3.29.41 \(\int \frac {5 x^2+8 x^3+e^{x^2} (-5+10 x^2+e (-1+2 x^2))}{5 x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {\frac {1}{5} e^{x^2} (5+e)-x+x \left (x+\frac {4 x^2}{5}\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 0.77, number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 14, 2288} \begin {gather*} \frac {4 x^2}{5}+\frac {(5+e) e^{x^2}}{5 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 + 8*x^3 + E^x^2*(-5 + 10*x^2 + E*(-1 + 2*x^2)))/(5*x^2),x]

[Out]

(E^x^2*(5 + E))/(5*x) + x + (4*x^2)/5

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]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {5 x^2+8 x^3+e^{x^2} \left (-5+10 x^2+e \left (-1+2 x^2\right )\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (5+8 x+\frac {e^{x^2} (5+e) \left (-1+2 x^2\right )}{x^2}\right ) \, dx\\ &=x+\frac {4 x^2}{5}+\frac {1}{5} (5+e) \int \frac {e^{x^2} \left (-1+2 x^2\right )}{x^2} \, dx\\ &=\frac {e^{x^2} (5+e)}{5 x}+x+\frac {4 x^2}{5}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 34, normalized size = 1.10 \begin {gather*} \frac {1}{5} \left (\frac {5 e^{x^2}}{x}+\frac {e^{1+x^2}}{x}+5 x+4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + 8*x^3 + E^x^2*(-5 + 10*x^2 + E*(-1 + 2*x^2)))/(5*x^2),x]

[Out]

((5*E^x^2)/x + E^(1 + x^2)/x + 5*x + 4*x^2)/5

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 25, normalized size = 0.81 \begin {gather*} \frac {4 \, x^{3} + 5 \, x^{2} + {\left (e + 5\right )} e^{\left (x^{2}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^2-1)*exp(1)+10*x^2-5)*exp(x^2)+8*x^3+5*x^2)/x^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.37, size = 28, normalized size = 0.90 \begin {gather*} \frac {4 \, x^{3} + 5 \, x^{2} + e^{\left (x^{2} + 1\right )} + 5 \, e^{\left (x^{2}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^2-1)*exp(1)+10*x^2-5)*exp(x^2)+8*x^3+5*x^2)/x^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 21, normalized size = 0.68

method	result	size
risch	\(\frac {4 x^{2}}{5}+x +\frac {\left ({\mathrm e}+5\right ) {\mathrm e}^{x^{2}}}{5 x}\)	\(21\)
norman	\(\frac {x^{2}+\left (\frac {{\mathrm e}}{5}+1\right ) {\mathrm e}^{x^{2}}+\frac {4 x^{3}}{5}}{x}\)	\(25\)
default	\(\frac {4 x^{2}}{5}+x +\frac {{\mathrm e} \sqrt {\pi }\, \erfi \relax (x )}{5}+\frac {{\mathrm e}^{x^{2}}}{x}-\frac {{\mathrm e} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \erfi \relax (x )\right )}{5}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((2*x^2-1)*exp(1)+10*x^2-5)*exp(x^2)+8*x^3+5*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [C]  time = 0.41, size = 67, normalized size = 2.16 \begin {gather*} -\frac {1}{5} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e + \frac {4}{5} \, x^{2} - i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) + \frac {\sqrt {-x^{2}} e \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{10 \, x} + x + \frac {\sqrt {-x^{2}} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x^2-1)*exp(1)+10*x^2-5)*exp(x^2)+8*x^3+5*x^2)/x^2,x, algorithm="maxima")

[Out]

-1/5*I*sqrt(pi)*erf(I*x)*e + 4/5*x^2 - I*sqrt(pi)*erf(I*x) + 1/10*sqrt(-x^2)*e*gamma(-1/2, -x^2)/x + x + 1/2*s
qrt(-x^2)*gamma(-1/2, -x^2)/x

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 20, normalized size = 0.65 \begin {gather*} x+\frac {4\,x^2}{5}+\frac {{\mathrm {e}}^{x^2}\,\left (\mathrm {e}+5\right )}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x^2)*(exp(1)*(2*x^2 - 1) + 10*x^2 - 5))/5 + x^2 + (8*x^3)/5)/x^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 20, normalized size = 0.65 \begin {gather*} \frac {4 x^{2}}{5} + x + \frac {\left (e + 5\right ) e^{x^{2}}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*x**2-1)*exp(1)+10*x**2-5)*exp(x**2)+8*x**3+5*x**2)/x**2,x)

[Out]

4*x**2/5 + x + (E + 5)*exp(x**2)/(5*x)

________________________________________________________________________________________