\(\int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx\) [9990]

Optimal result

Integrand size = 24, antiderivative size = 28 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=3 \left (-\frac {e^3}{x}+\frac {-\frac {2}{e}+\frac {x^5}{4}}{x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14} \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3 x^3}{4}-\frac {6}{e x^2}-\frac {3 e^3}{x} \]

[In]

Int[(48/E + 12*E^3*x + 9*x^5)/(4*x^3),x]

[Out]

-6/(E*x^2) - (3*E^3)/x + (3*x^3)/4

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{x^3} \, dx \\ & = \frac {1}{4} \int \left (\frac {48}{e x^3}+\frac {12 e^3}{x^2}+9 x^2\right ) \, dx \\ & = -\frac {6}{e x^2}-\frac {3 e^3}{x}+\frac {3 x^3}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3 \left (-\frac {8}{x^2}-\frac {4 e^4}{x}+e x^3\right )}{4 e} \]

[In]

Integrate[(48/E + 12*E^3*x + 9*x^5)/(4*x^3),x]

[Out]

(3*(-8/x^2 - (4*E^4)/x + E*x^3))/(4*E)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {3 \left (-x^{5}+8 \,{\mathrm e}^{-1}+4 x \,{\mathrm e}^{3}\right )}{4 x^{2}}\)	\(21\)
norman	\(\frac {\frac {3 x^{5}}{4}-6 \,{\mathrm e}^{-1}-3 x \,{\mathrm e}^{3}}{x^{2}}\)	\(22\)
gosper	\(-\frac {3 \left (-x^{5}+4 x \,{\mathrm e}^{3}+4 \,{\mathrm e}^{\ln \left (2\right )-1}\right )}{4 x^{2}}\)	\(24\)
default	\(\frac {3 x^{3}}{4}-\frac {3 \,{\mathrm e}^{3}}{x}-\frac {3 \,{\mathrm e}^{\ln \left (2\right )-1}}{x^{2}}\)	\(24\)
parallelrisch	\(-\frac {-3 x^{5}+12 x \,{\mathrm e}^{3}+12 \,{\mathrm e}^{\ln \left (2\right )-1}}{4 x^{2}}\)	\(24\)

[In]

int(1/4*(24*exp(ln(2)-1)+12*x*exp(3)+9*x^5)/x^3,x,method=_RETURNVERBOSE)

[Out]

-3/4*(-x^5+8*exp(-1)+4*x*exp(3))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3 \, {\left (x^{5} e^{\left (3 \, \log \left (2\right ) - 3\right )} - 32 \, x - 4 \, e^{\left (4 \, \log \left (2\right ) - 4\right )}\right )} e^{\left (-3 \, \log \left (2\right ) + 3\right )}}{4 \, x^{2}} \]

[In]

integrate(1/4*(24*exp(log(2)-1)+12*x*exp(3)+9*x^5)/x^3,x, algorithm="fricas")

[Out]

3/4*(x^5*e^(3*log(2) - 3) - 32*x - 4*e^(4*log(2) - 4))*e^(-3*log(2) + 3)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3 e x^{3} + \frac {- 12 x e^{4} - 24}{x^{2}}}{4 e} \]

[In]

integrate(1/4*(24*exp(ln(2)-1)+12*x*exp(3)+9*x**5)/x**3,x)

[Out]

(3*E*x**3 + (-12*x*exp(4) - 24)/x**2)*exp(-1)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3}{4} \, x^{3} - \frac {3 \, {\left (x e^{4} + 2\right )} e^{\left (-1\right )}}{x^{2}} \]

[In]

integrate(1/4*(24*exp(log(2)-1)+12*x*exp(3)+9*x^5)/x^3,x, algorithm="maxima")

[Out]

3/4*x^3 - 3*(x*e^4 + 2)*e^(-1)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3}{4} \, x^{3} - \frac {3 \, {\left (x e^{3} + e^{\left (\log \left (2\right ) - 1\right )}\right )}}{x^{2}} \]

[In]

integrate(1/4*(24*exp(log(2)-1)+12*x*exp(3)+9*x^5)/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\frac {48}{e}+12 e^3 x+9 x^5}{4 x^3} \, dx=\frac {3\,x^3}{4}-\frac {{\mathrm {e}}^{-1}\,\left (3\,x\,{\mathrm {e}}^4+6\right )}{x^2} \]

[In]

int((6*exp(log(2) - 1) + 3*x*exp(3) + (9*x^5)/4)/x^3,x)

[Out]

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