3.59.16 \(\int \frac {-20 x^2+10 x^3+e^{\frac {13+10 x^2-10 x^3}{10 x}} (13-13 x-10 x^2+30 x^3-20 x^4)}{-10 x^2+10 x^3} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {13}{10 x}-(-1+x) x}+x-\log (-1+x) \]

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Rubi [A]  time = 0.74, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 4, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {1593, 6742, 43, 6706} \begin {gather*} e^{-x^2+x+\frac {13}{10 x}}+x-\log (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x^2 + 10*x^3 + E^((13 + 10*x^2 - 10*x^3)/(10*x))*(13 - 13*x - 10*x^2 + 30*x^3 - 20*x^4))/(-10*x^2 + 1
0*x^3),x]

[Out]

E^(13/(10*x) + x - x^2) + x - Log[1 - x]

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 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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x^2+10 x^3+e^{\frac {13+10 x^2-10 x^3}{10 x}} \left (13-13 x-10 x^2+30 x^3-20 x^4\right )}{x^2 (-10+10 x)} \, dx\\ &=\int \left (\frac {-2+x}{-1+x}-\frac {e^{\frac {13}{10 x}+x-x^2} \left (13-10 x^2+20 x^3\right )}{10 x^2}\right ) \, dx\\ &=-\left (\frac {1}{10} \int \frac {e^{\frac {13}{10 x}+x-x^2} \left (13-10 x^2+20 x^3\right )}{x^2} \, dx\right )+\int \frac {-2+x}{-1+x} \, dx\\ &=e^{\frac {13}{10 x}+x-x^2}+\int \left (1+\frac {1}{1-x}\right ) \, dx\\ &=e^{\frac {13}{10 x}+x-x^2}+x-\log (1-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 34, normalized size = 1.42 \begin {gather*} \frac {1}{10} \left (10 e^{\frac {13}{10 x}+x-x^2}+10 x-10 \log (1-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x^2 + 10*x^3 + E^((13 + 10*x^2 - 10*x^3)/(10*x))*(13 - 13*x - 10*x^2 + 30*x^3 - 20*x^4))/(-10*x
^2 + 10*x^3),x]

[Out]

(10*E^(13/(10*x) + x - x^2) + 10*x - 10*Log[1 - x])/10

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fricas [A]  time = 0.56, size = 26, normalized size = 1.08 \begin {gather*} x + e^{\left (-\frac {10 \, x^{3} - 10 \, x^{2} - 13}{10 \, x}\right )} - \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+30*x^3-10*x^2-13*x+13)*exp(1/10*(-10*x^3+10*x^2+13)/x)+10*x^3-20*x^2)/(10*x^3-10*x^2),x, a
lgorithm="fricas")

[Out]

x + e^(-1/10*(10*x^3 - 10*x^2 - 13)/x) - log(x - 1)

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giac [A]  time = 0.21, size = 26, normalized size = 1.08 \begin {gather*} x + e^{\left (-\frac {10 \, x^{3} - 10 \, x^{2} - 13}{10 \, x}\right )} - \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+30*x^3-10*x^2-13*x+13)*exp(1/10*(-10*x^3+10*x^2+13)/x)+10*x^3-20*x^2)/(10*x^3-10*x^2),x, a
lgorithm="giac")

[Out]

x + e^(-1/10*(10*x^3 - 10*x^2 - 13)/x) - log(x - 1)

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maple [A]  time = 0.29, size = 27, normalized size = 1.12

method	result	size
risch	\(x -\ln \left (x -1\right )+{\mathrm e}^{-\frac {10 x^{3}-10 x^{2}-13}{10 x}}\)	\(27\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{\frac {-10 x^{3}+10 x^{2}+13}{10 x}}}{x}-\ln \left (x -1\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^4+30*x^3-10*x^2-13*x+13)*exp(1/10*(-10*x^3+10*x^2+13)/x)+10*x^3-20*x^2)/(10*x^3-10*x^2),x,method=_
RETURNVERBOSE)

[Out]

x-ln(x-1)+exp(-1/10*(10*x^3-10*x^2-13)/x)

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maxima [A]  time = 0.44, size = 21, normalized size = 0.88 \begin {gather*} x + e^{\left (-x^{2} + x + \frac {13}{10 \, x}\right )} - \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+30*x^3-10*x^2-13*x+13)*exp(1/10*(-10*x^3+10*x^2+13)/x)+10*x^3-20*x^2)/(10*x^3-10*x^2),x, a
lgorithm="maxima")

[Out]

x + e^(-x^2 + x + 13/10/x) - log(x - 1)

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mupad [B]  time = 4.05, size = 21, normalized size = 0.88 \begin {gather*} x-\ln \left (x-1\right )+{\mathrm {e}}^{x+\frac {13}{10\,x}-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^2 - x^3 + 13/10)/x)*(13*x + 10*x^2 - 30*x^3 + 20*x^4 - 13) + 20*x^2 - 10*x^3)/(10*x^2 - 10*x^3),x)

[Out]

x - log(x - 1) + exp(x + 13/(10*x) - x^2)

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sympy [A]  time = 0.18, size = 19, normalized size = 0.79 \begin {gather*} x + e^{\frac {- x^{3} + x^{2} + \frac {13}{10}}{x}} - \log {\left (x - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**4+30*x**3-10*x**2-13*x+13)*exp(1/10*(-10*x**3+10*x**2+13)/x)+10*x**3-20*x**2)/(10*x**3-10*x
**2),x)

[Out]

x + exp((-x**3 + x**2 + 13/10)/x) - log(x - 1)

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