3.57.91 \(\int \frac {-24 e^{-1+e^{\frac {2 (15+2 x)}{5 x}}+\frac {2 (15+2 x)}{5 x}}+9 x+3 x^2}{4 x^2} \, dx\)

Optimal. Leaf size=27 \[ 3+e^{-1+e^{\frac {4}{5}+\frac {6}{x}}}+\frac {3}{4} (x+3 \log (x)) \]

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Rubi [A]  time = 0.13, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 6715, 2282, 2194, 43} \begin {gather*} \frac {3 x}{4}+e^{e^{\frac {6}{x}+\frac {4}{5}}-1}+\frac {9 \log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24*E^(-1 + E^((2*(15 + 2*x))/(5*x)) + (2*(15 + 2*x))/(5*x)) + 9*x + 3*x^2)/(4*x^2),x]

[Out]

E^(-1 + E^(4/5 + 6/x)) + (3*x)/4 + (9*Log[x])/4

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-24 \exp \left (-1+e^{\frac {2 (15+2 x)}{5 x}}+\frac {2 (15+2 x)}{5 x}\right )+9 x+3 x^2}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {24 e^{-\frac {1}{5}+e^{\frac {4}{5}+\frac {6}{x}}+\frac {6}{x}}}{x^2}+\frac {3 (3+x)}{x}\right ) \, dx\\ &=\frac {3}{4} \int \frac {3+x}{x} \, dx-6 \int \frac {e^{-\frac {1}{5}+e^{\frac {4}{5}+\frac {6}{x}}+\frac {6}{x}}}{x^2} \, dx\\ &=\frac {3}{4} \int \left (1+\frac {3}{x}\right ) \, dx+6 \operatorname {Subst}\left (\int e^{-\frac {1}{5}+e^{\frac {4}{5}+6 x}+6 x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {3 x}{4}+\frac {9 \log (x)}{4}+\operatorname {Subst}\left (\int e^{-\frac {1}{5}+e^{4/5} x} \, dx,x,e^{6/x}\right )\\ &=e^{-1+e^{\frac {4}{5}+\frac {6}{x}}}+\frac {3 x}{4}+\frac {9 \log (x)}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 29, normalized size = 1.07 \begin {gather*} \frac {3}{4} \left (\frac {4}{3} e^{-1+e^{\frac {4}{5}+\frac {6}{x}}}+x+3 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24*E^(-1 + E^((2*(15 + 2*x))/(5*x)) + (2*(15 + 2*x))/(5*x)) + 9*x + 3*x^2)/(4*x^2),x]

[Out]

(3*((4*E^(-1 + E^(4/5 + 6/x)))/3 + x + 3*Log[x]))/4

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fricas [B]  time = 0.64, size = 70, normalized size = 2.59 \begin {gather*} \frac {1}{4} \, {\left (3 \, x e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} + 9 \, e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} \log \relax (x) + 4 \, e^{\left (\frac {5 \, x e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} - x + 30}{5 \, x}\right )}\right )} e^{\left (-\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*exp(1/5*(2*x+15)/x)^2*exp(exp(1/5*(2*x+15)/x)^2-1)+3*x^2+9*x)/x^2,x, algorithm="fricas")

[Out]

1/4*(3*x*e^(2/5*(2*x + 15)/x) + 9*e^(2/5*(2*x + 15)/x)*log(x) + 4*e^(1/5*(5*x*e^(2/5*(2*x + 15)/x) - x + 30)/x
))*e^(-2/5*(2*x + 15)/x)

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giac [B]  time = 0.20, size = 70, normalized size = 2.59 \begin {gather*} \frac {1}{4} \, {\left (3 \, x e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} + 9 \, e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} \log \relax (x) + 4 \, e^{\left (\frac {5 \, x e^{\left (\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} - x + 30}{5 \, x}\right )}\right )} e^{\left (-\frac {2 \, {\left (2 \, x + 15\right )}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*exp(1/5*(2*x+15)/x)^2*exp(exp(1/5*(2*x+15)/x)^2-1)+3*x^2+9*x)/x^2,x, algorithm="giac")

[Out]

1/4*(3*x*e^(2/5*(2*x + 15)/x) + 9*e^(2/5*(2*x + 15)/x)*log(x) + 4*e^(1/5*(5*x*e^(2/5*(2*x + 15)/x) - x + 30)/x
))*e^(-2/5*(2*x + 15)/x)

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maple [A]  time = 0.05, size = 23, normalized size = 0.85

method	result	size
risch	\(\frac {3 x}{4}+\frac {9 \ln \relax (x )}{4}+{\mathrm e}^{{\mathrm e}^{\frac {\frac {4 x}{5}+6}{x}}-1}\)	\(23\)
default	\(\frac {3 x}{4}+\frac {9 \ln \relax (x )}{4}+{\mathrm e}^{{\mathrm e}^{\frac {4}{5}} {\mathrm e}^{\frac {6}{x}}} {\mathrm e}^{-1}\)	\(24\)
norman	\(\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {\frac {4 x}{5}+6}{x}}-1}+\frac {3 x^{2}}{4}}{x}+\frac {9 \ln \relax (x )}{4}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-24*exp(1/5*(2*x+15)/x)^2*exp(exp(1/5*(2*x+15)/x)^2-1)+3*x^2+9*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

3/4*x+9/4*ln(x)+exp(exp(2/5*(2*x+15)/x)-1)

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maxima [A]  time = 0.35, size = 19, normalized size = 0.70 \begin {gather*} \frac {3}{4} \, x + e^{\left (e^{\left (\frac {6}{x} + \frac {4}{5}\right )} - 1\right )} + \frac {9}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*exp(1/5*(2*x+15)/x)^2*exp(exp(1/5*(2*x+15)/x)^2-1)+3*x^2+9*x)/x^2,x, algorithm="maxima")

[Out]

3/4*x + e^(e^(6/x + 4/5) - 1) + 9/4*log(x)

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mupad [B]  time = 3.78, size = 21, normalized size = 0.78 \begin {gather*} \frac {3\,x}{4}+\frac {9\,\ln \relax (x)}{4}+{\mathrm {e}}^{-1}\,{\mathrm {e}}^{{\mathrm {e}}^{4/5}\,{\mathrm {e}}^{6/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*x)/4 - 6*exp(exp((2*((2*x)/5 + 3))/x) - 1)*exp((2*((2*x)/5 + 3))/x) + (3*x^2)/4)/x^2,x)

[Out]

(3*x)/4 + (9*log(x))/4 + exp(-1)*exp(exp(4/5)*exp(6/x))

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sympy [A]  time = 0.30, size = 24, normalized size = 0.89 \begin {gather*} \frac {3 x}{4} + e^{e^{\frac {2 \left (\frac {2 x}{5} + 3\right )}{x}} - 1} + \frac {9 \log {\relax (x )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-24*exp(1/5*(2*x+15)/x)**2*exp(exp(1/5*(2*x+15)/x)**2-1)+3*x**2+9*x)/x**2,x)

[Out]

3*x/4 + exp(exp(2*(2*x/5 + 3)/x) - 1) + 9*log(x)/4

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