3.91.93 \(\int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {2 x^3}{3 \left (2-4 e^{-1-16 x^2} x^2\right )}\right ) \]

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Rubi [F]  time = 0.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*E^(1 + 16*x^2) - 2*x^2 - 64*x^4)/(E^(1 + 16*x^2)*x - 2*x^3),x]

[Out]

3*Log[x] + 2*Defer[Subst][Defer[Int][(E^(1 + 16*x) - 2*x)^(-1), x], x, x^2] - 32*Defer[Subst][Defer[Int][x/(E^
(1 + 16*x) - 2*x), x], x, x^2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {3}{x}-\frac {4 x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2}\right ) \, dx\\ &=3 \log (x)-4 \int \frac {x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2} \, dx\\ &=3 \log (x)-2 \operatorname {Subst}\left (\int \frac {-1+16 x}{e^{1+16 x}-2 x} \, dx,x,x^2\right )\\ &=3 \log (x)-2 \operatorname {Subst}\left (\int \left (-\frac {1}{e^{1+16 x}-2 x}+\frac {16 x}{e^{1+16 x}-2 x}\right ) \, dx,x,x^2\right )\\ &=3 \log (x)+2 \operatorname {Subst}\left (\int \frac {1}{e^{1+16 x}-2 x} \, dx,x,x^2\right )-32 \operatorname {Subst}\left (\int \frac {x}{e^{1+16 x}-2 x} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 28, normalized size = 1.08 \begin {gather*} 16 x^2+3 \log (x)-\log \left (e^{1+16 x^2}-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^(1 + 16*x^2) - 2*x^2 - 64*x^4)/(E^(1 + 16*x^2)*x - 2*x^3),x]

[Out]

16*x^2 + 3*Log[x] - Log[E^(1 + 16*x^2) - 2*x^2]

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fricas [A]  time = 0.57, size = 27, normalized size = 1.04 \begin {gather*} 16 \, x^{2} - \log \left (-2 \, x^{2} + e^{\left (16 \, x^{2} + 1\right )}\right ) + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="fricas")

[Out]

16*x^2 - log(-2*x^2 + e^(16*x^2 + 1)) + 3*log(x)

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giac [A]  time = 0.15, size = 34, normalized size = 1.31 \begin {gather*} 16 \, x^{2} + \frac {3}{2} \, \log \left (16 \, x^{2}\right ) - \log \left (16 \, x^{2} - 8 \, e^{\left (16 \, x^{2} + 1\right )}\right ) + 1 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="giac")

[Out]

16*x^2 + 3/2*log(16*x^2) - log(16*x^2 - 8*e^(16*x^2 + 1)) + 1

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maple [A]  time = 0.04, size = 29, normalized size = 1.12

method	result	size
risch	\(3 \ln \relax (x )+16 x^{2}+1-\ln \left (-2 x^{2}+{\mathrm e}^{16 x^{2}+1}\right )\)	\(29\)
norman	\(16 x^{2}+3 \ln \relax (x )-\ln \left (2 x^{2}-{\mathrm e}^{16 x^{2}+1}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x,method=_RETURNVERBOSE)

[Out]

3*ln(x)+16*x^2+1-ln(-2*x^2+exp(16*x^2+1))

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maxima [A]  time = 0.37, size = 33, normalized size = 1.27 \begin {gather*} 16 \, x^{2} - \log \left (-{\left (2 \, x^{2} - e^{\left (16 \, x^{2} + 1\right )}\right )} e^{\left (-1\right )}\right ) + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(16*x^2+1)-64*x^4-2*x^2)/(x*exp(16*x^2+1)-2*x^3),x, algorithm="maxima")

[Out]

16*x^2 - log(-(2*x^2 - e^(16*x^2 + 1))*e^(-1)) + 3*log(x)

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mupad [B]  time = 0.19, size = 29, normalized size = 1.12 \begin {gather*} 3\,\ln \relax (x)-\ln \left (2\,x^2-\mathrm {e}\,{\mathrm {e}}^{16\,x^2}\right )+16\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2 - 3*exp(16*x^2 + 1) + 64*x^4)/(x*exp(16*x^2 + 1) - 2*x^3),x)

[Out]

3*log(x) - log(2*x^2 - exp(1)*exp(16*x^2)) + 16*x^2

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sympy [A]  time = 0.14, size = 24, normalized size = 0.92 \begin {gather*} 16 x^{2} + 3 \log {\relax (x )} - \log {\left (- 2 x^{2} + e^{16 x^{2} + 1} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(16*x**2+1)-64*x**4-2*x**2)/(x*exp(16*x**2+1)-2*x**3),x)

[Out]

16*x**2 + 3*log(x) - log(-2*x**2 + exp(16*x**2 + 1))

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