3.47.81 \(\int \frac {(-5+x) x^6 (-250+60 x+(125-30 x) \log (4))}{-5 x^2+x^3} \, dx\)

Optimal. Leaf size=16 \[ 15+5 (-5+x) x^5 (2-\log (4)) \]

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Rubi [A]  time = 0.05, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1584, 186, 43} \begin {gather*} 5 x^6 (2-\log (4))-25 x^5 (2-\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-5 + x)*x^6*(-250 + 60*x + (125 - 30*x)*Log[4]))/(-5*x^2 + x^3),x]

[Out]

-25*x^5*(2 - Log[4]) + 5*x^6*(2 - Log[4])

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 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int x^4 (-250+60 x+(125-30 x) \log (4)) \, dx\\ &=\int x^4 (-125 (2-\log (4))+30 x (2-\log (4))) \, dx\\ &=\int \left (125 x^4 (-2+\log (4))-30 x^5 (-2+\log (4))\right ) \, dx\\ &=-25 x^5 (2-\log (4))+5 x^6 (2-\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.94 \begin {gather*} -5 \left (-5 x^5+x^6\right ) (-2+\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-5 + x)*x^6*(-250 + 60*x + (125 - 30*x)*Log[4]))/(-5*x^2 + x^3),x]

[Out]

-5*(-5*x^5 + x^6)*(-2 + Log[4])

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fricas [A]  time = 0.90, size = 24, normalized size = 1.50 \begin {gather*} 10 \, x^{6} - 50 \, x^{5} - 10 \, {\left (x^{6} - 5 \, x^{5}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-30*x+125)*log(2)+60*x-250)*exp(6*log(x)+log(x-5))/(x^3-5*x^2),x, algorithm="fricas")

[Out]

10*x^6 - 50*x^5 - 10*(x^6 - 5*x^5)*log(2)

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giac [A]  time = 0.14, size = 25, normalized size = 1.56 \begin {gather*} -10 \, x^{6} \log \relax (2) + 10 \, x^{6} + 50 \, x^{5} \log \relax (2) - 50 \, x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-30*x+125)*log(2)+60*x-250)*exp(6*log(x)+log(x-5))/(x^3-5*x^2),x, algorithm="giac")

[Out]

-10*x^6*log(2) + 10*x^6 + 50*x^5*log(2) - 50*x^5

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maple [A]  time = 0.08, size = 20, normalized size = 1.25

method	result	size
gosper	\(-\frac {10 \left (\ln \relax (2)-1\right ) {\mathrm e}^{6 \ln \relax (x )+\ln \left (x -5\right )}}{x}\)	\(20\)
default	\(2 \left (5 \ln \relax (2)-5\right ) \left (-x^{6}+5 x^{5}\right )\)	\(20\)
norman	\(\frac {\left (-50+50 \ln \relax (2)\right ) x^{6}+\left (-10 \ln \relax (2)+10\right ) x^{7}}{x}\)	\(26\)
risch	\(-10 x^{6} \ln \relax (2)+10 x^{6}+50 x^{5} \ln \relax (2)-50 x^{5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-30*x+125)*ln(2)+60*x-250)*exp(6*ln(x)+ln(x-5))/(x^3-5*x^2),x,method=_RETURNVERBOSE)

[Out]

-10/x*(ln(2)-1)*exp(6*ln(x)+ln(x-5))

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maxima [A]  time = 0.37, size = 19, normalized size = 1.19 \begin {gather*} -10 \, x^{6} {\left (\log \relax (2) - 1\right )} + 50 \, x^{5} {\left (\log \relax (2) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-30*x+125)*log(2)+60*x-250)*exp(6*log(x)+log(x-5))/(x^3-5*x^2),x, algorithm="maxima")

[Out]

-10*x^6*(log(2) - 1) + 50*x^5*(log(2) - 1)

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mupad [B]  time = 0.07, size = 12, normalized size = 0.75 \begin {gather*} -10\,x^5\,\left (\ln \relax (2)-1\right )\,\left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x - 5) + 6*log(x))*(2*log(2)*(30*x - 125) - 60*x + 250))/(5*x^2 - x^3),x)

[Out]

-10*x^5*(log(2) - 1)*(x - 5)

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sympy [A]  time = 0.08, size = 19, normalized size = 1.19 \begin {gather*} x^{6} \left (10 - 10 \log {\relax (2 )}\right ) + x^{5} \left (-50 + 50 \log {\relax (2 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-30*x+125)*ln(2)+60*x-250)*exp(6*ln(x)+ln(x-5))/(x**3-5*x**2),x)

[Out]

x**6*(10 - 10*log(2)) + x**5*(-50 + 50*log(2))

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