3.30.79 \(\int \frac {3600-27 x^3+9 e^5 x^3+e^5 (400-3 x^3) \log (\frac {1}{100} (-400+3 x^3))}{-400 x^2+3 x^5} \, dx\)

Optimal. Leaf size=20 \[ \frac {9+e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x} \]

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Rubi [B]  time = 0.53, antiderivative size = 200, normalized size of antiderivative = 10.00, number of steps used = 18, number of rules used = 11, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {6, 1593, 6725, 453, 292, 31, 634, 617, 204, 628, 2455} \begin {gather*} \frac {e^5 \log \left (\frac {3 x^3}{100}-4\right )}{x}-\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (3^{2/3} x^2+2\ 5^{2/3} \sqrt [3]{6} x+20\ 2^{2/3} \sqrt [3]{5}\right )}{4\ 5^{2/3}}+\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (30^{2/3} x^2+20 \sqrt [3]{15} x+200 \sqrt [3]{2}\right )}{4\ 5^{2/3}}+\frac {9}{x}+\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (2 \sqrt [3]{2} 5^{2/3}-\sqrt [3]{3} x\right )}{2\ 5^{2/3}}-\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (10\ 2^{2/3}-\sqrt [3]{30} x\right )}{2\ 5^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3600 - 27*x^3 + 9*E^5*x^3 + E^5*(400 - 3*x^3)*Log[(-400 + 3*x^3)/100])/(-400*x^2 + 3*x^5),x]

[Out]

9/x + ((3/2)^(1/3)*E^5*Log[2*2^(1/3)*5^(2/3) - 3^(1/3)*x])/(2*5^(2/3)) - ((3/2)^(1/3)*E^5*Log[10*2^(2/3) - 30^
(1/3)*x])/(2*5^(2/3)) - ((3/2)^(1/3)*E^5*Log[20*2^(2/3)*5^(1/3) + 2*5^(2/3)*6^(1/3)*x + 3^(2/3)*x^2])/(4*5^(2/
3)) + ((3/2)^(1/3)*E^5*Log[200*2^(1/3) + 20*15^(1/3)*x + 30^(2/3)*x^2])/(4*5^(2/3)) + (E^5*Log[-4 + (3*x^3)/10
0])/x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3600+\left (-27+9 e^5\right ) x^3+e^5 \left (400-3 x^3\right ) \log \left (\frac {1}{100} \left (-400+3 x^3\right )\right )}{-400 x^2+3 x^5} \, dx\\ &=\int \frac {3600+\left (-27+9 e^5\right ) x^3+e^5 \left (400-3 x^3\right ) \log \left (\frac {1}{100} \left (-400+3 x^3\right )\right )}{x^2 \left (-400+3 x^3\right )} \, dx\\ &=\int \left (\frac {9 \left (-400+\left (3-e^5\right ) x^3\right )}{x^2 \left (400-3 x^3\right )}-\frac {e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x^2}\right ) \, dx\\ &=9 \int \frac {-400+\left (3-e^5\right ) x^3}{x^2 \left (400-3 x^3\right )} \, dx-e^5 \int \frac {\log \left (-4+\frac {3 x^3}{100}\right )}{x^2} \, dx\\ &=\frac {9}{x}+\frac {e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x}-\frac {1}{100} \left (9 e^5\right ) \int \frac {x}{-4+\frac {3 x^3}{100}} \, dx-\left (9 e^5\right ) \int \frac {x}{400-3 x^3} \, dx\\ &=\frac {9}{x}+\frac {e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x}-\frac {\left (\left (\frac {3}{5}\right )^{2/3} e^5\right ) \int \frac {1}{2 \sqrt [3]{2} 5^{2/3}-\sqrt [3]{3} x} \, dx}{2 \sqrt [3]{2}}+\frac {\left (\left (\frac {3}{5}\right )^{2/3} e^5\right ) \int \frac {2 \sqrt [3]{2} 5^{2/3}-\sqrt [3]{3} x}{20\ 2^{2/3} \sqrt [3]{5}+2\ 5^{2/3} \sqrt [3]{6} x+3^{2/3} x^2} \, dx}{2 \sqrt [3]{2}}-\frac {\left (3^{2/3} e^5\right ) \int \frac {1}{-2^{2/3}+\frac {\sqrt [3]{3} x}{10^{2/3}}} \, dx}{20 \sqrt [3]{5}}+\frac {\left (3^{2/3} e^5\right ) \int \frac {-2^{2/3}+\frac {\sqrt [3]{3} x}{10^{2/3}}}{2 \sqrt [3]{2}+\frac {\sqrt [3]{3} x}{5^{2/3}}+\frac {3^{2/3} x^2}{10 \sqrt [3]{10}}} \, dx}{20 \sqrt [3]{5}}\\ &=\frac {9}{x}+\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (2 \sqrt [3]{2} 5^{2/3}-\sqrt [3]{3} x\right )}{2\ 5^{2/3}}-\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (10\ 2^{2/3}-\sqrt [3]{30} x\right )}{2\ 5^{2/3}}+\frac {e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x}+\frac {1}{2} \left (3\ 3^{2/3} e^5\right ) \int \frac {1}{20\ 2^{2/3} \sqrt [3]{5}+2\ 5^{2/3} \sqrt [3]{6} x+3^{2/3} x^2} \, dx-\frac {\left (\sqrt [3]{\frac {3}{2}} e^5\right ) \int \frac {2\ 5^{2/3} \sqrt [3]{6}+2\ 3^{2/3} x}{20\ 2^{2/3} \sqrt [3]{5}+2\ 5^{2/3} \sqrt [3]{6} x+3^{2/3} x^2} \, dx}{4\ 5^{2/3}}+\frac {\left (\sqrt [3]{\frac {3}{2}} e^5\right ) \int \frac {\frac {\sqrt [3]{3}}{5^{2/3}}+\frac {3^{2/3} x}{5 \sqrt [3]{10}}}{2 \sqrt [3]{2}+\frac {\sqrt [3]{3} x}{5^{2/3}}+\frac {3^{2/3} x^2}{10 \sqrt [3]{10}}} \, dx}{4\ 5^{2/3}}-\frac {\left (3\ 3^{2/3} e^5\right ) \int \frac {1}{2 \sqrt [3]{2}+\frac {\sqrt [3]{3} x}{5^{2/3}}+\frac {3^{2/3} x^2}{10 \sqrt [3]{10}}} \, dx}{20 \sqrt [3]{10}}\\ &=\frac {9}{x}+\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (2 \sqrt [3]{2} 5^{2/3}-\sqrt [3]{3} x\right )}{2\ 5^{2/3}}-\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (10\ 2^{2/3}-\sqrt [3]{30} x\right )}{2\ 5^{2/3}}-\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (20\ 2^{2/3} \sqrt [3]{5}+2\ 5^{2/3} \sqrt [3]{6} x+3^{2/3} x^2\right )}{4\ 5^{2/3}}+\frac {\sqrt [3]{\frac {3}{2}} e^5 \log \left (200 \sqrt [3]{2}+20 \sqrt [3]{15} x+30^{2/3} x^2\right )}{4\ 5^{2/3}}+\frac {e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 1.00 \begin {gather*} \frac {9+e^5 \log \left (-4+\frac {3 x^3}{100}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3600 - 27*x^3 + 9*E^5*x^3 + E^5*(400 - 3*x^3)*Log[(-400 + 3*x^3)/100])/(-400*x^2 + 3*x^5),x]

[Out]

(9 + E^5*Log[-4 + (3*x^3)/100])/x

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fricas [A]  time = 1.32, size = 17, normalized size = 0.85 \begin {gather*} \frac {e^{5} \log \left (\frac {3}{100} \, x^{3} - 4\right ) + 9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+400)*exp(5)*log(3/100*x^3-4)+9*x^3*exp(5)-27*x^3+3600)/(3*x^5-400*x^2),x, algorithm="fricas
")

[Out]

(e^5*log(3/100*x^3 - 4) + 9)/x

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giac [A]  time = 0.17, size = 17, normalized size = 0.85 \begin {gather*} \frac {e^{5} \log \left (\frac {3}{100} \, x^{3} - 4\right ) + 9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+400)*exp(5)*log(3/100*x^3-4)+9*x^3*exp(5)-27*x^3+3600)/(3*x^5-400*x^2),x, algorithm="giac")

[Out]

(e^5*log(3/100*x^3 - 4) + 9)/x

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maple [A]  time = 0.49, size = 18, normalized size = 0.90

method	result	size
norman	\(\frac {9+{\mathrm e}^{5} \ln \left (\frac {3 x^{3}}{100}-4\right )}{x}\)	\(18\)
risch	\(\frac {{\mathrm e}^{5} \ln \left (\frac {3 x^{3}}{100}-4\right )}{x}+\frac {9}{x}\)	\(21\)
default	\(\frac {9}{x}-\frac {2 \,{\mathrm e}^{5} \ln \left (10\right )}{x}+\frac {{\mathrm e}^{5} \ln \left (3 x^{3}-400\right )}{x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^3+400)*exp(5)*ln(3/100*x^3-4)+9*x^3*exp(5)-27*x^3+3600)/(3*x^5-400*x^2),x,method=_RETURNVERBOSE)

[Out]

(9+exp(5)*ln(3/100*x^3-4))/x

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maxima [B]  time = 1.12, size = 216, normalized size = 10.80 \begin {gather*} -\frac {1}{100} \cdot 50^{\frac {2}{3}} 3^{\frac {5}{6}} \arctan \left (\frac {1}{150} \cdot 50^{\frac {2}{3}} 3^{\frac {1}{6}} {\left (3^{\frac {2}{3}} x + 50^{\frac {1}{3}} 3^{\frac {1}{3}}\right )}\right ) e^{5} + \frac {1}{200} \cdot 50^{\frac {2}{3}} 3^{\frac {1}{3}} e^{5} \log \left (3^{\frac {2}{3}} x^{2} + 2 \cdot 50^{\frac {1}{3}} 3^{\frac {1}{3}} x + 4 \cdot 50^{\frac {2}{3}}\right ) - \frac {1}{100} \cdot 50^{\frac {2}{3}} 3^{\frac {1}{3}} e^{5} \log \left (\frac {1}{3} \cdot 3^{\frac {2}{3}} {\left (3^{\frac {1}{3}} x - 2 \cdot 50^{\frac {1}{3}}\right )}\right ) + \frac {1}{200} \, {\left (2 \cdot 50^{\frac {2}{3}} 3^{\frac {5}{6}} \arctan \left (\frac {1}{150} \cdot 50^{\frac {2}{3}} 3^{\frac {1}{6}} {\left (3^{\frac {2}{3}} x + 50^{\frac {1}{3}} 3^{\frac {1}{3}}\right )}\right ) - 50^{\frac {2}{3}} 3^{\frac {1}{3}} \log \left (3^{\frac {2}{3}} x^{2} + 2 \cdot 50^{\frac {1}{3}} 3^{\frac {1}{3}} x + 4 \cdot 50^{\frac {2}{3}}\right ) + 2 \cdot 50^{\frac {2}{3}} 3^{\frac {1}{3}} \log \left (\frac {1}{3} \cdot 3^{\frac {2}{3}} {\left (3^{\frac {1}{3}} x - 2 \cdot 50^{\frac {1}{3}}\right )}\right )\right )} e^{5} - \frac {2 \, {\left (\log \relax (5) + \log \relax (2)\right )} e^{5} - e^{5} \log \left (3 \, x^{3} - 400\right )}{x} + \frac {9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+400)*exp(5)*log(3/100*x^3-4)+9*x^3*exp(5)-27*x^3+3600)/(3*x^5-400*x^2),x, algorithm="maxima
")

[Out]

-1/100*50^(2/3)*3^(5/6)*arctan(1/150*50^(2/3)*3^(1/6)*(3^(2/3)*x + 50^(1/3)*3^(1/3)))*e^5 + 1/200*50^(2/3)*3^(
1/3)*e^5*log(3^(2/3)*x^2 + 2*50^(1/3)*3^(1/3)*x + 4*50^(2/3)) - 1/100*50^(2/3)*3^(1/3)*e^5*log(1/3*3^(2/3)*(3^
(1/3)*x - 2*50^(1/3))) + 1/200*(2*50^(2/3)*3^(5/6)*arctan(1/150*50^(2/3)*3^(1/6)*(3^(2/3)*x + 50^(1/3)*3^(1/3)
)) - 50^(2/3)*3^(1/3)*log(3^(2/3)*x^2 + 2*50^(1/3)*3^(1/3)*x + 4*50^(2/3)) + 2*50^(2/3)*3^(1/3)*log(1/3*3^(2/3
)*(3^(1/3)*x - 2*50^(1/3))))*e^5 - (2*(log(5) + log(2))*e^5 - e^5*log(3*x^3 - 400))/x + 9/x

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mupad [B]  time = 0.21, size = 17, normalized size = 0.85 \begin {gather*} \frac {\ln \left (\frac {3\,x^3}{100}-4\right )\,{\mathrm {e}}^5+9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x^3*exp(5) - 27*x^3 - log((3*x^3)/100 - 4)*exp(5)*(3*x^3 - 400) + 3600)/(400*x^2 - 3*x^5),x)

[Out]

(log((3*x^3)/100 - 4)*exp(5) + 9)/x

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sympy [A]  time = 0.16, size = 17, normalized size = 0.85 \begin {gather*} \frac {e^{5} \log {\left (\frac {3 x^{3}}{100} - 4 \right )}}{x} + \frac {9}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**3+400)*exp(5)*ln(3/100*x**3-4)+9*x**3*exp(5)-27*x**3+3600)/(3*x**5-400*x**2),x)

[Out]

exp(5)*log(3*x**3/100 - 4)/x + 9/x

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