3.21.33 \(\int \frac {e^{\frac {15 x}{12+58 x+3 x^2+3 x^3}} (180-45 x^2-90 x^3)}{144+1392 x+3436 x^2+420 x^3+357 x^4+18 x^5+9 x^6} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {5}{-2+4 \left (\frac {16}{3}+\frac {1}{x}\right )+x+x^2}} \]

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Rubi [A]  time = 0.36, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6688, 12, 6706} \begin {gather*} e^{\frac {15 x}{3 x^3+3 x^2+58 x+12}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((15*x)/(12 + 58*x + 3*x^2 + 3*x^3))*(180 - 45*x^2 - 90*x^3))/(144 + 1392*x + 3436*x^2 + 420*x^3 + 357*
x^4 + 18*x^5 + 9*x^6),x]

[Out]

E^((15*x)/(12 + 58*x + 3*x^2 + 3*x^3))

Rule 12

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {45 e^{\frac {15 x}{12+58 x+3 x^2+3 x^3}} \left (4-x^2-2 x^3\right )}{\left (12+58 x+3 x^2+3 x^3\right )^2} \, dx\\ &=45 \int \frac {e^{\frac {15 x}{12+58 x+3 x^2+3 x^3}} \left (4-x^2-2 x^3\right )}{\left (12+58 x+3 x^2+3 x^3\right )^2} \, dx\\ &=e^{\frac {15 x}{12+58 x+3 x^2+3 x^3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 22, normalized size = 1.05 \begin {gather*} e^{\frac {15 x}{12+58 x+3 x^2+3 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((15*x)/(12 + 58*x + 3*x^2 + 3*x^3))*(180 - 45*x^2 - 90*x^3))/(144 + 1392*x + 3436*x^2 + 420*x^3
+ 357*x^4 + 18*x^5 + 9*x^6),x]

[Out]

E^((15*x)/(12 + 58*x + 3*x^2 + 3*x^3))

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fricas [A]  time = 0.92, size = 21, normalized size = 1.00 \begin {gather*} e^{\left (\frac {15 \, x}{3 \, x^{3} + 3 \, x^{2} + 58 \, x + 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-90*x^3-45*x^2+180)*exp(15*x/(3*x^3+3*x^2+58*x+12))/(9*x^6+18*x^5+357*x^4+420*x^3+3436*x^2+1392*x+1
44),x, algorithm="fricas")

[Out]

e^(15*x/(3*x^3 + 3*x^2 + 58*x + 12))

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giac [A]  time = 0.18, size = 21, normalized size = 1.00 \begin {gather*} e^{\left (\frac {15 \, x}{3 \, x^{3} + 3 \, x^{2} + 58 \, x + 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-90*x^3-45*x^2+180)*exp(15*x/(3*x^3+3*x^2+58*x+12))/(9*x^6+18*x^5+357*x^4+420*x^3+3436*x^2+1392*x+1
44),x, algorithm="giac")

[Out]

e^(15*x/(3*x^3 + 3*x^2 + 58*x + 12))

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maple [A]  time = 0.07, size = 22, normalized size = 1.05

method	result	size
gosper	\({\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}\)	\(22\)
risch	\({\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}\)	\(22\)
norman	\(\frac {58 x \,{\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}+3 x^{2} {\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}+3 x^{3} {\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}+12 \,{\mathrm e}^{\frac {15 x}{3 x^{3}+3 x^{2}+58 x +12}}}{3 x^{3}+3 x^{2}+58 x +12}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-90*x^3-45*x^2+180)*exp(15*x/(3*x^3+3*x^2+58*x+12))/(9*x^6+18*x^5+357*x^4+420*x^3+3436*x^2+1392*x+144),x,
method=_RETURNVERBOSE)

[Out]

exp(15*x/(3*x^3+3*x^2+58*x+12))

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maxima [A]  time = 0.80, size = 21, normalized size = 1.00 \begin {gather*} e^{\left (\frac {15 \, x}{3 \, x^{3} + 3 \, x^{2} + 58 \, x + 12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-90*x^3-45*x^2+180)*exp(15*x/(3*x^3+3*x^2+58*x+12))/(9*x^6+18*x^5+357*x^4+420*x^3+3436*x^2+1392*x+1
44),x, algorithm="maxima")

[Out]

e^(15*x/(3*x^3 + 3*x^2 + 58*x + 12))

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mupad [B]  time = 1.30, size = 21, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{\frac {15\,x}{3\,x^3+3\,x^2+58\,x+12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((15*x)/(58*x + 3*x^2 + 3*x^3 + 12))*(45*x^2 + 90*x^3 - 180))/(1392*x + 3436*x^2 + 420*x^3 + 357*x^4
+ 18*x^5 + 9*x^6 + 144),x)

[Out]

exp((15*x)/(58*x + 3*x^2 + 3*x^3 + 12))

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sympy [A]  time = 0.23, size = 19, normalized size = 0.90 \begin {gather*} e^{\frac {15 x}{3 x^{3} + 3 x^{2} + 58 x + 12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-90*x**3-45*x**2+180)*exp(15*x/(3*x**3+3*x**2+58*x+12))/(9*x**6+18*x**5+357*x**4+420*x**3+3436*x**2
+1392*x+144),x)

[Out]

exp(15*x/(3*x**3 + 3*x**2 + 58*x + 12))

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