3.52.70 \(\int \frac {-1176+172 x-172 x^3+24 x^4+e^5 (43 x^3-6 x^4)}{5 x^3} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{5} \left (4-e^5+\frac {4}{x^2}\right ) \left (3 (7-x)^2-x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 14} \begin {gather*} \frac {3}{5} \left (4-e^5\right ) x^2+\frac {588}{5 x^2}-\frac {43}{5} \left (4-e^5\right ) x-\frac {172}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1176 + 172*x - 172*x^3 + 24*x^4 + E^5*(43*x^3 - 6*x^4))/(5*x^3),x]

[Out]

588/(5*x^2) - 172/(5*x) - (43*(4 - E^5)*x)/5 + (3*(4 - E^5)*x^2)/5

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-1176+172 x-172 x^3+24 x^4+e^5 \left (43 x^3-6 x^4\right )}{x^3} \, dx\\ &=\frac {1}{5} \int \left (43 \left (-4+e^5\right )-\frac {1176}{x^3}+\frac {172}{x^2}-6 \left (-4+e^5\right ) x\right ) \, dx\\ &=\frac {588}{5 x^2}-\frac {172}{5 x}-\frac {43}{5} \left (4-e^5\right ) x+\frac {3}{5} \left (4-e^5\right ) x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.14 \begin {gather*} \frac {1}{5} \left (\frac {588}{x^2}-\frac {172}{x}+43 \left (-4+e^5\right ) x-3 \left (-4+e^5\right ) x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1176 + 172*x - 172*x^3 + 24*x^4 + E^5*(43*x^3 - 6*x^4))/(5*x^3),x]

[Out]

(588/x^2 - 172/x + 43*(-4 + E^5)*x - 3*(-4 + E^5)*x^2)/5

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fricas [A]  time = 1.33, size = 35, normalized size = 1.21 \begin {gather*} \frac {12 \, x^{4} - 172 \, x^{3} - {\left (3 \, x^{4} - 43 \, x^{3}\right )} e^{5} - 172 \, x + 588}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^4+43*x^3)*exp(5)+24*x^4-172*x^3+172*x-1176)/x^3,x, algorithm="fricas")

[Out]

1/5*(12*x^4 - 172*x^3 - (3*x^4 - 43*x^3)*e^5 - 172*x + 588)/x^2

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giac [A]  time = 0.19, size = 31, normalized size = 1.07 \begin {gather*} -\frac {3}{5} \, x^{2} e^{5} + \frac {12}{5} \, x^{2} + \frac {43}{5} \, x e^{5} - \frac {172}{5} \, x - \frac {4 \, {\left (43 \, x - 147\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^4+43*x^3)*exp(5)+24*x^4-172*x^3+172*x-1176)/x^3,x, algorithm="giac")

[Out]

-3/5*x^2*e^5 + 12/5*x^2 + 43/5*x*e^5 - 172/5*x - 4/5*(43*x - 147)/x^2

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maple [A]  time = 0.05, size = 30, normalized size = 1.03

method	result	size
norman	\(\frac {\frac {588}{5}+\left (-\frac {3 \,{\mathrm e}^{5}}{5}+\frac {12}{5}\right ) x^{4}+\left (\frac {43 \,{\mathrm e}^{5}}{5}-\frac {172}{5}\right ) x^{3}-\frac {172 x}{5}}{x^{2}}\)	\(30\)
default	\(-\frac {3 x^{2} {\mathrm e}^{5}}{5}+\frac {43 x \,{\mathrm e}^{5}}{5}+\frac {12 x^{2}}{5}-\frac {172 x}{5}-\frac {172}{5 x}+\frac {588}{5 x^{2}}\)	\(32\)
risch	\(-\frac {3 x^{2} {\mathrm e}^{5}}{5}+\frac {43 x \,{\mathrm e}^{5}}{5}+\frac {12 x^{2}}{5}-\frac {172 x}{5}+\frac {-172 x +588}{5 x^{2}}\)	\(32\)
gosper	\(-\frac {3 x^{4} {\mathrm e}^{5}-43 x^{3} {\mathrm e}^{5}-12 x^{4}+172 x^{3}+172 x -588}{5 x^{2}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-6*x^4+43*x^3)*exp(5)+24*x^4-172*x^3+172*x-1176)/x^3,x,method=_RETURNVERBOSE)

[Out]

(588/5+(-3/5*exp(5)+12/5)*x^4+(43/5*exp(5)-172/5)*x^3-172/5*x)/x^2

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maxima [A]  time = 0.35, size = 27, normalized size = 0.93 \begin {gather*} -\frac {3}{5} \, x^{2} {\left (e^{5} - 4\right )} + \frac {43}{5} \, x {\left (e^{5} - 4\right )} - \frac {4 \, {\left (43 \, x - 147\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x^4+43*x^3)*exp(5)+24*x^4-172*x^3+172*x-1176)/x^3,x, algorithm="maxima")

[Out]

-3/5*x^2*(e^5 - 4) + 43/5*x*(e^5 - 4) - 4/5*(43*x - 147)/x^2

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mupad [B]  time = 0.06, size = 30, normalized size = 1.03 \begin {gather*} x\,\left (\frac {43\,{\mathrm {e}}^5}{5}-\frac {172}{5}\right )-\frac {\frac {172\,x}{5}-\frac {588}{5}}{x^2}-x^2\,\left (\frac {3\,{\mathrm {e}}^5}{5}-\frac {12}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((172*x)/5 + (exp(5)*(43*x^3 - 6*x^4))/5 - (172*x^3)/5 + (24*x^4)/5 - 1176/5)/x^3,x)

[Out]

x*((43*exp(5))/5 - 172/5) - ((172*x)/5 - 588/5)/x^2 - x^2*((3*exp(5))/5 - 12/5)

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sympy [A]  time = 0.10, size = 31, normalized size = 1.07 \begin {gather*} \frac {x^{2} \left (12 - 3 e^{5}\right )}{5} + \frac {x \left (-172 + 43 e^{5}\right )}{5} + \frac {588 - 172 x}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-6*x**4+43*x**3)*exp(5)+24*x**4-172*x**3+172*x-1176)/x**3,x)

[Out]

x**2*(12 - 3*exp(5))/5 + x*(-172 + 43*exp(5))/5 + (588 - 172*x)/(5*x**2)

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