3.69.53 \(\int \frac {-6 x-16 x^3+e^{10} (1-6 x+4 x^2-16 x^3)+e^5 (12 x+32 x^3)}{1-2 e^5+e^{10}} \, dx\)

Optimal. Leaf size=26 \[ \left (x^2+\frac {4 x^4}{3}\right ) \left (-3+\frac {x}{\left (x-\frac {x}{e^5}\right )^2}\right ) \]

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Rubi [B]  time = 0.03, antiderivative size = 130, normalized size of antiderivative = 5.00, number of steps used = 4, number of rules used = 1, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {12} \begin {gather*} -\frac {4 e^{10} x^4}{\left (1-e^5\right )^2}+\frac {8 e^5 x^4}{\left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {4 e^{10} x^3}{3 \left (1-e^5\right )^2}-\frac {3 e^{10} x^2}{\left (1-e^5\right )^2}+\frac {6 e^5 x^2}{\left (1-e^5\right )^2}-\frac {3 x^2}{\left (1-e^5\right )^2}+\frac {e^{10} x}{\left (1-e^5\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x - 16*x^3 + E^10*(1 - 6*x + 4*x^2 - 16*x^3) + E^5*(12*x + 32*x^3))/(1 - 2*E^5 + E^10),x]

[Out]

(E^10*x)/(1 - E^5)^2 - (3*x^2)/(1 - E^5)^2 + (6*E^5*x^2)/(1 - E^5)^2 - (3*E^10*x^2)/(1 - E^5)^2 + (4*E^10*x^3)
/(3*(1 - E^5)^2) - (4*x^4)/(1 - E^5)^2 + (8*E^5*x^4)/(1 - E^5)^2 - (4*E^10*x^4)/(1 - E^5)^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )\right ) \, dx}{1-2 e^5+e^{10}}\\ &=-\frac {3 x^2}{\left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {e^5 \int \left (12 x+32 x^3\right ) \, dx}{\left (1-e^5\right )^2}+\frac {e^{10} \int \left (1-6 x+4 x^2-16 x^3\right ) \, dx}{\left (1-e^5\right )^2}\\ &=\frac {e^{10} x}{\left (1-e^5\right )^2}-\frac {3 x^2}{\left (1-e^5\right )^2}+\frac {6 e^5 x^2}{\left (1-e^5\right )^2}-\frac {3 e^{10} x^2}{\left (1-e^5\right )^2}+\frac {4 e^{10} x^3}{3 \left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {8 e^5 x^4}{\left (1-e^5\right )^2}-\frac {4 e^{10} x^4}{\left (1-e^5\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 66, normalized size = 2.54 \begin {gather*} \frac {e^{10} x-3 x^2+6 e^5 x^2-3 e^{10} x^2+\frac {4 e^{10} x^3}{3}-4 x^4+8 e^5 x^4-4 e^{10} x^4}{\left (-1+e^5\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x - 16*x^3 + E^10*(1 - 6*x + 4*x^2 - 16*x^3) + E^5*(12*x + 32*x^3))/(1 - 2*E^5 + E^10),x]

[Out]

(E^10*x - 3*x^2 + 6*E^5*x^2 - 3*E^10*x^2 + (4*E^10*x^3)/3 - 4*x^4 + 8*E^5*x^4 - 4*E^10*x^4)/(-1 + E^5)^2

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fricas [B]  time = 0.60, size = 60, normalized size = 2.31 \begin {gather*} -\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3+4*x^2-6*x+1)*exp(5)^2+(32*x^3+12*x)*exp(5)-16*x^3-6*x)/(exp(5)^2-2*exp(5)+1),x, algorithm=
"fricas")

[Out]

-1/3*(12*x^4 + 9*x^2 + (12*x^4 - 4*x^3 + 9*x^2 - 3*x)*e^10 - 6*(4*x^4 + 3*x^2)*e^5)/(e^10 - 2*e^5 + 1)

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giac [B]  time = 0.15, size = 60, normalized size = 2.31 \begin {gather*} -\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3+4*x^2-6*x+1)*exp(5)^2+(32*x^3+12*x)*exp(5)-16*x^3-6*x)/(exp(5)^2-2*exp(5)+1),x, algorithm=
"giac")

[Out]

-1/3*(12*x^4 + 9*x^2 + (12*x^4 - 4*x^3 + 9*x^2 - 3*x)*e^10 - 6*(4*x^4 + 3*x^2)*e^5)/(e^10 - 2*e^5 + 1)

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maple [B]  time = 0.04, size = 56, normalized size = 2.15

method	result	size
norman	\(\frac {\left (-4 \,{\mathrm e}^{5}+4\right ) x^{4}+\left (-3 \,{\mathrm e}^{5}+3\right ) x^{2}+\frac {{\mathrm e}^{10} x}{{\mathrm e}^{5}-1}+\frac {4 \,{\mathrm e}^{10} x^{3}}{3 \left ({\mathrm e}^{5}-1\right )}}{{\mathrm e}^{5}-1}\)	\(56\)
default	\(\frac {{\mathrm e}^{10} \left (-4 x^{4}+\frac {4}{3} x^{3}-3 x^{2}+x \right )+{\mathrm e}^{5} \left (8 x^{4}+6 x^{2}\right )-4 x^{4}-3 x^{2}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}\)	\(61\)
gosper	\(-\frac {x \left (12 x^{3} {\mathrm e}^{10}-4 x^{2} {\mathrm e}^{10}-24 x^{3} {\mathrm e}^{5}+9 x \,{\mathrm e}^{10}+12 x^{3}-3 \,{\mathrm e}^{10}-18 x \,{\mathrm e}^{5}+9 x \right )}{3 \left ({\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1\right )}\)	\(68\)
risch	\(-\frac {4 x^{4} {\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {4 x^{3} {\mathrm e}^{10}}{3 \left ({\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1\right )}+\frac {8 x^{4} {\mathrm e}^{5}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {3 x^{2} {\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {4 x^{4}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {x \,{\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {6 x^{2} {\mathrm e}^{5}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {3 x^{2}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^3+4*x^2-6*x+1)*exp(5)^2+(32*x^3+12*x)*exp(5)-16*x^3-6*x)/(exp(5)^2-2*exp(5)+1),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [B]  time = 0.36, size = 60, normalized size = 2.31 \begin {gather*} -\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3+4*x^2-6*x+1)*exp(5)^2+(32*x^3+12*x)*exp(5)-16*x^3-6*x)/(exp(5)^2-2*exp(5)+1),x, algorithm=
"maxima")

[Out]

-1/3*(12*x^4 + 9*x^2 + (12*x^4 - 4*x^3 + 9*x^2 - 3*x)*e^10 - 6*(4*x^4 + 3*x^2)*e^5)/(e^10 - 2*e^5 + 1)

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mupad [B]  time = 4.10, size = 34, normalized size = 1.31 \begin {gather*} -4\,x^4+\frac {4\,{\mathrm {e}}^{10}\,x^3}{3\,{\left ({\mathrm {e}}^5-1\right )}^2}-3\,x^2+\frac {{\mathrm {e}}^{10}\,x}{{\left ({\mathrm {e}}^5-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x - exp(5)*(12*x + 32*x^3) + exp(10)*(6*x - 4*x^2 + 16*x^3 - 1) + 16*x^3)/(exp(10) - 2*exp(5) + 1),x)

[Out]

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

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sympy [B]  time = 0.08, size = 44, normalized size = 1.69 \begin {gather*} - 4 x^{4} + \frac {4 x^{3} e^{10}}{- 6 e^{5} + 3 + 3 e^{10}} - 3 x^{2} + \frac {x e^{10}}{- 2 e^{5} + 1 + e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**3+4*x**2-6*x+1)*exp(5)**2+(32*x**3+12*x)*exp(5)-16*x**3-6*x)/(exp(5)**2-2*exp(5)+1),x)

[Out]

-4*x**4 + 4*x**3*exp(10)/(-6*exp(5) + 3 + 3*exp(10)) - 3*x**2 + x*exp(10)/(-2*exp(5) + 1 + exp(10))

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