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

Optimal. Leaf size=23 \[ -2+5 \left (-5+e^{x^6}-\frac {3}{x}-x+\log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {14, 2209} \begin {gather*} 5 e^{x^6}-5 x-\frac {15}{x}+10 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 2209

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.87 \begin {gather*} 5 \left (e^{x^6}-\frac {3}{x}-x+2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.59, size = 22, normalized size = 0.96 \begin {gather*} -\frac {5 \, {\left (x^{2} - x e^{\left (x^{6}\right )} - 2 \, x \log \relax (x) + 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.22, size = 22, normalized size = 0.96 \begin {gather*} -\frac {5 \, {\left (x^{2} - x e^{\left (x^{6}\right )} - 2 \, x \log \relax (x) + 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 20, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.37, size = 19, normalized size = 0.83 \begin {gather*} -5 \, x - \frac {15}{x} + 5 \, e^{\left (x^{6}\right )} + 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.18, size = 24, normalized size = 1.04 \begin {gather*} 10\,\ln \relax (x)-\frac {5\,x^2-5\,x\,{\mathrm {e}}^{x^6}+15}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.12, size = 17, normalized size = 0.74 \begin {gather*} - 5 x + 5 e^{x^{6}} + 10 \log {\relax (x )} - \frac {15}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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