3.84.10 \(\int \frac {4+6 x+e^{x^2} (4+6 x+16 x^2+12 x^3)}{4 x+3 x^2+e^{x^2} (4 x+3 x^2)} \, dx\)

Optimal. Leaf size=28 \[ \log \left (\frac {\left (x+e^{x^2} x\right )^2 \left (4 x+3 x^2\right )}{5 x^2}\right ) \]

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Rubi [A]  time = 0.66, antiderivative size = 19, normalized size of antiderivative = 0.68, number of steps used = 10, number of rules used = 8, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.151, Rules used = {6741, 6742, 6715, 2282, 36, 29, 31, 1620} \begin {gather*} 2 \log \left (e^{x^2}+1\right )+\log (x)+\log (3 x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 6*x + E^x^2*(4 + 6*x + 16*x^2 + 12*x^3))/(4*x + 3*x^2 + E^x^2*(4*x + 3*x^2)),x]

[Out]

2*Log[1 + E^x^2] + Log[x] + Log[4 + 3*x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6741

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

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 \frac {4+6 x+e^{x^2} \left (4+6 x+16 x^2+12 x^3\right )}{\left (1+e^{x^2}\right ) x (4+3 x)} \, dx\\ &=\int \left (-\frac {4 x}{1+e^{x^2}}+\frac {2 \left (2+3 x+8 x^2+6 x^3\right )}{x (4+3 x)}\right ) \, dx\\ &=2 \int \frac {2+3 x+8 x^2+6 x^3}{x (4+3 x)} \, dx-4 \int \frac {x}{1+e^{x^2}} \, dx\\ &=2 \int \left (\frac {1}{2 x}+2 x+\frac {3}{2 (4+3 x)}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {1}{1+e^x} \, dx,x,x^2\right )\\ &=2 x^2+\log (x)+\log (4+3 x)-2 \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,e^{x^2}\right )\\ &=2 x^2+\log (x)+\log (4+3 x)-2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{x^2}\right )+2 \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,e^{x^2}\right )\\ &=2 \log \left (1+e^{x^2}\right )+\log (x)+\log (4+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 19, normalized size = 0.68 \begin {gather*} 2 \log \left (1+e^{x^2}\right )+\log (x)+\log (4+3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 6*x + E^x^2*(4 + 6*x + 16*x^2 + 12*x^3))/(4*x + 3*x^2 + E^x^2*(4*x + 3*x^2)),x]

[Out]

2*Log[1 + E^x^2] + Log[x] + Log[4 + 3*x]

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fricas [A]  time = 0.68, size = 20, normalized size = 0.71 \begin {gather*} \log \left (3 \, x^{2} + 4 \, x\right ) + 2 \, \log \left (e^{\left (x^{2}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3+16*x^2+6*x+4)*exp(x^2)+6*x+4)/((3*x^2+4*x)*exp(x^2)+3*x^2+4*x),x, algorithm="fricas")

[Out]

log(3*x^2 + 4*x) + 2*log(e^(x^2) + 1)

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giac [A]  time = 0.19, size = 18, normalized size = 0.64 \begin {gather*} \log \left (3 \, x + 4\right ) + \log \relax (x) + 2 \, \log \left (e^{\left (x^{2}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3+16*x^2+6*x+4)*exp(x^2)+6*x+4)/((3*x^2+4*x)*exp(x^2)+3*x^2+4*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 19, normalized size = 0.68

method	result	size
norman	\(2 \ln \left ({\mathrm e}^{x^{2}}+1\right )+\ln \relax (x )+\ln \left (4+3 x \right )\)	\(19\)
risch	\(\ln \left (3 x^{2}+4 x \right )+2 \ln \left ({\mathrm e}^{x^{2}}+1\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^3+16*x^2+6*x+4)*exp(x^2)+6*x+4)/((3*x^2+4*x)*exp(x^2)+3*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(exp(x^2)+1)+ln(x)+ln(4+3*x)

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maxima [A]  time = 0.41, size = 18, normalized size = 0.64 \begin {gather*} \log \left (3 \, x + 4\right ) + \log \relax (x) + 2 \, \log \left (e^{\left (x^{2}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^3+16*x^2+6*x+4)*exp(x^2)+6*x+4)/((3*x^2+4*x)*exp(x^2)+3*x^2+4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.11, size = 18, normalized size = 0.64 \begin {gather*} \ln \relax (x)+\ln \left (\left (3\,x+4\right )\,{\left ({\mathrm {e}}^{x^2}+1\right )}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + exp(x^2)*(6*x + 16*x^2 + 12*x^3 + 4) + 4)/(4*x + exp(x^2)*(4*x + 3*x^2) + 3*x^2),x)

[Out]

log(x) + log((3*x + 4)*(exp(x^2) + 1)^2)

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sympy [A]  time = 0.18, size = 19, normalized size = 0.68 \begin {gather*} \log {\left (3 x^{2} + 4 x \right )} + 2 \log {\left (e^{x^{2}} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**3+16*x**2+6*x+4)*exp(x**2)+6*x+4)/((3*x**2+4*x)*exp(x**2)+3*x**2+4*x),x)

[Out]

log(3*x**2 + 4*x) + 2*log(exp(x**2) + 1)

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