3.26.29 \(\int \frac {-15 x+e^{\frac {9}{5 x^3}} (-54-27 x+10 x^3+10 x^4)}{5 x^3} \, dx\)

Optimal. Leaf size=23 \[ \frac {3}{x}+e^{\frac {9}{5 x^3}} \left (2 x+x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14, 2288} \begin {gather*} e^{\frac {9}{5 x^3}} x (x+2)+\frac {3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x + E^(9/(5*x^3))*(-54 - 27*x + 10*x^3 + 10*x^4))/(5*x^3),x]

[Out]

3/x + E^(9/(5*x^3))*x*(2 + x)

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-15 x+e^{\frac {9}{5 x^3}} \left (-54-27 x+10 x^3+10 x^4\right )}{x^3} \, dx\\ &=\frac {1}{5} \int \left (-\frac {15}{x^2}+\frac {e^{\frac {9}{5 x^3}} \left (-54-27 x+10 x^3+10 x^4\right )}{x^3}\right ) \, dx\\ &=\frac {3}{x}+\frac {1}{5} \int \frac {e^{\frac {9}{5 x^3}} \left (-54-27 x+10 x^3+10 x^4\right )}{x^3} \, dx\\ &=\frac {3}{x}+e^{\frac {9}{5 x^3}} x (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.09 \begin {gather*} \frac {1}{5} \left (\frac {15}{x}+5 e^{\frac {9}{5 x^3}} x (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x + E^(9/(5*x^3))*(-54 - 27*x + 10*x^3 + 10*x^4))/(5*x^3),x]

[Out]

(15/x + 5*E^(9/(5*x^3))*x*(2 + x))/5

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fricas [A]  time = 0.92, size = 22, normalized size = 0.96 \begin {gather*} \frac {{\left (x^{3} + 2 \, x^{2}\right )} e^{\left (\frac {9}{5 \, x^{3}}\right )} + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^4+10*x^3-27*x-54)*exp(9/5/x^3)-15*x)/x^3,x, algorithm="fricas")

[Out]

((x^3 + 2*x^2)*e^(9/5/x^3) + 3)/x

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giac [A]  time = 0.44, size = 27, normalized size = 1.17 \begin {gather*} \frac {x^{3} e^{\left (\frac {9}{5 \, x^{3}}\right )} + 2 \, x^{2} e^{\left (\frac {9}{5 \, x^{3}}\right )} + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^4+10*x^3-27*x-54)*exp(9/5/x^3)-15*x)/x^3,x, algorithm="giac")

[Out]

(x^3*e^(9/5/x^3) + 2*x^2*e^(9/5/x^3) + 3)/x

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maple [A]  time = 0.09, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {3}{x}+\frac {\left (5 x^{2}+10 x \right ) {\mathrm e}^{\frac {9}{5 x^{3}}}}{5}\)	\(24\)
norman	\(\frac {{\mathrm e}^{\frac {9}{5 x^{3}}} x^{4}+3 x +2 \,{\mathrm e}^{\frac {9}{5 x^{3}}} x^{3}}{x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((10*x^4+10*x^3-27*x-54)*exp(9/5/x^3)-15*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

3/x+1/5*(5*x^2+10*x)*exp(9/5/x^3)

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maxima [C]  time = 0.45, size = 92, normalized size = 4.00 \begin {gather*} \frac {2}{3} \, \left (\frac {9}{5}\right )^{\frac {2}{3}} x^{2} \left (-\frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (-\frac {2}{3}, -\frac {9}{5 \, x^{3}}\right ) + \frac {2}{3} \, \left (\frac {9}{5}\right )^{\frac {1}{3}} x \left (-\frac {1}{x^{3}}\right )^{\frac {1}{3}} \Gamma \left (-\frac {1}{3}, -\frac {9}{5 \, x^{3}}\right ) - \frac {\left (\frac {9}{5}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {9}{5 \, x^{3}}\right )}{x \left (-\frac {1}{x^{3}}\right )^{\frac {1}{3}}} + \frac {3}{x} - \frac {2 \, \left (\frac {9}{5}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {9}{5 \, x^{3}}\right )}{x^{2} \left (-\frac {1}{x^{3}}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^4+10*x^3-27*x-54)*exp(9/5/x^3)-15*x)/x^3,x, algorithm="maxima")

[Out]

2/3*(9/5)^(2/3)*x^2*(-1/x^3)^(2/3)*gamma(-2/3, -9/5/x^3) + 2/3*(9/5)^(1/3)*x*(-1/x^3)^(1/3)*gamma(-1/3, -9/5/x
^3) - (9/5)^(2/3)*gamma(1/3, -9/5/x^3)/(x*(-1/x^3)^(1/3)) + 3/x - 2*(9/5)^(1/3)*gamma(2/3, -9/5/x^3)/(x^2*(-1/
x^3)^(2/3))

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mupad [B]  time = 1.46, size = 25, normalized size = 1.09 \begin {gather*} 2\,x\,{\mathrm {e}}^{\frac {9}{5\,x^3}}+x^2\,{\mathrm {e}}^{\frac {9}{5\,x^3}}+\frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + (exp(9/(5*x^3))*(27*x - 10*x^3 - 10*x^4 + 54))/5)/x^3,x)

[Out]

2*x*exp(9/(5*x^3)) + x^2*exp(9/(5*x^3)) + 3/x

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sympy [A]  time = 0.13, size = 17, normalized size = 0.74 \begin {gather*} \left (x^{2} + 2 x\right ) e^{\frac {9}{5 x^{3}}} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x**4+10*x**3-27*x-54)*exp(9/5/x**3)-15*x)/x**3,x)

[Out]

(x**2 + 2*x)*exp(9/(5*x**3)) + 3/x

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