\(\int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx\) [6661]

Optimal result

Integrand size = 24, antiderivative size = 23 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=-2+5 \left (-5+e^{x^6}-\frac {3}{x}-x+\log \left (x^2\right )\right ) \]

[Out]

5*exp(x^6)-5*x-27+5*ln(x^2)-15/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {14, 2240} \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=5 e^{x^6}-5 x-\frac {15}{x}+10 \log (x) \]

[In]

Int[(15 + 10*x - 5*x^2 + 30*E^x^6*x^7)/x^2,x]

[Out]

5*E^x^6 - 15/x - 5*x + 10*Log[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 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (30 e^{x^6} x^5-\frac {5 \left (-3-2 x+x^2\right )}{x^2}\right ) \, dx \\ & = -\left (5 \int \frac {-3-2 x+x^2}{x^2} \, dx\right )+30 \int e^{x^6} x^5 \, dx \\ & = 5 e^{x^6}-5 \int \left (1-\frac {3}{x^2}-\frac {2}{x}\right ) \, dx \\ & = 5 e^{x^6}-\frac {15}{x}-5 x+10 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=5 \left (e^{x^6}-\frac {3}{x}-x+2 \log (x)\right ) \]

[In]

Integrate[(15 + 10*x - 5*x^2 + 30*E^x^6*x^7)/x^2,x]

[Out]

5*(E^x^6 - 3/x - x + 2*Log[x])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-5 x -\frac {15}{x}+10 \ln \left (x \right )+5 \,{\mathrm e}^{x^{6}}\)	\(20\)
risch	\(-5 x -\frac {15}{x}+10 \ln \left (x \right )+5 \,{\mathrm e}^{x^{6}}\)	\(20\)
parts	\(-5 x -\frac {15}{x}+10 \ln \left (x \right )+5 \,{\mathrm e}^{x^{6}}\)	\(20\)
norman	\(\frac {-15-5 x^{2}+5 x \,{\mathrm e}^{x^{6}}}{x}+10 \ln \left (x \right )\)	\(24\)
parallelrisch	\(\frac {10 x \ln \left (x \right )-5 x^{2}+5 x \,{\mathrm e}^{x^{6}}-15}{x}\)	\(24\)

[In]

int((30*x^7*exp(x^6)-5*x^2+10*x+15)/x^2,x,method=_RETURNVERBOSE)

[Out]

-5*x-15/x+10*ln(x)+5*exp(x^6)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=-\frac {5 \, {\left (x^{2} - x e^{\left (x^{6}\right )} - 2 \, x \log \left (x\right ) + 3\right )}}{x} \]

[In]

integrate((30*x^7*exp(x^6)-5*x^2+10*x+15)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=- 5 x + 5 e^{x^{6}} + 10 \log {\left (x \right )} - \frac {15}{x} \]

[In]

integrate((30*x**7*exp(x**6)-5*x**2+10*x+15)/x**2,x)

[Out]

-5*x + 5*exp(x**6) + 10*log(x) - 15/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=-5 \, x - \frac {15}{x} + 5 \, e^{\left (x^{6}\right )} + 10 \, \log \left (x\right ) \]

[In]

integrate((30*x^7*exp(x^6)-5*x^2+10*x+15)/x^2,x, algorithm="maxima")

[Out]

-5*x - 15/x + 5*e^(x^6) + 10*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=-\frac {5 \, {\left (x^{2} - x e^{\left (x^{6}\right )} - 2 \, x \log \left (x\right ) + 3\right )}}{x} \]

[In]

integrate((30*x^7*exp(x^6)-5*x^2+10*x+15)/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.63 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {15+10 x-5 x^2+30 e^{x^6} x^7}{x^2} \, dx=10\,\ln \left (x\right )-\frac {5\,x^2-5\,x\,{\mathrm {e}}^{x^6}+15}{x} \]

[In]

int((10*x + 30*x^7*exp(x^6) - 5*x^2 + 15)/x^2,x)

[Out]

10*log(x) - (5*x^2 - 5*x*exp(x^6) + 15)/x