3.87.58 \(\int \frac {5 x^2+e^x (1225 x^2+245 x^3+e^{5/x} (-1225+245 x^2))}{x^2} \, dx\)

Optimal. Leaf size=19 \[ 5 \left (x+49 e^x \left (4+e^{5/x}+x\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.36, antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 9, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 6688, 6742, 2194, 2176, 6706} \begin {gather*} 245 e^x x+5 x+980 e^x+245 e^{x+\frac {5}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 + E^x*(1225*x^2 + 245*x^3 + E^(5/x)*(-1225 + 245*x^2)))/x^2,x]

[Out]

980*E^x + 245*E^(5/x + x) + 5*x + 245*E^x*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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (5+\frac {245 e^x \left (-5 e^{5/x}+5 x^2+e^{5/x} x^2+x^3\right )}{x^2}\right ) \, dx\\ &=5 x+245 \int \frac {e^x \left (-5 e^{5/x}+5 x^2+e^{5/x} x^2+x^3\right )}{x^2} \, dx\\ &=5 x+245 \int e^x \left (5+e^{5/x} \left (1-\frac {5}{x^2}\right )+x\right ) \, dx\\ &=5 x+245 \int \left (5 e^x+e^x x+\frac {e^{\frac {5}{x}+x} \left (-5+x^2\right )}{x^2}\right ) \, dx\\ &=5 x+245 \int e^x x \, dx+245 \int \frac {e^{\frac {5}{x}+x} \left (-5+x^2\right )}{x^2} \, dx+1225 \int e^x \, dx\\ &=1225 e^x+245 e^{\frac {5}{x}+x}+5 x+245 e^x x-245 \int e^x \, dx\\ &=980 e^x+245 e^{\frac {5}{x}+x}+5 x+245 e^x x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 23, normalized size = 1.21 \begin {gather*} 245 e^{\frac {5}{x}+x}+5 x+245 e^x (4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + E^x*(1225*x^2 + 245*x^3 + E^(5/x)*(-1225 + 245*x^2)))/x^2,x]

[Out]

245*E^(5/x + x) + 5*x + 245*E^x*(4 + x)

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 17, normalized size = 0.89 \begin {gather*} 245 \, {\left (x + e^{\frac {5}{x}} + 4\right )} e^{x} + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((245*x^2-1225)*exp(5/x)+245*x^3+1225*x^2)*exp(x)+5*x^2)/x^2,x, algorithm="fricas")

[Out]

245*(x + e^(5/x) + 4)*e^x + 5*x

________________________________________________________________________________________

giac [A]  time = 0.20, size = 25, normalized size = 1.32 \begin {gather*} 245 \, x e^{x} + 5 \, x + 980 \, e^{x} + 245 \, e^{\left (\frac {x^{2} + 5}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((245*x^2-1225)*exp(5/x)+245*x^3+1225*x^2)*exp(x)+5*x^2)/x^2,x, algorithm="giac")

[Out]

245*x*e^x + 5*x + 980*e^x + 245*e^((x^2 + 5)/x)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 25, normalized size = 1.32

method	result	size
risch	\(5 x +\left (980+245 x \right ) {\mathrm e}^{x}+245 \,{\mathrm e}^{\frac {x^{2}+5}{x}}\)	\(25\)
norman	\(\frac {5 x^{2}+980 \,{\mathrm e}^{x} x +245 \,{\mathrm e}^{x} x^{2}+245 \,{\mathrm e}^{x} x \,{\mathrm e}^{\frac {5}{x}}}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((245*x^2-1225)*exp(5/x)+245*x^3+1225*x^2)*exp(x)+5*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

5*x+(980+245*x)*exp(x)+245*exp((x^2+5)/x)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 25, normalized size = 1.32 \begin {gather*} 245 \, {\left (x - 1\right )} e^{x} + 5 \, x + 245 \, e^{\left (x + \frac {5}{x}\right )} + 1225 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((245*x^2-1225)*exp(5/x)+245*x^3+1225*x^2)*exp(x)+5*x^2)/x^2,x, algorithm="maxima")

[Out]

245*(x - 1)*e^x + 5*x + 245*e^(x + 5/x) + 1225*e^x

________________________________________________________________________________________

mupad [B]  time = 5.29, size = 23, normalized size = 1.21 \begin {gather*} 5\,x+245\,{\mathrm {e}}^{x+\frac {5}{x}}+980\,{\mathrm {e}}^x+245\,x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(exp(5/x)*(245*x^2 - 1225) + 1225*x^2 + 245*x^3) + 5*x^2)/x^2,x)

[Out]

5*x + 245*exp(x + 5/x) + 980*exp(x) + 245*x*exp(x)

________________________________________________________________________________________

sympy [A]  time = 2.30, size = 17, normalized size = 0.89 \begin {gather*} 5 x + \left (245 x + 245 e^{\frac {5}{x}} + 980\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((245*x**2-1225)*exp(5/x)+245*x**3+1225*x**2)*exp(x)+5*x**2)/x**2,x)

[Out]

5*x + (245*x + 245*exp(5/x) + 980)*exp(x)

________________________________________________________________________________________