x*(6 + 5*x) + E^4*(450 + 147*x - 280*x^2 - 75*x^3) + E^8*(-360 + 450*x + 447*x^2 - 90*x^3 - 75*x^4))/(25*
E^8*(x*(6 + 5*x) + E^4*(-30 - 25*x + 6*x^2 + 5*x^3)))

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fricas [B]  time = 0.73, size = 86, normalized size = 2.61 \begin {gather*} -\frac {75 \, x^{2} + 3 \, {\left (25 \, x^{4} + 30 \, x^{3} - 149 \, x^{2} - 150 \, x + 120\right )} e^{8} + {\left (75 \, x^{3} + 280 \, x^{2} - 147 \, x - 450\right )} e^{4} + 90 \, x}{25 \, {\left ({\left (5 \, x^{3} + 6 \, x^{2} - 25 \, x - 30\right )} e^{12} + {\left (5 \, x^{2} + 6 \, x\right )} e^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^6-36*x^5+114*x^4+360*x^3-15*x^2-900*x-900)*exp(4)-30*x^5-34*x^4+48*x^3+262*x^2+600*x+360)/((
25*x^6+60*x^5-214*x^4-600*x^3+265*x^2+1500*x+900)*exp(4)^2+(50*x^5+120*x^4-178*x^3-600*x^2-360*x)*exp(4)+25*x^
4+60*x^3+36*x^2),x, algorithm="fricas")

[Out]

-1/25*(75*x^2 + 3*(25*x^4 + 30*x^3 - 149*x^2 - 150*x + 120)*e^8 + (75*x^3 + 280*x^2 - 147*x - 450)*e^4 + 90*x)
/((5*x^3 + 6*x^2 - 25*x - 30)*e^12 + (5*x^2 + 6*x)*e^8)

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giac [B]  time = 1.03, size = 95, normalized size = 2.88 \begin {gather*} -\frac {3}{5} \, x e^{\left (-4\right )} - \frac {-9.27347088374000 \times 10^{50} \, \log \left (x + 2.24524454980000\right ) + 2.29089301024000 \times 10^{50} \, \log \left (x - 2.22692891091000\right )}{4 \, {\left (3936588805702081 \, e^{56} + 10615520374926960 \, e^{52} + 12523928532217200 \, e^{48} + 8443097886888000 \, e^{44} + 3557485064700000 \, e^{40} + 959321815200000 \, e^{36} + 161683452000000 \, e^{32} + 15571440000000 \, e^{28} + 656100000000 \, e^{24}\right )}} + \frac {6408 \, {\left (704969 \, e^{12} + 712890 \, e^{8} + 240300 \, e^{4} + 27000\right )}}{25 \, {\left (5 \, x + 6\right )} {\left (7921 \, e^{8} + 5340 \, e^{4} + 900\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^6-36*x^5+114*x^4+360*x^3-15*x^2-900*x-900)*exp(4)-30*x^5-34*x^4+48*x^3+262*x^2+600*x+360)/((
25*x^6+60*x^5-214*x^4-600*x^3+265*x^2+1500*x+900)*exp(4)^2+(50*x^5+120*x^4-178*x^3-600*x^2-360*x)*exp(4)+25*x^
4+60*x^3+36*x^2),x, algorithm="giac")

[Out]

-3/5*x*e^(-4) - 1/4*(-9.27347088374000e50*log(x + 2.24524454980000) + 2.29089301024000e50*log(x - 2.2269289109
1000))/(3936588805702081*e^56 + 10615520374926960*e^52 + 12523928532217200*e^48 + 8443097886888000*e^44 + 3557
485064700000*e^40 + 959321815200000*e^36 + 161683452000000*e^32 + 15571440000000*e^28 + 656100000000*e^24) + 6
408/25*(704969*e^12 + 712890*e^8 + 240300*e^4 + 27000)/((5*x + 6)*(7921*e^8 + 5340*e^4 + 900)^2)

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maple [A]  time = 0.28, size = 52, normalized size = 1.58

method	result	size
norman	\(\frac {-3 x^{4}-\frac {24 x \,{\mathrm e}^{-4}}{5}+\frac {{\mathrm e}^{-4} \left (-20+111 \,{\mathrm e}^{4}\right ) x^{2}}{5}-36}{\left (5 x +6\right ) \left (x^{2} {\mathrm e}^{4}-5 \,{\mathrm e}^{4}+x \right )}\)	\(52\)
gosper	\(-\frac {\left (15 x^{4} {\mathrm e}^{4}-111 x^{2} {\mathrm e}^{4}+20 x^{2}+180 \,{\mathrm e}^{4}+24 x \right ) {\mathrm e}^{-4}}{5 \left (5 x^{3} {\mathrm e}^{4}+6 x^{2} {\mathrm e}^{4}-25 x \,{\mathrm e}^{4}+5 x^{2}-30 \,{\mathrm e}^{4}+6 x \right )}\)	\(68\)
risch	\(-\frac {3 x \,{\mathrm e}^{-4}}{5}+\frac {\left (\frac {{\mathrm e}^{-4} \left (72 \,{\mathrm e}^{8}-190 \,{\mathrm e}^{4}-75\right ) x^{2}}{25}+\frac {3 \left (49 \,{\mathrm e}^{4}-30\right ) {\mathrm e}^{-4} x}{25}-\frac {72 \,{\mathrm e}^{4}}{5}+18\right ) {\mathrm e}^{-4}}{5 x^{3} {\mathrm e}^{4}+6 x^{2} {\mathrm e}^{4}-25 x \,{\mathrm e}^{4}+5 x^{2}-30 \,{\mathrm e}^{4}+6 x}\)	\(76\)
default	\(-\frac {3 \,{\mathrm e}^{4} {\mathrm e}^{-8} x}{5}-\frac {-1539842400 \,{\mathrm e}^{4}-4568199120 \,{\mathrm e}^{8}-4517441352 \,{\mathrm e}^{12}-173016000}{25 \left (7921 \,{\mathrm e}^{8}+5340 \,{\mathrm e}^{4}+900\right )^{2} \left (5 x +6\right )}+\frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4} {\mathrm e}^{8}+2 \textit {\_Z}^{3} {\mathrm e}^{4}+\left (-10 \,{\mathrm e}^{8}+1\right ) \textit {\_Z}^{2}-10 \textit {\_Z} \,{\mathrm e}^{4}+25 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (8100000+\left (486000 \,{\mathrm e}^{-4}[phone] \,{\mathrm e}^{16}[phone] \,{\mathrm e}^{4}[phone] \,{\mathrm e}^{8}[phone] \,{\mathrm e}^{12}+6998400\right ) \textit {\_R}^{2}+30 \left (-162000-9164597 \,{\mathrm e}^{16}-1792800 \,{\mathrm e}^{4}-7401240 \,{\mathrm e}^{8}-13497384 \,{\mathrm e}^{12}\right ) \textit {\_R} [phone] \,{\mathrm e}^{16}[phone] \,{\mathrm e}^{4}[phone] \,{\mathrm e}^{8}[phone] \,{\mathrm e}^{12}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} {\mathrm e}^{8}+3 \textit {\_R}^{2} {\mathrm e}^{4}-10 \textit {\_R} \,{\mathrm e}^{8}-5 \,{\mathrm e}^{4}+\textit {\_R}}}{2 \left (7921 \,{\mathrm e}^{8}+5340 \,{\mathrm e}^{4}+900\right )^{2}}\)	\(196\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^6-36*x^5+114*x^4+360*x^3-15*x^2-900*x-900)*exp(4)-30*x^5-34*x^4+48*x^3+262*x^2+600*x+360)/((25*x^6
+60*x^5-214*x^4-600*x^3+265*x^2+1500*x+900)*exp(4)^2+(50*x^5+120*x^4-178*x^3-600*x^2-360*x)*exp(4)+25*x^4+60*x
^3+36*x^2),x,method=_RETURNVERBOSE)

[Out]

(-3*x^4-24/5*x/exp(4)+1/5/exp(4)*(-20+111*exp(4))*x^2-36)/(5*x+6)/(x^2*exp(4)-5*exp(4)+x)

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maxima [B]  time = 0.49, size = 79, normalized size = 2.39 \begin {gather*} -\frac {3}{5} \, x e^{\left (-4\right )} + \frac {x^{2} {\left (72 \, e^{8} - 190 \, e^{4} - 75\right )} + 3 \, x {\left (49 \, e^{4} - 30\right )} - 360 \, e^{8} + 450 \, e^{4}}{25 \, {\left (5 \, x^{3} e^{12} + x^{2} {\left (6 \, e^{12} + 5 \, e^{8}\right )} - x {\left (25 \, e^{12} - 6 \, e^{8}\right )} - 30 \, e^{12}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^6-36*x^5+114*x^4+360*x^3-15*x^2-900*x-900)*exp(4)-30*x^5-34*x^4+48*x^3+262*x^2+600*x+360)/((
25*x^6+60*x^5-214*x^4-600*x^3+265*x^2+1500*x+900)*exp(4)^2+(50*x^5+120*x^4-178*x^3-600*x^2-360*x)*exp(4)+25*x^
4+60*x^3+36*x^2),x, algorithm="maxima")

[Out]

-3/5*x*e^(-4) + 1/25*(x^2*(72*e^8 - 190*e^4 - 75) + 3*x*(49*e^4 - 30) - 360*e^8 + 450*e^4)/(5*x^3*e^12 + x^2*(
6*e^12 + 5*e^8) - x*(25*e^12 - 6*e^8) - 30*e^12)

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mupad [B]  time = 0.45, size = 80, normalized size = 2.42 \begin {gather*} -\frac {3\,x\,{\mathrm {e}}^{-4}}{5}-\frac {\frac {{\mathrm {e}}^{-4}\,\left (190\,{\mathrm {e}}^4-72\,{\mathrm {e}}^8+75\right )\,x^2}{5}-\frac {3\,{\mathrm {e}}^{-4}\,\left (49\,{\mathrm {e}}^4-30\right )\,x}{5}+72\,{\mathrm {e}}^4-90}{25\,{\mathrm {e}}^8\,x^3+\left (25\,{\mathrm {e}}^4+30\,{\mathrm {e}}^8\right )\,x^2+\left (30\,{\mathrm {e}}^4-125\,{\mathrm {e}}^8\right )\,x-150\,{\mathrm {e}}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((600*x - exp(4)*(900*x + 15*x^2 - 360*x^3 - 114*x^4 + 36*x^5 + 15*x^6 + 900) + 262*x^2 + 48*x^3 - 34*x^4 -
30*x^5 + 360)/(exp(8)*(1500*x + 265*x^2 - 600*x^3 - 214*x^4 + 60*x^5 + 25*x^6 + 900) - exp(4)*(360*x + 600*x^
2 + 178*x^3 - 120*x^4 - 50*x^5) + 36*x^2 + 60*x^3 + 25*x^4),x)

[Out]

- (3*x*exp(-4))/5 - (72*exp(4) - (3*x*exp(-4)*(49*exp(4) - 30))/5 + (x^2*exp(-4)*(190*exp(4) - 72*exp(8) + 75)
)/5 - 90)/(x^2*(25*exp(4) + 30*exp(8)) - 150*exp(8) + 25*x^3*exp(8) + x*(30*exp(4) - 125*exp(8)))

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sympy [B]  time = 3.50, size = 82, normalized size = 2.48 \begin {gather*} - \frac {3 x}{5 e^{4}} - \frac {x^{2} \left (- 72 e^{8} + 75 + 190 e^{4}\right ) + x \left (90 - 147 e^{4}\right ) - 450 e^{4} + 360 e^{8}}{125 x^{3} e^{12} + x^{2} \left (125 e^{8} + 150 e^{12}\right ) + x \left (- 625 e^{12} + 150 e^{8}\right ) - 750 e^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**6-36*x**5+114*x**4+360*x**3-15*x**2-900*x-900)*exp(4)-30*x**5-34*x**4+48*x**3+262*x**2+600*
x+360)/((25*x**6+60*x**5-214*x**4-600*x**3+265*x**2+1500*x+900)*exp(4)**2+(50*x**5+120*x**4-178*x**3-600*x**2-
360*x)*exp(4)+25*x**4+60*x**3+36*x**2),x)

[Out]

-3*x*exp(-4)/5 - (x**2*(-72*exp(8) + 75 + 190*exp(4)) + x*(90 - 147*exp(4)) - 450*exp(4) + 360*exp(8))/(125*x*
*3*exp(12) + x**2*(125*exp(8) + 150*exp(12)) + x*(-625*exp(12) + 150*exp(8)) - 750*exp(12))

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