3.43.97 \(\int \frac {x^3+e^{2 (4 \log (3)+5 e^x x \log (3))} (16-2 x+e^x (-80 x-70 x^2+10 x^3) \log (3))}{-8 x^3+x^4} \, dx\)

Optimal. Leaf size=21 \[ \frac {3^{8+10 e^x x}}{x^2}+\log (8-x) \]

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Rubi [B]  time = 0.69, antiderivative size = 45, normalized size of antiderivative = 2.14, number of steps used = 4, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1593, 6688, 2288} \begin {gather*} \frac {2\ 9^{5 e^x x+4} e^x (x+1) \log (3)}{x^2 \left (e^x x+e^x\right ) \log (9)}+\log (8-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3 + E^(2*(4*Log[3] + 5*E^x*x*Log[3]))*(16 - 2*x + E^x*(-80*x - 70*x^2 + 10*x^3)*Log[3]))/(-8*x^3 + x^4)
,x]

[Out]

(2*9^(4 + 5*E^x*x)*E^x*(1 + x)*Log[3])/(x^2*(E^x + E^x*x)*Log[9]) + Log[8 - x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^3+e^{2 \left (4 \log (3)+5 e^x x \log (3)\right )} \left (16-2 x+e^x \left (-80 x-70 x^2+10 x^3\right ) \log (3)\right )}{(-8+x) x^3} \, dx\\ &=\int \left (\frac {1}{-8+x}+\frac {2\ 9^{4+5 e^x x} \left (-1+5 e^x x (1+x) \log (3)\right )}{x^3}\right ) \, dx\\ &=\log (8-x)+2 \int \frac {9^{4+5 e^x x} \left (-1+5 e^x x (1+x) \log (3)\right )}{x^3} \, dx\\ &=\frac {2\ 9^{4+5 e^x x} e^x (1+x) \log (3)}{x^2 \left (e^x+e^x x\right ) \log (9)}+\log (8-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 19, normalized size = 0.90 \begin {gather*} \frac {9^{4+5 e^x x}}{x^2}+\log (-8+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + E^(2*(4*Log[3] + 5*E^x*x*Log[3]))*(16 - 2*x + E^x*(-80*x - 70*x^2 + 10*x^3)*Log[3]))/(-8*x^3
+ x^4),x]

[Out]

9^(4 + 5*E^x*x)/x^2 + Log[-8 + x]

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fricas [A]  time = 0.67, size = 26, normalized size = 1.24 \begin {gather*} \frac {x^{2} \log \left (x - 8\right ) + e^{\left (10 \, x e^{x} \log \relax (3) + 8 \, \log \relax (3)\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-70*x^2-80*x)*log(3)*exp(x)+16-2*x)*exp(5/2*x*log(3)*exp(x)+2*log(3))^4+x^3)/(x^4-8*x^3),x,
algorithm="fricas")

[Out]

(x^2*log(x - 8) + e^(10*x*e^x*log(3) + 8*log(3)))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} + 2 \, {\left (5 \, {\left (x^{3} - 7 \, x^{2} - 8 \, x\right )} e^{x} \log \relax (3) - x + 8\right )} e^{\left (10 \, x e^{x} \log \relax (3) + 8 \, \log \relax (3)\right )}}{x^{4} - 8 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-70*x^2-80*x)*log(3)*exp(x)+16-2*x)*exp(5/2*x*log(3)*exp(x)+2*log(3))^4+x^3)/(x^4-8*x^3),x,
algorithm="giac")

[Out]

integrate((x^3 + 2*(5*(x^3 - 7*x^2 - 8*x)*e^x*log(3) - x + 8)*e^(10*x*e^x*log(3) + 8*log(3)))/(x^4 - 8*x^3), x
)

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maple [A]  time = 0.10, size = 20, normalized size = 0.95

method	result	size
risch	\(\ln \left (-8+x \right )+\frac {6561 \,3^{10 \,{\mathrm e}^{x} x}}{x^{2}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^3-70*x^2-80*x)*ln(3)*exp(x)+16-2*x)*exp(5/2*x*ln(3)*exp(x)+2*ln(3))^4+x^3)/(x^4-8*x^3),x,method=_R
ETURNVERBOSE)

[Out]

ln(-8+x)+6561/x^2*(3^(5/2*exp(x)*x))^4

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maxima [A]  time = 0.49, size = 17, normalized size = 0.81 \begin {gather*} \frac {6561 \cdot 3^{10 \, x e^{x}}}{x^{2}} + \log \left (x - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-70*x^2-80*x)*log(3)*exp(x)+16-2*x)*exp(5/2*x*log(3)*exp(x)+2*log(3))^4+x^3)/(x^4-8*x^3),x,
algorithm="maxima")

[Out]

6561*3^(10*x*e^x)/x^2 + log(x - 8)

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mupad [B]  time = 3.42, size = 17, normalized size = 0.81 \begin {gather*} \ln \left (x-8\right )+\frac {6561\,3^{10\,x\,{\mathrm {e}}^x}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(8*log(3) + 10*x*exp(x)*log(3))*(2*x + exp(x)*log(3)*(80*x + 70*x^2 - 10*x^3) - 16) - x^3)/(8*x^3 - x^
4),x)

[Out]

log(x - 8) + (6561*3^(10*x*exp(x)))/x^2

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sympy [A]  time = 0.18, size = 20, normalized size = 0.95 \begin {gather*} \log {\left (x - 8 \right )} + \frac {6561 e^{10 x e^{x} \log {\relax (3 )}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**3-70*x**2-80*x)*ln(3)*exp(x)+16-2*x)*exp(5/2*x*ln(3)*exp(x)+2*ln(3))**4+x**3)/(x**4-8*x**3)
,x)

[Out]

log(x - 8) + 6561*exp(10*x*exp(x)*log(3))/x**2

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