3.46.75 \(\int \frac {-120 e^x+4 x-x^3+e^x (60 x-15 x^3) \log (\frac {4-x^2}{x^2})}{-60 x+15 x^3} \, dx\)

Optimal. Leaf size=28 \[ 2+\frac {1}{15} \left (-e^2-x\right )-e^x \log \left (-1+\frac {4}{x^2}\right ) \]

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Rubi [A]  time = 0.54, antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1593, 6725, 2288} \begin {gather*} -\frac {e^x \left (4 x \log \left (\frac {4}{x^2}-1\right )-x^3 \log \left (\frac {4}{x^2}-1\right )\right )}{\left (4-x^2\right ) x}-\frac {x}{15} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-120*E^x + 4*x - x^3 + E^x*(60*x - 15*x^3)*Log[(4 - x^2)/x^2])/(-60*x + 15*x^3),x]

[Out]

-1/15*x - (E^x*(4*x*Log[-1 + 4/x^2] - x^3*Log[-1 + 4/x^2]))/(x*(4 - x^2))

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 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]

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 {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{x \left (-60+15 x^2\right )} \, dx\\ &=\int \left (-\frac {1}{15}-\frac {e^x \left (8-4 x \log \left (-1+\frac {4}{x^2}\right )+x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (-4+x^2\right )}\right ) \, dx\\ &=-\frac {x}{15}-\int \frac {e^x \left (8-4 x \log \left (-1+\frac {4}{x^2}\right )+x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (-4+x^2\right )} \, dx\\ &=-\frac {x}{15}-\frac {e^x \left (4 x \log \left (-1+\frac {4}{x^2}\right )-x^3 \log \left (-1+\frac {4}{x^2}\right )\right )}{x \left (4-x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 21, normalized size = 0.75 \begin {gather*} \frac {1}{15} \left (-x-15 e^x \log \left (-1+\frac {4}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-120*E^x + 4*x - x^3 + E^x*(60*x - 15*x^3)*Log[(4 - x^2)/x^2])/(-60*x + 15*x^3),x]

[Out]

(-x - 15*E^x*Log[-1 + 4/x^2])/15

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fricas [A]  time = 0.55, size = 19, normalized size = 0.68 \begin {gather*} -e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-60*x),x, algorithm="fricas")

[Out]

-e^x*log(-(x^2 - 4)/x^2) - 1/15*x

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giac [A]  time = 0.16, size = 19, normalized size = 0.68 \begin {gather*} -e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-60*x),x, algorithm="giac")

[Out]

-e^x*log(-(x^2 - 4)/x^2) - 1/15*x

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maple [A]  time = 0.22, size = 21, normalized size = 0.75

method	result	size
default	\(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\)	\(21\)
norman	\(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\)	\(21\)
risch	\(-{\mathrm e}^{x} \ln \left (x^{2}-4\right )+2 \,{\mathrm e}^{x} \ln \relax (x )-\frac {x}{15}-\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}+\frac {i {\mathrm e}^{x} \pi \,\mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}{2}-\frac {i {\mathrm e}^{x} \pi \,\mathrm {csgn}\left (i \left (x^{2}-4\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )^{2}}{2}+i {\mathrm e}^{x} \pi \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )^{2}-\frac {i {\mathrm e}^{x} \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )^{2}}{2}-\frac {i {\mathrm e}^{x} \pi \mathrm {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )^{3}}{2}-i {\mathrm e}^{x} \pi \)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*x^3+60*x)*exp(x)*ln((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-60*x),x,method=_RETURNVERBOSE)

[Out]

-1/15*x-ln((-x^2+4)/x^2)*exp(x)

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maxima [A]  time = 0.40, size = 28, normalized size = 1.00 \begin {gather*} -e^{x} \log \left (x + 2\right ) + 2 \, e^{x} \log \relax (x) - e^{x} \log \left (-x + 2\right ) - \frac {1}{15} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-60*x),x, algorithm="maxima")

[Out]

-e^x*log(x + 2) + 2*e^x*log(x) - e^x*log(-x + 2) - 1/15*x

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mupad [B]  time = 3.38, size = 19, normalized size = 0.68 \begin {gather*} -\frac {x}{15}-{\mathrm {e}}^x\,\ln \left (-\frac {x^2-4}{x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 120*exp(x) - x^3 + exp(x)*log(-(x^2 - 4)/x^2)*(60*x - 15*x^3))/(60*x - 15*x^3),x)

[Out]

- x/15 - exp(x)*log(-(x^2 - 4)/x^2)

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sympy [A]  time = 0.40, size = 17, normalized size = 0.61 \begin {gather*} - \frac {x}{15} - e^{x} \log {\left (\frac {4 - x^{2}}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*x**3+60*x)*exp(x)*ln((-x**2+4)/x**2)-120*exp(x)-x**3+4*x)/(15*x**3-60*x),x)

[Out]

-x/15 - exp(x)*log((4 - x**2)/x**2)

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