3.63.41 \(\int \frac {e^{\frac {1}{5} (-15 x+10 x^2+4 x^3)} (-15+20 x+12 x^2)}{-25+5 e^{\frac {1}{5} (-15 x+10 x^2+4 x^3)}} \, dx\)

Optimal. Leaf size=21 \[ 1+\log \left (-5+e^{x \left (-3+2 x+\frac {4 x^2}{5}\right )}\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 33, normalized size of antiderivative = 1.57, number of steps used = 1, number of rules used = 1, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6684} \begin {gather*} \log \left (-e^{-3 x} \left (5 e^{3 x}-e^{\frac {4 x^3}{5}+2 x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-15*x + 10*x^2 + 4*x^3)/5)*(-15 + 20*x + 12*x^2))/(-25 + 5*E^((-15*x + 10*x^2 + 4*x^3)/5)),x]

[Out]

Log[-((5*E^(3*x) - E^(2*x^2 + (4*x^3)/5))/E^(3*x))]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (-e^{-3 x} \left (5 e^{3 x}-e^{2 x^2+\frac {4 x^3}{5}}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 34, normalized size = 1.62 \begin {gather*} \frac {1}{5} \left (-15 x+5 \log \left (-5 e^{3 x}+e^{2 x^2+\frac {4 x^3}{5}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-15*x + 10*x^2 + 4*x^3)/5)*(-15 + 20*x + 12*x^2))/(-25 + 5*E^((-15*x + 10*x^2 + 4*x^3)/5)),x]

[Out]

(-15*x + 5*Log[-5*E^(3*x) + E^(2*x^2 + (4*x^3)/5)])/5

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fricas [A]  time = 0.50, size = 18, normalized size = 0.86 \begin {gather*} \log \left (e^{\left (\frac {4}{5} \, x^{3} + 2 \, x^{2} - 3 \, x\right )} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+20*x-15)*exp(4/5*x^3+2*x^2-3*x)/(5*exp(4/5*x^3+2*x^2-3*x)-25),x, algorithm="fricas")

[Out]

log(e^(4/5*x^3 + 2*x^2 - 3*x) - 5)

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giac [A]  time = 0.22, size = 18, normalized size = 0.86 \begin {gather*} \log \left (e^{\left (\frac {4}{5} \, x^{3} + 2 \, x^{2} - 3 \, x\right )} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+20*x-15)*exp(4/5*x^3+2*x^2-3*x)/(5*exp(4/5*x^3+2*x^2-3*x)-25),x, algorithm="giac")

[Out]

log(e^(4/5*x^3 + 2*x^2 - 3*x) - 5)

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maple [A]  time = 0.05, size = 18, normalized size = 0.86

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {x \left (4 x^{2}+10 x -15\right )}{5}}-5\right )\)	\(18\)
norman	\(\ln \left (5 \,{\mathrm e}^{\frac {4}{5} x^{3}+2 x^{2}-3 x}-25\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x^2+20*x-15)*exp(4/5*x^3+2*x^2-3*x)/(5*exp(4/5*x^3+2*x^2-3*x)-25),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/5*x*(4*x^2+10*x-15))-5)

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maxima [A]  time = 0.42, size = 36, normalized size = 1.71 \begin {gather*} 2 \, x^{2} - 3 \, x + \log \left ({\left (e^{\left (\frac {4}{5} \, x^{3} + 2 \, x^{2}\right )} - 5 \, e^{\left (3 \, x\right )}\right )} e^{\left (-2 \, x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+20*x-15)*exp(4/5*x^3+2*x^2-3*x)/(5*exp(4/5*x^3+2*x^2-3*x)-25),x, algorithm="maxima")

[Out]

2*x^2 - 3*x + log((e^(4/5*x^3 + 2*x^2) - 5*e^(3*x))*e^(-2*x^2))

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mupad [B]  time = 0.13, size = 18, normalized size = 0.86 \begin {gather*} \ln \left ({\mathrm {e}}^{\frac {4\,x^3}{5}+2\,x^2-3\,x}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x^2 - 3*x + (4*x^3)/5)*(20*x + 12*x^2 - 15))/(5*exp(2*x^2 - 3*x + (4*x^3)/5) - 25),x)

[Out]

log(exp(2*x^2 - 3*x + (4*x^3)/5) - 5)

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sympy [B]  time = 0.17, size = 39, normalized size = 1.86 \begin {gather*} \frac {16 x^{3}}{25} + \frac {8 x^{2}}{5} - \frac {12 x}{5} + \frac {\log {\left (e^{\frac {4 x^{3}}{5} + 2 x^{2} - 3 x} - 5 \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x**2+20*x-15)*exp(4/5*x**3+2*x**2-3*x)/(5*exp(4/5*x**3+2*x**2-3*x)-25),x)

[Out]

16*x**3/25 + 8*x**2/5 - 12*x/5 + log(exp(4*x**3/5 + 2*x**2 - 3*x) - 5)/5

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