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

Optimal. Leaf size=19 \[ 4+4 e^x \log (3)+\log \left (\frac {16+x^2}{x}\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1593, 6725, 446, 72, 2194} \begin {gather*} \log \left (x^2+16\right )-\log (x)+e^x \log (81) \end {gather*}

Antiderivative was successfully verified.

[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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

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 1593

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 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 6725

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 {gather*} \begin {aligned} \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} \operatorname {Subst}\left (\int \frac {-16+x}{x (16+x)} \, dx,x,x^2\right )\\ &=e^x \log (81)+\frac {1}{2} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 17, normalized size = 0.89 \begin {gather*} e^x \log (81)-\log (x)+\log \left (16+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.62, size = 17, normalized size = 0.89 \begin {gather*} 4 \, e^{x} \log \relax (3) + \log \left (x^{2} + 16\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.14, size = 17, normalized size = 0.89 \begin {gather*} 4 \, e^{x} \log \relax (3) + \log \left (x^{2} + 16\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.26, size = 18, normalized size = 0.95

method	result	size
default	\(\ln \left (x^{2}+16\right )-\ln \relax (x )+4 \ln \relax (3) {\mathrm e}^{x}\)	\(18\)
norman	\(\ln \left (x^{2}+16\right )-\ln \relax (x )+4 \ln \relax (3) {\mathrm e}^{x}\)	\(18\)
risch	\(\ln \left (x^{2}+16\right )-\ln \relax (x )+4 \ln \relax (3) {\mathrm e}^{x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 17, normalized size = 0.89 \begin {gather*} 4 \, e^{x} \log \relax (3) + \log \left (x^{2} + 16\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.11, size = 17, normalized size = 0.89 \begin {gather*} \ln \left (x^2+16\right )-\ln \relax (x)+4\,{\mathrm {e}}^x\,\ln \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.15, size = 17, normalized size = 0.89 \begin {gather*} 4 e^{x} \log {\relax (3 )} - \log {\relax (x )} + \log {\left (x^{2} + 16 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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