\(\int \frac {-10 e^4-5 x+6 x^3+e^x (10-5 x+5 x^3)-5 x \log (2)}{x^3} \, dx\) [980]

Optimal result

Integrand size = 37, antiderivative size = 38 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=x+5 \left (1+e^x-\frac {-1+\frac {-e^4+e^x}{x}-x^2-\log (2)}{x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6, 14, 2230, 2225, 2208, 2209} \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=-\frac {5 e^x}{x^2}+\frac {5 e^4}{x^2}+6 x+5 e^x+\frac {5 (1+\log (2))}{x} \]

[In]

Int[(-10*E^4 - 5*x + 6*x^3 + E^x*(10 - 5*x + 5*x^3) - 5*x*Log[2])/x^3,x]

[Out]

5*E^x + (5*E^4)/x^2 - (5*E^x)/x^2 + 6*x + (5*(1 + Log[2]))/x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-10 e^4+6 x^3+e^x \left (10-5 x+5 x^3\right )+x (-5-5 \log (2))}{x^3} \, dx \\ & = \int \left (\frac {5 e^x \left (2-x+x^3\right )}{x^3}+\frac {-10 e^4+6 x^3-5 x (1+\log (2))}{x^3}\right ) \, dx \\ & = 5 \int \frac {e^x \left (2-x+x^3\right )}{x^3} \, dx+\int \frac {-10 e^4+6 x^3-5 x (1+\log (2))}{x^3} \, dx \\ & = 5 \int \left (e^x+\frac {2 e^x}{x^3}-\frac {e^x}{x^2}\right ) \, dx+\int \left (6-\frac {10 e^4}{x^3}-\frac {5 (1+\log (2))}{x^2}\right ) \, dx \\ & = \frac {5 e^4}{x^2}+6 x+\frac {5 (1+\log (2))}{x}+5 \int e^x \, dx-5 \int \frac {e^x}{x^2} \, dx+10 \int \frac {e^x}{x^3} \, dx \\ & = 5 e^x+\frac {5 e^4}{x^2}-\frac {5 e^x}{x^2}+\frac {5 e^x}{x}+6 x+\frac {5 (1+\log (2))}{x}+5 \int \frac {e^x}{x^2} \, dx-5 \int \frac {e^x}{x} \, dx \\ & = 5 e^x+\frac {5 e^4}{x^2}-\frac {5 e^x}{x^2}+6 x-5 \operatorname {ExpIntegralEi}(x)+\frac {5 (1+\log (2))}{x}+5 \int \frac {e^x}{x} \, dx \\ & = 5 e^x+\frac {5 e^4}{x^2}-\frac {5 e^x}{x^2}+6 x+\frac {5 (1+\log (2))}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=\frac {5 e^4+5 e^x \left (-1+x^2\right )+x \left (5+6 x^2+\log (32)\right )}{x^2} \]

[In]

Integrate[(-10*E^4 - 5*x + 6*x^3 + E^x*(10 - 5*x + 5*x^3) - 5*x*Log[2])/x^3,x]

[Out]

(5*E^4 + 5*E^x*(-1 + x^2) + x*(5 + 6*x^2 + Log[32]))/x^2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87

method	result	size
parts	\(6 x +\frac {5 \ln \left (2\right )+5}{x}+\frac {5 \,{\mathrm e}^{4}}{x^{2}}-\frac {5 \,{\mathrm e}^{x}}{x^{2}}+5 \,{\mathrm e}^{x}\)	\(33\)
norman	\(\frac {\left (5 \ln \left (2\right )+5\right ) x +6 x^{3}+5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{4}-5 \,{\mathrm e}^{x}}{x^{2}}\)	\(34\)
risch	\(6 x +\frac {\left (5 \ln \left (2\right )+5\right ) x +5 \,{\mathrm e}^{4}}{x^{2}}+\frac {5 \left (x^{2}-1\right ) {\mathrm e}^{x}}{x^{2}}\)	\(34\)
parallelrisch	\(\frac {5 \,{\mathrm e}^{x} x^{2}+6 x^{3}+5 x \ln \left (2\right )+5 \,{\mathrm e}^{4}-5 \,{\mathrm e}^{x}+5 x}{x^{2}}\)	\(34\)
default	\(6 x +\frac {5}{x}+\frac {5 \,{\mathrm e}^{4}}{x^{2}}-\frac {5 \,{\mathrm e}^{x}}{x^{2}}+\frac {5 \ln \left (2\right )}{x}+5 \,{\mathrm e}^{x}\)	\(35\)

[In]

int(((5*x^3-5*x+10)*exp(x)-5*x*ln(2)-10*exp(4)+6*x^3-5*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

6*x+(5*ln(2)+5)/x+5*exp(4)/x^2-5*exp(x)/x^2+5*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=\frac {6 \, x^{3} + 5 \, {\left (x^{2} - 1\right )} e^{x} + 5 \, x \log \left (2\right ) + 5 \, x + 5 \, e^{4}}{x^{2}} \]

[In]

integrate(((5*x^3-5*x+10)*exp(x)-5*x*log(2)-10*exp(4)+6*x^3-5*x)/x^3,x, algorithm="fricas")

[Out]

(6*x^3 + 5*(x^2 - 1)*e^x + 5*x*log(2) + 5*x + 5*e^4)/x^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=6 x + \frac {\left (5 x^{2} - 5\right ) e^{x}}{x^{2}} + \frac {x \left (5 \log {\left (2 \right )} + 5\right ) + 5 e^{4}}{x^{2}} \]

[In]

integrate(((5*x**3-5*x+10)*exp(x)-5*x*ln(2)-10*exp(4)+6*x**3-5*x)/x**3,x)

[Out]

6*x + (5*x**2 - 5)*exp(x)/x**2 + (x*(5*log(2) + 5) + 5*exp(4))/x**2

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=6 \, x + \frac {5 \, \log \left (2\right )}{x} + \frac {5}{x} + \frac {5 \, e^{4}}{x^{2}} + 5 \, e^{x} - 5 \, \Gamma \left (-1, -x\right ) - 10 \, \Gamma \left (-2, -x\right ) \]

[In]

integrate(((5*x^3-5*x+10)*exp(x)-5*x*log(2)-10*exp(4)+6*x^3-5*x)/x^3,x, algorithm="maxima")

[Out]

6*x + 5*log(2)/x + 5/x + 5*e^4/x^2 + 5*e^x - 5*gamma(-1, -x) - 10*gamma(-2, -x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=\frac {6 \, x^{3} + 5 \, x^{2} e^{x} + 5 \, x \log \left (2\right ) + 5 \, x + 5 \, e^{4} - 5 \, e^{x}}{x^{2}} \]

[In]

integrate(((5*x^3-5*x+10)*exp(x)-5*x*log(2)-10*exp(4)+6*x^3-5*x)/x^3,x, algorithm="giac")

[Out]

(6*x^3 + 5*x^2*e^x + 5*x*log(2) + 5*x + 5*e^4 - 5*e^x)/x^2

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {-10 e^4-5 x+6 x^3+e^x \left (10-5 x+5 x^3\right )-5 x \log (2)}{x^3} \, dx=6\,x+5\,{\mathrm {e}}^x+\frac {5\,{\mathrm {e}}^4-5\,{\mathrm {e}}^x+x\,\left (\ln \left (32\right )+5\right )}{x^2} \]

[In]

int(-(5*x + 10*exp(4) + 5*x*log(2) - exp(x)*(5*x^3 - 5*x + 10) - 6*x^3)/x^3,x)

[Out]

6*x + 5*exp(x) + (5*exp(4) - 5*exp(x) + x*(log(32) + 5))/x^2