3.84.96 \(\int \frac {-9 x^4-2 x^5+e^{15} (288 x-152 x^2-48 x^3)+e^{20} (-148+184 x-33 x^2-18 x^3)+e^{10} (-215 x^2+6 x^3+12 x^4)+e^5 (72 x^3+16 x^4)}{-8 e^5 x^3+x^4+e^{20} (16-24 x+9 x^2)+e^{15} (-32 x+24 x^2)+e^{10} (24 x^2-6 x^3)} \, dx\)

Optimal. Leaf size=28 \[ x-(5+x)^2+\frac {x}{3 x-\left (-2+\frac {x}{e^5}\right )^2} \]

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Rubi [B]  time = 0.41, antiderivative size = 67, normalized size of antiderivative = 2.39, number of steps used = 4, number of rules used = 3, integrand size = 140, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {1680, 1814, 1586} \begin {gather*} \frac {e^{10} x}{-x^2+e^5 \left (4+3 e^5\right ) x-4 e^{10}}-\left (x+\frac {1}{4} \left (-8 e^5-6 e^{10}\right )\right )^2-\left (9+4 e^5+3 e^{10}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*x^4 - 2*x^5 + E^15*(288*x - 152*x^2 - 48*x^3) + E^20*(-148 + 184*x - 33*x^2 - 18*x^3) + E^10*(-215*x^2
+ 6*x^3 + 12*x^4) + E^5*(72*x^3 + 16*x^4))/(-8*E^5*x^3 + x^4 + E^20*(16 - 24*x + 9*x^2) + E^15*(-32*x + 24*x^
2) + E^10*(24*x^2 - 6*x^3)),x]

[Out]

-((9 + 4*E^5 + 3*E^10)*x) - ((-8*E^5 - 6*E^10)/4 + x)^2 + (E^10*x)/(-4*E^10 + E^5*(4 + 3*E^5)*x - x^2)

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {-1620 e^{45} \left (1+\frac {3 e^5}{20}\right )-64 e^5 x^4-16 x^4 (9+2 x)-144 e^{35} (43+6 x)-27 e^{40} (155+6 x)+36 e^{30} \left (-143-32 x+6 x^2\right )+64 e^{15} x \left (1+27 x+6 x^2\right )+24 e^{20} x \left (2+59 x+6 x^2\right )+96 e^{25} \left (1+9 x^2\right )+e^{10} \left (16 x^2-48 x^4\right )}{\left (24 e^{15}+9 e^{20}-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (-8 e^5-6 e^{10}\right )+x\right )\\ &=\frac {e^{10} x}{-4 e^{10}+e^5 \left (4+3 e^5\right ) x-x^2}-\frac {\operatorname {Subst}\left (\int \frac {18 e^{30} \left (8+3 e^5\right )^2 \left (9+4 e^5+3 e^{10}\right )+36 e^{30} \left (8+3 e^5\right )^2 x-24 e^{15} \left (72+59 e^5+36 e^{10}+9 e^{15}\right ) x^2-48 e^{15} \left (8+3 e^5\right ) x^3}{24 e^{15}+9 e^{20}-4 x^2} \, dx,x,\frac {1}{4} \left (-8 e^5-6 e^{10}\right )+x\right )}{6 e^{15} \left (8+3 e^5\right )}\\ &=\frac {e^{10} x}{-4 e^{10}+e^5 \left (4+3 e^5\right ) x-x^2}-\frac {\operatorname {Subst}\left (\int \left (432 e^{15}+354 e^{20}+216 e^{25}+54 e^{30}+\left (96 e^{15}+36 e^{20}\right ) x\right ) \, dx,x,\frac {1}{4} \left (-8 e^5-6 e^{10}\right )+x\right )}{6 e^{15} \left (8+3 e^5\right )}\\ &=-\left (\left (9+4 e^5+3 e^{10}\right ) x\right )-\left (-\frac {1}{2} e^5 \left (4+3 e^5\right )+x\right )^2+\frac {e^{10} x}{-4 e^{10}+e^5 \left (4+3 e^5\right ) x-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 34, normalized size = 1.21 \begin {gather*} x \left (-9-x+\frac {e^{10}}{4 e^5 x-x^2+e^{10} (-4+3 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*x^4 - 2*x^5 + E^15*(288*x - 152*x^2 - 48*x^3) + E^20*(-148 + 184*x - 33*x^2 - 18*x^3) + E^10*(-2
15*x^2 + 6*x^3 + 12*x^4) + E^5*(72*x^3 + 16*x^4))/(-8*E^5*x^3 + x^4 + E^20*(16 - 24*x + 9*x^2) + E^15*(-32*x +
24*x^2) + E^10*(24*x^2 - 6*x^3)),x]

[Out]

x*(-9 - x + E^10/(4*E^5*x - x^2 + E^10*(-4 + 3*x)))

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fricas [B]  time = 0.79, size = 62, normalized size = 2.21 \begin {gather*} -\frac {x^{4} + 9 \, x^{3} - {\left (3 \, x^{3} + 23 \, x^{2} - 37 \, x\right )} e^{10} - 4 \, {\left (x^{3} + 9 \, x^{2}\right )} e^{5}}{x^{2} - {\left (3 \, x - 4\right )} e^{10} - 4 \, x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-33*x^2+184*x-148)*exp(5)^4+(-48*x^3-152*x^2+288*x)*exp(5)^3+(12*x^4+6*x^3-215*x^2)*exp(5)^
2+(16*x^4+72*x^3)*exp(5)-2*x^5-9*x^4)/((9*x^2-24*x+16)*exp(5)^4+(24*x^2-32*x)*exp(5)^3+(-6*x^3+24*x^2)*exp(5)^
2-8*x^3*exp(5)+x^4),x, algorithm="fricas")

[Out]

-(x^4 + 9*x^3 - (3*x^3 + 23*x^2 - 37*x)*e^10 - 4*(x^3 + 9*x^2)*e^5)/(x^2 - (3*x - 4)*e^10 - 4*x*e^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, x^{5} + 9 \, x^{4} + {\left (18 \, x^{3} + 33 \, x^{2} - 184 \, x + 148\right )} e^{20} + 8 \, {\left (6 \, x^{3} + 19 \, x^{2} - 36 \, x\right )} e^{15} - {\left (12 \, x^{4} + 6 \, x^{3} - 215 \, x^{2}\right )} e^{10} - 8 \, {\left (2 \, x^{4} + 9 \, x^{3}\right )} e^{5}}{x^{4} - 8 \, x^{3} e^{5} + {\left (9 \, x^{2} - 24 \, x + 16\right )} e^{20} + 8 \, {\left (3 \, x^{2} - 4 \, x\right )} e^{15} - 6 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-33*x^2+184*x-148)*exp(5)^4+(-48*x^3-152*x^2+288*x)*exp(5)^3+(12*x^4+6*x^3-215*x^2)*exp(5)^
2+(16*x^4+72*x^3)*exp(5)-2*x^5-9*x^4)/((9*x^2-24*x+16)*exp(5)^4+(24*x^2-32*x)*exp(5)^3+(-6*x^3+24*x^2)*exp(5)^
2-8*x^3*exp(5)+x^4),x, algorithm="giac")

[Out]

integrate(-(2*x^5 + 9*x^4 + (18*x^3 + 33*x^2 - 184*x + 148)*e^20 + 8*(6*x^3 + 19*x^2 - 36*x)*e^15 - (12*x^4 +
6*x^3 - 215*x^2)*e^10 - 8*(2*x^4 + 9*x^3)*e^5)/(x^4 - 8*x^3*e^5 + (9*x^2 - 24*x + 16)*e^20 + 8*(3*x^2 - 4*x)*e
^15 - 6*(x^3 - 4*x^2)*e^10), x)

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maple [A]  time = 0.25, size = 36, normalized size = 1.29

method	result	size
risch	\(-x^{2}-9 x +\frac {x \,{\mathrm e}^{10}}{3 x \,{\mathrm e}^{10}-4 \,{\mathrm e}^{10}+4 x \,{\mathrm e}^{5}-x^{2}}\)	\(36\)
norman	\(\frac {x^{4}+\left (9-3 \,{\mathrm e}^{10}-4 \,{\mathrm e}^{5}\right ) x^{3}+\left (-69 \,{\mathrm e}^{20}-200 \,{\mathrm e}^{15}-107 \,{\mathrm e}^{10}\right ) x +4 \,{\mathrm e}^{15} \left (23 \,{\mathrm e}^{5}+36\right )}{3 x \,{\mathrm e}^{10}-4 \,{\mathrm e}^{10}+4 x \,{\mathrm e}^{5}-x^{2}}\)	\(81\)
gosper	\(-\frac {69 x \,{\mathrm e}^{20}+3 x^{3} {\mathrm e}^{10}-92 \,{\mathrm e}^{20}+200 x \,{\mathrm e}^{15}+4 x^{3} {\mathrm e}^{5}-x^{4}-144 \,{\mathrm e}^{15}+107 x \,{\mathrm e}^{10}-9 x^{3}}{3 x \,{\mathrm e}^{10}-4 \,{\mathrm e}^{10}+4 x \,{\mathrm e}^{5}-x^{2}}\)	\(89\)
default	\(-x^{2}-9 x -\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+\left (-8 \,{\mathrm e}^{5}-6 \,{\mathrm e}^{10}\right ) \textit {\_Z}^{3}+\left (24 \,{\mathrm e}^{10}+9 \,{\mathrm e}^{20}+24 \,{\mathrm e}^{15}\right ) \textit {\_Z}^{2}+\left (-24 \,{\mathrm e}^{20}-32 \,{\mathrm e}^{15}\right ) \textit {\_Z} +16 \,{\mathrm e}^{20}\right )}{\sum }\frac {\left ({\mathrm e}^{10} \textit {\_R}^{2}-4 \,{\mathrm e}^{20}\right ) \ln \left (x -\textit {\_R} \right )}{12 \textit {\_R}^{2} {\mathrm e}^{5}+9 \,{\mathrm e}^{10} \textit {\_R}^{2}-2 \textit {\_R}^{3}-9 \textit {\_R} \,{\mathrm e}^{20}-24 \,{\mathrm e}^{15} \textit {\_R} -24 \,{\mathrm e}^{10} \textit {\_R} +12 \,{\mathrm e}^{20}+16 \,{\mathrm e}^{15}}\right )}{2}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*x^3-33*x^2+184*x-148)*exp(5)^4+(-48*x^3-152*x^2+288*x)*exp(5)^3+(12*x^4+6*x^3-215*x^2)*exp(5)^2+(16*
x^4+72*x^3)*exp(5)-2*x^5-9*x^4)/((9*x^2-24*x+16)*exp(5)^4+(24*x^2-32*x)*exp(5)^3+(-6*x^3+24*x^2)*exp(5)^2-8*x^
3*exp(5)+x^4),x,method=_RETURNVERBOSE)

[Out]

-x^2-9*x+1/3*x*exp(10)/(x*exp(10)-4/3*exp(10)+4/3*x*exp(5)-1/3*x^2)

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maxima [A]  time = 0.36, size = 36, normalized size = 1.29 \begin {gather*} -x^{2} - 9 \, x - \frac {x e^{10}}{x^{2} - x {\left (3 \, e^{10} + 4 \, e^{5}\right )} + 4 \, e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x^3-33*x^2+184*x-148)*exp(5)^4+(-48*x^3-152*x^2+288*x)*exp(5)^3+(12*x^4+6*x^3-215*x^2)*exp(5)^
2+(16*x^4+72*x^3)*exp(5)-2*x^5-9*x^4)/((9*x^2-24*x+16)*exp(5)^4+(24*x^2-32*x)*exp(5)^3+(-6*x^3+24*x^2)*exp(5)^
2-8*x^3*exp(5)+x^4),x, algorithm="maxima")

[Out]

-x^2 - 9*x - x*e^10/(x^2 - x*(3*e^10 + 4*e^5) + 4*e^10)

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mupad [B]  time = 5.98, size = 55, normalized size = 1.96 \begin {gather*} x\,\left (16\,{\mathrm {e}}^5+12\,{\mathrm {e}}^{10}-4\,{\mathrm {e}}^5\,\left (3\,{\mathrm {e}}^5+4\right )-9\right )-x^2-\frac {x\,{\mathrm {e}}^{10}}{x^2+\left (-4\,{\mathrm {e}}^5-3\,{\mathrm {e}}^{10}\right )\,x+4\,{\mathrm {e}}^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(20)*(33*x^2 - 184*x + 18*x^3 + 148) + exp(15)*(152*x^2 - 288*x + 48*x^3) - exp(5)*(72*x^3 + 16*x^4)
- exp(10)*(6*x^3 - 215*x^2 + 12*x^4) + 9*x^4 + 2*x^5)/(exp(20)*(9*x^2 - 24*x + 16) - exp(15)*(32*x - 24*x^2) +
exp(10)*(24*x^2 - 6*x^3) - 8*x^3*exp(5) + x^4),x)

[Out]

x*(16*exp(5) + 12*exp(10) - 4*exp(5)*(3*exp(5) + 4) - 9) - x^2 - (x*exp(10))/(4*exp(10) - x*(4*exp(5) + 3*exp(
10)) + x^2)

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sympy [A]  time = 0.62, size = 34, normalized size = 1.21 \begin {gather*} - x^{2} - 9 x - \frac {x e^{10}}{x^{2} + x \left (- 3 e^{10} - 4 e^{5}\right ) + 4 e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x**3-33*x**2+184*x-148)*exp(5)**4+(-48*x**3-152*x**2+288*x)*exp(5)**3+(12*x**4+6*x**3-215*x**2
)*exp(5)**2+(16*x**4+72*x**3)*exp(5)-2*x**5-9*x**4)/((9*x**2-24*x+16)*exp(5)**4+(24*x**2-32*x)*exp(5)**3+(-6*x
**3+24*x**2)*exp(5)**2-8*x**3*exp(5)+x**4),x)

[Out]

-x**2 - 9*x - x*exp(10)/(x**2 + x*(-3*exp(10) - 4*exp(5)) + 4*exp(10))

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