\(\int \frac {-16+x^2+e^x (64 x+4 x^3) \log (3)}{16 x+x^3} \, dx\) [1553]

Optimal result

Integrand size = 30, antiderivative size = 19 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4+4 e^x \log (3)+\log \left (\frac {16+x^2}{x}\right ) \]

[Out]

4+4*ln(3)*exp(x)+ln((x^2+16)/x)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1607, 6857, 457, 78, 2225} \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=\log \left (x^2+16\right )-\log (x)+e^x \log (81) \]

[In]

Int[(-16 + x^2 + E^x*(64*x + 4*x^3)*Log[3])/(16*x + x^3),x]

[Out]

E^x*Log[81] - Log[x] + Log[16 + x^2]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2225

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 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{x \left (16+x^2\right )} \, dx \\ & = \int \left (\frac {-16+x^2}{x \left (16+x^2\right )}+e^x \log (81)\right ) \, dx \\ & = \log (81) \int e^x \, dx+\int \frac {-16+x^2}{x \left (16+x^2\right )} \, dx \\ & = e^x \log (81)+\frac {1}{2} \text {Subst}\left (\int \frac {-16+x}{x (16+x)} \, dx,x,x^2\right ) \\ & = e^x \log (81)+\frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{x}+\frac {2}{16+x}\right ) \, dx,x,x^2\right ) \\ & = e^x \log (81)-\log (x)+\log \left (16+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=e^x \log (81)-\log (x)+\log \left (16+x^2\right ) \]

[In]

Integrate[(-16 + x^2 + E^x*(64*x + 4*x^3)*Log[3])/(16*x + x^3),x]

[Out]

E^x*Log[81] - Log[x] + Log[16 + x^2]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
norman	\(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\)	\(18\)
risch	\(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\)	\(18\)
parallelrisch	\(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\)	\(18\)
parts	\(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\)	\(18\)
default	\(\ln \left (x^{2}+16\right )-\ln \left (x \right )+8 i \ln \left (3\right ) {\mathrm e}^{4 i} \operatorname {Ei}_{1}\left (-x +4 i\right )-8 i \ln \left (3\right ) {\mathrm e}^{-4 i} \operatorname {Ei}_{1}\left (-x -4 i\right )+4 \ln \left (3\right ) \left ({\mathrm e}^{x}-2 i {\mathrm e}^{4 i} \operatorname {Ei}_{1}\left (-x +4 i\right )+2 i {\mathrm e}^{-4 i} \operatorname {Ei}_{1}\left (-x -4 i\right )\right )\)	\(79\)

[In]

int(((4*x^3+64*x)*ln(3)*exp(x)+x^2-16)/(x^3+16*x),x,method=_RETURNVERBOSE)

[Out]

4*ln(3)*exp(x)+ln(x^2+16)-ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]

[In]

integrate(((4*x^3+64*x)*log(3)*exp(x)+x^2-16)/(x^3+16*x),x, algorithm="fricas")

[Out]

4*e^x*log(3) + log(x^2 + 16) - log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 e^{x} \log {\left (3 \right )} - \log {\left (x \right )} + \log {\left (x^{2} + 16 \right )} \]

[In]

integrate(((4*x**3+64*x)*ln(3)*exp(x)+x**2-16)/(x**3+16*x),x)

[Out]

4*exp(x)*log(3) - log(x) + log(x**2 + 16)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]

[In]

integrate(((4*x^3+64*x)*log(3)*exp(x)+x^2-16)/(x^3+16*x),x, algorithm="maxima")

[Out]

4*e^x*log(3) + log(x^2 + 16) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]

[In]

integrate(((4*x^3+64*x)*log(3)*exp(x)+x^2-16)/(x^3+16*x),x, algorithm="giac")

[Out]

4*e^x*log(3) + log(x^2 + 16) - log(x)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=\ln \left (x^2+16\right )-\ln \left (x\right )+4\,{\mathrm {e}}^x\,\ln \left (3\right ) \]

[In]

int((x^2 + exp(x)*log(3)*(64*x + 4*x^3) - 16)/(16*x + x^3),x)

[Out]

log(x^2 + 16) - log(x) + 4*exp(x)*log(3)